Download Additional Problems, Often with Answers Reasoned Out

Additional Problems, Often with Answers Reasoned Out and/or
Calculations Shown
A Supplement and Learning Aid for:
Statistics and Data Interpretation for the Helping Professions
James A. Rosenthal
University of Oklahoma, Norman
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Given that the problems and questions presented here typically provide
the reasoning that underlies answers and, also, calculation steps,
students may want to complete them prior to completing the problems
and questions at each Chapter’s end. (The chapter-end problems do not
provide reasoning and calculations.) Supplemental problems and
questions are presented here for all Chapters. As some problems
reference information presented in the text, one should have the text
with them as they proceed.
To save disk space, graphical images are used sparingly. Where you
encounter the square root symbol “”, that symbol should be
interpreted as “spanning” the number that follows it. For instance
“100" conveys the square root of 100.
Note: This document uses Courier, font size 11. If it does not format
in this font for you, you may want to set it to this.
Chapter 1
1.
A researcher conducts in-depth interviews, carefully recording
the extended, open-ended, and lengthy responses of delinquent
youth regarding their experiences in a wilderness adventure
program. (T or F). This study uses qualitative methods rather
than quantitative methods.
True. If the researcher had formulated questions that required
participants to choose the best response from a list of responses
– yes/no, agree/disagree, etc. – the study would have been
characterized as quantitative. But as the researcher goes beyond
1
this and encourages extended replies in the respondents own
words, the study is qualitative.
2.
Think about the fact that the number “2" always has the same
value (that being “2"). (T or F) The number “2" is a variable
rather than a constant.
False. As “2" always has the same value it (and any other
number) is a constant.
3.
In a given class, 29 students have brown eyes and 1 has blue
eyes. (T or F) In this class, eye color is a variable rather
than a constant.
True. As eye color takes on different values (brown and blue),
it is a variable. One could say that it is almost a constant as
almost everyone has the same eye color.
4.
A market survey asks respondents to choose the magazine they
prefer from the following list: Newsweek, Time, U.S. News and
World Report.
a.
b.
(T or F) Newsweek, Time, and U.S. News and World Report are
variables rather than values.
(T or F) In this example, type of magazine is a value
rather than a variable.
Answers follow:
a.
b.
5.
False (These magazines are values.)
False (Type of magazine is a variable.)
Indicate the level of measurement used for each of the following.
a.
b.
c.
d.
The number of books in children’s houses is counted.
Attendees rate a workshop as: Excellent, Good, Fair, or
Poor.
At a track meet, the distance that competitors throw the
javelin is recorded.
Seniors at a high school vote on whether they would prefer
a senior trip to: Washington, DC, New York City, or Boston.
Answers follow:
a.
b.
c.
d.
interval/ratio
ordinal
interval/ratio
nominal
2
6.
(T or F) As it is usually measured (female or male), the variable
sex is dichotomous variable.
True.
7.
A dichotomous variable is one with (exactly) two values.
Respondents answer 10 questions about their self-esteem,
responding “strongly agree”, “agree”, “not sure”, “disagree”, or
“strongly disagree” to each question. Each response choice
generates points and points are added for the 10 questions to
generate a self-esteem scale score. Respond to the following
questions:
a.
b.
c.
d.
(T or F) This scale is an example of what the text means
when it uses the term “multi-item attitudinal scale” (see
page 8 near bottom.)
(T or F) Strictly and formally, the level of measurement of
the scale is interval/ratio rather than ordinal. (Hint:
there may not be an easy answer; see page 8)
(T or F) The level of measurement comes extremely close to
being interval/ratio.
(T or F) The text recommends against using statistical
procedures designed for interval/ratio level variables with
multi-item scales such as that presented in this example.
Answers follow:
a.
b.
c.
d.
8.
True.
Hard to say if true or false. Scholars disagree. My view
is that the measurement here is not strictly interval/ratio
in the same sense as are physical quantities such as
height, weight, etc. (see page 8)
True. And this is the key point. Even if it may be
difficult to get scholarly concurrence that, in the
strictest sense, multi-item scales reflect interval/ratio
level measurement, pragmatically speaking, scholars treat
such scales as being at this level as they carry out their
everyday work.
False. Do use these procedures with multi-item scales such
as this.
Indicate whether you feel each pair of variables is likely to be
related or not to be related. (Technically, as no data is
presented, one can’t know assuredly whether or not the variables
are related; but, hopefully, the examples are intuitive.)
a.
b.
c.
Level of education and income.
Whether there are children in a family (yes or no) and
whether that family owns a minivan (yes or no).
How much students like brussel sprouts and the grade they
earn on a statistics test.
3
d.
e.
Shoe size and height.
Whether or not there are clouds in the sky and whether or
not persons bring rain gear with them (umbrella, raincoat,
etc.)
Answers follow:
a.
b.
c.
d.
e.
9.
Related. On balance, the higher one’s education level the
higher their income.
Related. (I presume that) families with kids are more
likely to own minivans than are those without.
Can’t imagine that there is any relationship here. Some
students who like brussel sprouts will likely do well on
the test and others won’t. The same is likely true of
those who don’t like brussel sprouts. There is no reason
to think that those who like brussel sprouts will tend to
do either better or worse on the test than will those who
don’t like them.
Related. Those who are tall tend to have bigger feet than
those who are short.
Related. People are more likely to bring rain gear when
there are clouds than when there are not.
Suppose that you read that those who eat brussel sprouts have
lower cancer rates than those who do not. Why should you be
somewhat hesitant to conclude that eating brussel sprouts causes
reduced risk for cancer? (By the way, brussel sprouts are a small
green vegetable in the cabbage family.) Hint: consider
confounding variables such as those presented in the text at the
bottom of page 9.
Perhaps those who eat brussel sprouts differ from those who do
not in other food preferences as well. Perhaps one of these
differences rather than eating brussel sprouts explains the
reduced cancer risk.
10.
For each pair of variables below, indicate which is the
independent variable and which is the dependent variable. Be
alert for one pair of variables where it is not possible to
easily categorize one as independent and the other as dependent.
To help in identifying the variables, they are in italics.
a.
b.
c.
Hours spent studying for a test and grade earned on that
test.
Adolescents at risk for being placed out of their homes are
randomly assigned to Treatment A or to Treatment B. A
research study then tracks whether or not the adolescent
actually requires placement (yes vs. no).
How high students high jump and how far students long jump.
4
d.
Whether social work graduate student have prior work
experience in social work and their degree of satisfaction
with their graduate program.
Answers follow:
a.
b.
c.
d.
11.
Hours spent studying is the independent variable and test
grade is the dependent variable. Presumably studying
affects test grades.
Type of treatment (A or B) is the independent variable.
Placement (yes, placed versus no, not placed) is the
dependent variable. Presumably, the effectiveness of the
program that serves adolescent (A versus B) affects their
risk of subsequent placement.
Here it is not clear which variable would be the
independent and which would be the independent. In a
situation such as this, one would not bother to attempt to
classify the variables as independent or dependent. In
essence, neither would be independent and neither would be
dependent. These terms don’t make any sense in this
example as neither variable would be readily viewed as
causing the other.
Here prior experience is the independent which (presumably)
affects satisfaction, the dependent. (Clearly, it is
nonsensical to think of satisfaction with school as
affecting prior work experience; something that occurs
later in time cannot affect something that occurs before.)
At a given school of social work, members of the student social
work association fill out a questionnaire indicating their
willingness to participate in a service project. (Some students
at the school are members of the association and others are not.
Each student decides on their own whether or not to join the
association.)
a.
b.
c.
d.
e.
f.
Are the students in the association a sample of students in
the school?
Are the students in the association a random sample of
students in the school?
Do you know assuredly that the willingness of students in
the student association to participate in the service
project differs systematically from – that is, is biased
relative to-- that of all students at the school?
Can you rule out bias, that is, can you be sure that the
opinion of those in the association does not differ
systematically from that of all students at the school?
Do you suspect systematic bias? Stated differently, are
you concerned about possible bias?
Presuming that you suspect bias, what is the nature of the
bias that you suspect? Stated differently, in what
5
g.
h.
direction (more willing versus less) do you suspect that
the willingness of student association members to
participate in the project differs from that of the full
population of students in the school?
What type of a sample of students at the school should be
taken to eliminate concerns about systematic bias?
If you took a random sample of students at the School,
would you be assured that the willingness of these students
to participate in the project would be precisely the same
as that of the full population of students at the School?
Here follow answers:
a.
b.
c.
d.
e.
f.
g.
h.
12.
Yes, they are a sample.
No, they are not a random sample.
Do not know this for sure. (Note: this question takes you a
bit beyond information presented in Chapter. Whenever a
sample is nonrandom, you are concerned about bias, but you
cannot be 100% sure that bias is present.)
This is the key idea. While you don’t know for sure that
bias is present, you are concerned that such is the case.
Yes, whenever sample is not random, one should be concerned
about (should suspect) bias.
Given that the members of the association did indeed join
at least one association (the student association), one
suspects that they would be more likely to participate in
the service project than would students who did not join
the association. So, one suspects, that, relative to the
full school population, the study sample is biased in the
direction of greater willingness to participate.
Though not a magic pill (and presuming that everyone
selected for the sample did fill out their questionnaire)
taking a random sample eliminates the systematic bias that
could emerge from a nonrandom sampling method.
No. Selecting a random sample eliminates systematic bias.
However, due to the luck of the draw, the opinions in the
study sample would likely differ at least to some degree
from those in the full population.
(T or F) One may appropriately use inferential statistical
procedures to draw conclusions about a population even when the
study sample has not been randomly selected from that population.
False. The study sample must be a random sample from the
population.
13.
For each of the following, indicate whether the researcher is
engaging in descriptive or inferential statistical procedures.
If the researcher is engaging in inferential statistical
procedures, indicate whether these are being used appropriately.
6
(Hint: one may only use inferential procedures to draw
conclusions about the population from which the study sample was
randomly selected.)
a.
b.
c.
A researcher reports that 27% of women in her study sample
had experienced physical abuse by their partners.
A researcher finds that 17% of female social work students
at a particular university have experienced physical abuse
by their partners. S/he concludes that a similar percentage
of all female students at the university have experienced
such abuse.
A researcher takes a random sample of female students at a
given university. Nineteen percent in this sample have
experienced physical abuse by their partners. The
researcher states that 19% is his/her best estimate of the
percentage of all female university students who have
experienced such abuse.
Answers follow:
a.
b.
c.
14.
This is simply descriptive statistics, describing the study
sample.
This is an (incredibly sloppy) attempt at inferential
statistics. It can be viewed as an attempt at inferential
statistics because the researcher is indeed drawing a
statistically-based conclusion about a group/population
that is broader/larger than her study sample. This is not
an appropriate/correct use of inferential statistics
because the study sample was not a random sample from the
population of students at the university. The conclusion
drawn is not a valid one.
This is an appropriate use of inferential statistics.
(Chapters 12 and 13 will demonstrate this kind of reasoning
in much greater depth.)
Indicate: 1) whether each study uses random assignment, 2)
whether (possible) bias from confounding variables raises
questions with the conclusion that is drawn, and 3) if
confounding variables are suspected, what these are and the
nature of the bias that they may cause. (Two points: 1) if no
mention is made of using methods of chance to assign to persons
groups, assume that assignment is nonrandom and 2) one needs to
be concerned about confounding variables only in studies that do
not use random assignment.)
a.
By a random process, a computer assigns patients to Medical
Treatment A or Treatment B. Those in A have much better
outcomes. The researcher concludes that A is a more
effective treatment than B.
7
b.
c.
Via a survey, a researcher finds that youth who watch more
violent television are more aggressive in school than are
those who watch less. She concludes that violent
television causes aggressive behavior in school.
A researcher finds that children served in family foster
homes are much more likely to reunite with their families
of origin than are children served in group homes. The
researcher concludes that the beneficial effect(s) of
family foster homes – individualized attention, interaction
in a family setting, bonding, etc. – are the cause of the
greater likelihood of reuniting.
Answers follow:
a.
b.
c.
15.
Assignment to groups is random and thus one need not be
concerned about bias from confounding variables. Though
the researcher’s conclusion oversimplifies, it is
essentially correct. (When you study statistical
significance tests, you will learn more about how chance
(the luck of the draw) affects study results, but we will
hold that idea for later.)
There has been no random assignment to “watch” or “not
watch” violent television. Perhaps youth who watch violent
television are more likely to live in violent homes than
are youth who do not watch such television. Perhaps higher
levels of violence in the home, not watching violent
television, is the cause of more aggressive behavior at
school.
Children were not randomly assigned to family foster versus
group homes. Perhaps children who were in foster family
care tended to have less serious problems than did those
who were in group homes. Perhaps these less serious
problems, not the benefits of foster care over group care,
explain why children served in foster homes are more likely
to reunite with their families.
What one factor do researchers tend to look at more than any
other as they attempt to decide whether study results obtained in
one setting will generalize to another setting?
Degree of similarity between the settings. The greater the
similarity, the greater the expected generalizability. Or, stated
differently, the greater the similarity of the settings the
greater the expected similarity of study results.
8
Chapter 2
16.
There are 20 women and 5 men in a classroom.
women?
What proportion are
Divide the number of objects with characteristics by the TOTAL
number of objects, and don’t forget the decimal point:
20/25 = .80
17.
For the prior problem, find the percentage of women.
Multiply the proportion by 100
.80  100 = 80%
18.
In table 2.3 on page 24, what is the cumulative frequency of
perpetrators aged 26 or below?
Sum the frequencies for all ages 26 and below:
2 + 1 + 2 + 2 + 1 + 1 = 9
(Alternatively, you could get the answer by looking in the
Cumulative Frequency column which lists cumulative frequencies.)
19.
In Table 2.3, what is the cumulative percentage of perpetrators
aged 26 and below?
Divide the cumulative frequency by the sample size and multiply
by 100:
9/20 = .45  100 = 45%
20.
In Table 2.3, what is the percentile rank of perpetrators 26 and
younger?
Percentile rank and cumulative percentage are basically the same
thing. Hence, the percentile rank was calculated in the prior
problem and is 45 or 45%.
21.
The stem of a given value (score) is 8 and the leaf is 3.
is that value?
What
Presuming that the value in question has two digits, the stem is
the first digit and the leaf is the second.
The value is 83.
9
22.
In Table 2.6 on page 27, how many persons scored 28 on the
Cohesion scale?
A score of 28 would have a stem of 2 and a leaf of 8.
In the row in which the stem equals 2, the only digits are a 2
(conveying a score of 22) and a 9 (29). Hence, no persons scored
28.
23.
In Table 2.6 on page 27, how many scored 39?
Find the stem equals 3 row and count the number of 9's.
There are 5 9's.
24.
Hence, 5 persons scored 39?
In Table 2.6 on page 27, what is the frequency of scores of 39?
Frequency is the number of scores of a given value.
determined in the prior problem.
This was
The frequency of 39's is five.
Chapter 3
1.
Find the mode of the observations: 2,7,2,2,4,5,6,4
The mode is the most frequently occurring value.
The value 2 occurs three times, more than any other.
mode is 2.
2.
Hence, the
Find the median of the observations in the prior problem.
The median is the middle value in the ordering, or where the
number of values is even it is the mean of the two “middlemost”
values. For this problem, values must be ordered and then the
mean of the two middlemost values must be calculated.
2,2,2,4,4,5,6,7
The two middlemost values are both 4's.
values is also 4: (4 + 4)/2 = 4
3.
The mean of these two
What is the mean of the observations in problem 1 above.
10
To calculate a mean, sum and then divide by the number of
observations.
2+7+2+2+4+5+6+4 = 32
There are eight observations (N = 8): 32/8 = 4
4.
Find the median of the observations 2,119,2,2,4,5,6,4.
compare to observations in problem 1.)
(Hint
Changing the highest value to a more extreme value will not
affect the value of the middle value (or of the middlemost
values) and hence will not affect the median.
The median, thus, is 4.
5.
Find the mean of the values in the prior question.
Changing any value will affect the mean.
(2+119+2+2+4+5+6+4)/8 = 18
6.
Which measure of central tendency, the median or the mean, is
preferred for the data presented in the prior two problems?
The mean is a misleading measure when there are outliers or
extreme values (e.g., 119).
The median is preferred.
7.
Consider the following frequency table:
Value
Frequency
Percent
poor
5
7
fair
10
14
good
40
57
excellent
15
21
What is the mode? median? mean?
The mode is the value with the greatest frequency. The median in a
table is the first value with a cumulative percentage greater than or
equal to 50. The mean requires interval/ratio level data.
The mode is “good” which occurs more than any other value.
11
Cumulative percentages can be derived by summing percentages: the
first cumulative percentage greater than 50 is for “good”. The
cumulative percentage for good is: 7+14+57=78. The median is “good”.
The mean cannot be computed as the data is only at the ordinal level
of measurement.
8.
At a given mental health clinic, 22 persons are diagnosed as
having a depression-related condition, 14 as having a personality
disorder-related condition, and 5 as having a schizophrenic
condition. What is the mode? median? mean?
To determine whether a given measure of central tendency can be
calculated, one should focus on a variable’s level of
measurement. The mental health condition variable is at the
nominal level. Only the mode can be calculated at this level. (A
nominal-level variable has no mean and no median.)
The mode is depression-related condition, which occurs most
often. There is no mean. There is no median.
Chapter 4
1.
Sample 1 consists of 20 women and 80 men. Sample 2 consists of
200 women and 800 men. In which sample is the variability of
gender greater?
Percentages rather than frequencies are used to assess
variability.
In both samples, percentages are identical – 20% women and 80%
men. Hence, variability is equal in the two samples.
2.
What is the range of the following data: 88,21,33,16,52,99,71
To calculate the range, subtract the lowest score from the
highest.
99 - 16 = 83
The following data will be used for the next several problems:
12, 18, 15, 16, 14.
3.
Calculate the mean deviation of the just-presented data:
12
To do so, carry out the steps as presented on page 61:
1)
The mean is (12+18+15+16+14)/5 = 15
2)
Deviation scores are:
12-15=-3; 18-15=3; 15-15=0; 16-15=1; 14-15=-1
3)
Absolute values of deviation scores are: 3, 3, 0, 1, 1
4)
Sum of absolute values is: 3+3+0+1+1=8
5)
Dividing by N yields the mean deviation: 8/5 = 1.6
4.
Via formula 4.5 on page 63, calculate the standard deviation
(s)of the just-presented data.
Calculate the mean. Then use of a grid as on page 62 helps.
As calculated in prior problem, the mean is 5.
X
‾
X
X - X
‾
(X - X
‾)2
12
15
-3
9
18
15
3
9
15
15
0
0
16
15
1
1
14
15
-1
1
sum = 20
The next step is to divide by (N - 1): 20/(5 - 1) = 20/4 = 5
Finally take the square root: the square root of 5 is 2.24 which is
the standard deviation.
5.
Via formula 4.7 on page 65, calculate the variance of the justpresented data.
The next to the last step of the standard deviation formula (just
prior to taking the square root) is the variance.
The variance is 5.00.
6.
Here are the scores in Group 1: 77, 56, 93, 84, 37
13
Here are the scores in Group 2: 75, 54, 91, 82, 35
Is the standard deviation of scores:
a) Greater in Group 1 than in Group 2?
b) Equal in the two Groups?
c) Greater in Group 2 than in Group 1?
Hint: compare the two sets of scores closely.
Adding or subtracting a constant does not affect the standard
deviation.
All scores in Group 2 are exactly two points lower than their
corresponding score in Group 1. One could think of the scores in
Group 2 as having been formed by subtracting (the constant) “2"
from each score in Group 1. Subtracting the constant makes the
mean of Group 2 lower, but does not affect the spread of scores
around the mean. In other words, the standard deviations in the
two Groups are equal. Choice “b” is correct.
6.
Suppose that 97 of 100 (97%) of students pass a given test. Why
would it be difficult to assess the possible association of
gender to passing (versus failing) the test?
Where variability is very limited it is difficult to study
relationships.
Almost everyone has passed the test, and, thus, it will be
difficult to study the possible relationship of gender to
passing. (If virtually no one fails then the variable pass versus
fail is, in effect, almost not a variable at all – it is very
nearly a constant).
Chapter 5
1.
Name several “real world” variables that precisely follow a
normal distribution.
The normal distribution is a mathematical, abstract distribution.
No real variables precisely follow a normal distribution.
2.
What do you think is the shape of the distribution of calories
consumed by you each day last year?
Here is a response that probably applies to most:
14
Well, it is perhaps pretty close to normal, that is, on
most days I eat an intermediate, “middle” amount of
calories, and on fewer days I eat less or more. Thinking
further, on some days I really pig out, that is I eat huge
numbers of calories. These few huge number of calorie days
are, essentially, outliers that would stretch out the
positive tail a long way. As the positive tail is extended
more than the negative, I’d say the shape of the
distribution is positively skewed, at least to a modest
degree.
3.
What do you think is the shape of the distribution you would get
from throwing a dice, say, 1000 times and each time recording the
value thrown (a 1,2,3,4,5 or 6).
Where each value has (about) the same frequency, a distribution
is said to be flat or rectangular.
Though luck would cause some variation in the numbers thrown,
1000 throws is a lot of throws, so luck would tend to “even out”
and you should throw about the same number (though with some
variation) of each. Hence, the shape of the distribution would
be fairly close to rectangular. (It certainly would not be close
to normal.)
4.
What do you think is the shape of the distribution of the number
of movies seen in the past month by students at your
college/university?
Most folks watch just a few movies. Some watch none. On the
other hand, a few folks are movie junkies and watch tons and tons
of movies.
Those few who watch tons of movies are positive outliers. They
stretch out the positive (right) tail of the distribution. Thus,
most likely, the distribution is positively skewed.
5.
What percentage of cases in a normal distribution are below the
mean?
In a normal distribution, the mean, median, and mode all have the
same value. By definition 50% of cases are below the median.
50% of cases in a normal distribution are below the mean.
6.
In a normal distribution, what percentage of cases have z scores
of -1.00 or below?
15
In a normal distribution: 1) 50% of cases are below the mean and
2) 34% are between the mean and 1 standard deviation below the
mean (see figure 5.13 on page 86)
50% - 34% = 16%
7.
Ann scored 74 on a spelling test on which the mean was 84 and the
standard deviation was 10. The shape of the distribution of
scores on the test was negatively skewed. What is her z score?
A z score indicates how many standard deviations a score is above
or below the mean. One may calculate a z score for any
interval/ratio level variable, even those with skewed
distributions. (What one can’t do when distributions are not
normal is figure out percentile ranks, percentages, etc.)
Ann’s score is 10 points below the mean. The standard deviation
is 10 points. Therefore Ann scored 1.00 standard deviation below
the mean. This is the same thing as saying that her z score is
1.00.
8.
On the same spelling test, Shawnda scored 89. What is her z
score?
One can simply use the z score formula on page 88.
z = (89 - 84)/10 = 0.50
9.
A distribution is negatively skewed.
or below the mean.
Is a z score of 0.32 above
Even for a nonnormally distributed deviation, z scores above 0.00
are above the mean.
It is above the mean.
10.
What is the percentile rank of a z score of 0.32 in a negatively
skewed distribution?
The normal distribution table (page 448) can only be used when
variables are normally distributed.
As the distribution described has a negative skew, the asked for
percentile rank cannot be determined.
16
11.
Given a normal distribution, what is the percentile rank of a z
score of 0.32?
Given that the shape is normal, the percentile rank can be
determined.
Via the normal distribution table on page 448, 12.6% of cases are
between the mean and a z score of 0.32. In a normal
distribution, 50% of cases are below the mean:
50 + 12.6 = 62.6 is the percentile rank
12.
Lucia scored 83 on her self-esteem inventory.
self-esteem relative to her peers?
Does she have good
Where variables are abstract, one needs z score (or means and
standard deviations so that z scores can be determined) to
interpret a score.
Without the just-mentioned data, one cannot respond to the
question; one does not know whether she has good self-esteem.
13.
The mean of the just presented self-esteem inventory is 100 with
a standard deviation of 10 points. Does Lucia have good or poor
self-esteem relative to her peers?
Now one can respond. Lucia’s self-esteem is well below the mean:
(z = (83 - 100)/10 = -1.7) She has reasonably poor self-esteem.
14.
Given the information presented in the prior problem, what is the
percentile rank of Lucia’s self-esteem score?
To compute percentile ranks, one needs to know that the
distribution is normally distributed.
As the shape of the distribution is not stated (and thus one does
not know that it is normal or nearly so), one cannot determine
Lucia’s percentile rank.
15.
Presuming a normal shape, what is Lucia’s percentile rank?
Now this can be computed via the normal distribution table on
page 448.
Via the third column, 4.5% of cases are more extreme than -1.7.
All of these more extreme cases have lower values (are more
extreme in a negative direction). Hence, the percentile rank is
4.5%.
17
Chapter 6
1.
Indicate whether you believe each pair of variables is related:
a)
How loud one partner snores and how much sleep the other
gets.
Yes, the louder the snoring, the less the sleep.
b)
Birth order and grades in school.
Probably so, though the relationship is likely weak. Some
studies (and common wisdom) suggest that first-borns strive
harder to achieve and, thus, on average, earn slightly
higher grades than their siblings.
c)
How late students stay out at a party and level of
concentration in class.
Yes, the later one stays out (on average) the less they
concentrate in class.
d)
Ethnicity of actors in TV shows and whether or not they
play a lead role.
My experience is that minority actors may be more likely
than white actors to be assigned to a supporting role
rather than to a lead one. If so, the variables are
related.
e)
Left-handedness
Ridiculous question. Only one variable, so we can’t assess
whether there is a relationship.
f)
How much one likes to dance and how much they like green
beans.
Not sure but presumably whatever relationship there is is
quite weak. So, let’s just say no, these variables are not
related.
18
Observe the following contingency table:
Table 1
Gender and Choice of Social Work Concentration
Female
Male
Direct Practice
(DP)
100
84%
40
67%
Admin. and
Community
Practice
(ACP)
20
16%
20
33%
120
67%
60
33%
2.
Respond to the following:
a.
b.
c.
d.
140
78%
40
22%
140
[answers provided to the right]
What is the frequency of men in ACP?
[20]
What are the column margin frequencies?
[120,60]
Does the table present row or column percentages? [column]
Which concentration do men choose most often?
[DP]
Note: the fact that men choose DP more often than they
choose ACP is irrelevant to the assessment of relationship.
e.
f.
Are the column percentages for women and men identical?
[no]
Are women or men more likely to choose DP?
[women]
This is the key to the assessment of relationship, that the
two groups differ in the pattern of their choices, that
women are more likely than men to choose DP.
g.
h.
Which variable is the dependent variable?
[concentration]
What percentage of women choose DP? Men? [84% versus 67%]
The fact that percentages differ is the key.
do, gender and concentration are related.
Because they
Comment: Frequencies can differ just simply due to
numbers. For instance, presume a situation in which: 1)
there are twice as many female students as male students 2)
the same percentages of women and men choose DP. In this
situation, more women than men choose DP but this greater
number is simply due to the fact that women outnumber men.
In this situation, the variables are unrelated.
i.
Are gender and concentration choice related?
19
[yes]
i.
In your own words, describe that relationship.
Women are more likely than men to choose DP. But that is
just one of many ways to describe the relationship (see
page 104) For instance, one could also say: Women are less
likely than men to choose ACP.
j.
Via the column percentages, what is the D%?
[84%67%=17%]
k.
Via Table 6.4 on page 109, interpret size of association.
The relationship is medium in size.
3.
m.
Does the table follow the text’s tip (from the bottom of
page 104) to make the independent variable the column
variable?
[yes]
n.
Are the presented percentages in accord with the text’s
recommendation (in middle of page 102) to use column
percentages when the independent variable is the column
variable?
[yes]
Eighty-one percent of non-minority college students versus 71% of
minority students indicate that they have a personal computer.
Are minority versus majority status and computer ownership
related? If so, how strong is that association?
To determine whether variables are associated, see whether
percentages differ. To gage size of association, compute the D%
and use table 6.4 on page 109 as a guide.
Percentages differ, so the variables are related. The D% is 10%,
which most would view as indicate of a reasonably weak
relationship.
4.
Ninety-five percent of men versus 90% of women know the name of
at least one player on their college’s football team. What’s
wrong with simply dividing 95% by 90% to determine that the ratio
of percentages is 1.056.? Calculate the R% in a better way.
When computing the R%, one should use percentages closer to 0%
rather than those closer to 1.
The better calculation follows: If 90% of women know a name, then
10% do not. For men, 5% do not know: 10%/5% = 2.00 = R%. Hence,
women are twice as likely as men not to know a name.
5.
24% of social work students versus 8% of psychology students
participate in a volunteer work day. What is the ratio of
20
percentages with social work students considered to be the first
group? With psychology students considered to be the first
group? Which ratio is easier to communicate about and
understand?
To calculate, divide the percentage in the first group by that in
the second.
24%/8% = 3.00 with social work students as the first group
8%/24% = 0.33 with psychology students as the first group
Probably the first ratio is easier to communicate. One can say:
“Social work students are three times more likely to volunteer
than are psychology students.” On the other hand, one could also
say: “Psychology students are one-third as likely to volunteer as
social work students.”, which is perhaps only a bit more
difficult to understand.
Chapter 7
For convenience Table 1 from the problem set for the prior chapter is
repeated here:
Table 1
Gender and Choice of Social Work Concentration
Female
Male
Direct Practice
(DP)
100
84%
40
67%
Admin. and
Community
Practice
(ACP)
20
16%
20
33%
120
67%
60
33%
1.
140
78%
40
22%
140
What are the odds that a woman will choose DP?
To determine odds divide the number that experiences event by
number that does not:
100/20 = 5.00
21
2.
3.
4.
What are the odds that a woman will choose ACP? [20/100=0.20]
What are the odds that a man will choose DP?
[40/20=2.00]
What is the odds ratio with women considered as the first group
and DP as the event?
To determine an odds ratio, divide odds for the first group by
those for the second.
Odds for both groups have already been calculated:
5.00/2.00 = 2.50
5.
Now compute the odds of choosing DP with men considered to be the
first group.
One can use the method described in the prior question to answer
this question Alternatively, when group designations change, the
resulting odds ratio is a reciprocal. (To calculate a number’s
reciprocal, divide 1.00 by that number.)
Via the first method: 2.00/5.00 = 0.40
Calculating the reciprocal odds ratio: 1/(2.50) = 0.40
6.
Which odds ratio conveys a stronger relationship, 2.50 or 0.40?
Hopefully, it is intuitive to you that the strength of the
relationship does not change simply because we change which group
is designated as the first group. Further, odds ratios that are
reciprocals convey the same strength of relationship.
The two odds ratios convey the same strength of relationship.
7.
Calculate the odds ratio for choosing ACP with women considered
to be the first group.
Simply divide one odds ratio by the other, or, alternatively work
directly from the numbers in the contingency table.
Via the method of dividing one odds ratio by the other:
The odds for women were computed earlier to be 0.20 For
men these are : 20/40=0.50. Thus, the odds ratio is:
0.20/0.50 = 0.40
Via the method of using the numbers in the table:
(20/100)(20/40) = (20/100)(40/20) = 800/2000 = 0.40
22
8.
Using the word likely, interpret (communicate about) the odds
ratio in the table.
Generally, odds ratios greater than 1.00 are easier to
communicate about than are those less than 1.00. Hence, I will
communicate using the odds ratio 2.50 rather than 0.40.
Two good ways to communicate results are:
Men are two and one-half times more likely than women to
choose ACP. (Or, to be more formal: the odds that men will
choose ACP are two and one-half times greater than are
those for women.)
Women are two and one-half times more likely than men to
choose DP. (Or, the odds that women will choose DP are two
and one-half times greater than are those for women.)
9.
Via Table 7.4 on page 118, interpret the size of the association
between gender and concentration choice.
An odds ratio of 2.5 conveys a relationship that most would
regard as being of medium strength.
10.
Which odds ratio conveys the stronger association between
variables, 0.1 or 7.0?
Odds ratios that are reciprocals convey the same size of
association. So, one way to answer this question is to find the
reciprocal of 0.1 and see if that is larger or smaller than 7.0.
To find a number’s reciprocal, divide 1.0 by that number).
reciprocal of .1 = 1.0/.1 = 10.0
As the reciprocal of 0.1 is 10 and 10 is larger than 7, one may
conclude that an odds ratio of 0.1 conveys a larger association
than does one of 7.0.
11.
(T or F) As the odds ratio gets closer and closer to 1.0, this
conveys that the relationship is weakening rather than
strengthening.
An odds ratio of 1.0 conveys the absence of association.
as the odds ratio approaches 1.0, this conveys weakening
relationship.
The correct answer is True.
23
Hence,
12.
Presume that an odds ratio is less than 1.0. (T or F) As this
odds ratio gets closer and closer to 0.00, this conveys
strengthening of rather than weakening of relationship.
Presuming that an odds ratio is less than 1.0, the closer that it
gets to 0.00, the stronger the association. For instance, an
odds ratio of .1 conveys a stronger association than does one of
.2.
The correct answer is true.
13.
Where one or more variables in a nondirectional relationship has
more than two categories, this text recommends using what measure
of size of association?
In general, it recommends using Cramer’s V.
14.
Among those who study less than one hour, 60% pass a test. Among
those who study one to two hours, 80% do so. Among those who
study more than two hours 70% pass the test. Is there a
relationship between studying and passing? Is this relationship
directional?
Whenever there is a difference in percentages, there is a
relationship. For the relationship, to be directional, the
percentages must have a directional (always increasing or always
decreasing) pattern.
There is a relationship, but the relationship is not directional.
15.
Among those who study less than one hour, 60%
those who study one to two hours, 80% do so.
study more than two hours 90% pass the test.
relationship between studying and passing? Is
directional?
pass a test. Among
Among those who
Is there a
this relationship
There is a relationship and the relationship is directional. (The
first paragraph in Section 8.13 in Chapter 8 on page 150 is a
note on dichotomous variables that is relevant to this problem.)
Chapter 8
1.
Find the lowest (vertical) marker in the scatterplot on page 131.
How far did this case long jump and how high did it high jump.
24
Trace left from the marker to determine the score on the vertical
axis variable and down to determine that on the horizontal axis
variable.
Tracing left, the case high jumped about 3.2 feet.
s/he long jumped about 10.0 feet.
2.
Tracing down,
Here are z scores for five persons for two statistics tests.
Person
z score on test 1
z score on test 2
Sally
.272
.996
Juan
-1.30
-1.54
Yi
.795
-.325
Jeremy
-.774
.183
Mary
1.00
.692
What is the correlation of scores on the two tests?
As formula for r (8.1 on page 133) indicates, to compute r one
multiplies the z score on the first variable by that on the
second and then divides by N - 1.
Person
z score on test 1
z score on test 2
z on 1  z on 2
Sally
.272
.996
0.27
Juan
-1.30
-1.54
2.01
Yi
.795
-.325
-0.26
Jeremy
-.774
.183
-0.14
Mary
1.00
.692
0.69
sum = 2.57
r = 2.57/(5 - 1) = 2.57/4 = .643
3.
The independent variable is plotted on the ___________ axis and
the dependent variable is plotted on the ___________ axis.
Independent = x (horizontal) axis
Dependent = y (vertical) axis
25
4.
The correlation between number of visits of family members and
reported happiness (measured on a standardized scale) of nursing
home clients is home is 0.53. Would you characterize this as a
strong or a weak association?
Via Table 8.3 on page 137, this relationship is a reasonably
strong one.
5.
Timothy’s z score in self-esteem is -0.80. The correlation of
self-esteem scores and scores on a satisfaction with work scale
is .44. What is Timothy’s predicted z score on the work scale.
Formula 8.2 on page 138 directs one to multiply the z score on
the first variable by the correlation.
predicted z score on satisfaction = .44(-0.80) = -.352
6.
The correlation of X and Y is 0.25. As the z score on X
increases by 1.00 unit, what is the change in the predicted z
score on Y?
As discussed in Section 8.6.2 on pages 139-140 and also on page
148 (last two paragraphs), r conveys the predicted change in
standard deviation units on one variable as the other increases
by one standard deviation. An increase of 1.0 (one) z score is
the same as an increase of one standard deviation.
As the z score on X increases by 1.00, the predicted z score on Y
increases by 0.25.
7.
The correlation of score on a family functioning measure and a
child’s grades in school is .23. Following calculation of this
correlation, the researcher realizes that she mistakenly scored
each case 10 points too high on the family functioning measure.
What will be the correlation of family functioning scores and
grades after she subtracts 10 points from each family functioning
score to correct for the error?
As discussed in Section 8.7 on page 140, adding or subtracting a
constant from each score (that is from all scores) does not
affect the value of r.
The correlation will be .23.
8.
The correlation between two variables is -.70. What is the
proportion of explained variance? What is the coefficient of
determination, r2? What is the percentage of explained variance?
As discussed on page 141, to determine the proportion of
explained variance – this is what r2 measures – one squares r:
26
r2 = -.7  -.7 = .49;
9.
49% of the variance is explained
A given regression equation is:
Predicted Y = -4 + (-5)X
What is ...
a.
b.
c.
d.
the
the
The
The
constant
[-4]
regression coefficient
[-5]
slope of the regression line
[-5]
change in predicted Y as X increases by 1.00.
[-5]
Comment: the regression coefficient, the slope, and the
change in predicted Y as X increases by 1.00 always have
the same value – they are, in essence, the same thing. The
coefficient B conveys all of these.
e.
Predicted Y when X = 3
Carry out the math: -4 + (-5)(3)= -4 +(-15) = -4-15 = -19
f.
Predicted Y when X = 0
Here the predicted Y is simply the constant, -4:
-4 + (-5)(0) = -4 + 0 = -4
g.
Predicted Y when X = -3
-4 + (-5)(-3) = -4 + 15 = 11
10.
A given scatterplot displays two variables using z scores. The
correlation between these variables is 0.33. What is the slope
of the regression line?
Referring to page 148, the slope of the regression line for a
standardized regression equation (that is, for a regression
equation that uses z scores equals r (the correlation
coefficient).
The slope of the regression line is 0.33.
11.
A given regression equation examines the relationship between the
number of school absences (dependent variable) and family income
(independent variable) measured in dollars. Would the
standardized regression equation or the unstandardized regression
equation be preferred for communicating about this relationship?
Or might both be helpful?
To some degree this is a matter of judgment.
27
As both variables are tangible and in everyday, easy to
understand units of measure, the unstandardized regression
equation would have easily interpreted meaning. It would convey
the predicted change in absences as income increased by $1.00.
On the other hand, the regression coefficient in the standardized
equation is, simply, r, which conveys strength of association.
Hence, both equations would yield useful information. (Where
both variables are not tangible/familiar, the unstandardized
equation is, generally speaking, less useful.)
12.
Consider an upcoming statistics test that covers the past four
weeks of work. Suppose that I developed a scatterplot showing
the relationship between time spent studying (independent
variable, X axis) and grade earned on the test (dependent
variable, Y axis). Suppose further that I could measure time
studying all the way out to 500 hours (let’s presume, say, that
this was an experiment, where for the purposes of science, some
persons agreed to study for 500 hours). What do you think would
be the shape of the line/curve (or scatterplot dots) that would
depict the relationship between time spent studying and grade
earned? (Hint: would it be straight?)
Here is my thinking: Presumably for the first 50 hours of study,
the more one studied, the better, on average, they would do on
the test. After this amount of time, they would reach a point of
diminishing returns. Pretty soon, extra study would only make
them sleepy, frazzled, and nonfunctional. Hence, after 100 hours
or so, I would say that the line would slope downward, the more
the hours studied, the worse the grade. Those who studied 500
hours would all be asleep and miss the exam and score 0. If my
thinking is correct, the relationship between study and grade
will be a curvilinear one. (In a four-week period there are only
672 hours to begin with so 500 hours of study leaves little time
for anything else, including sleep.)
Chapter 9
1.
The mean number of days of school missed in the school year is
7.0 at School 1 and 14.0 at School 2. The standard deviation is
5 days at both schools. What is the standardized difference
between means?
Here the standard deviation of the two groups is assumed to be
the same. To find the SDM, divide the difference in means by the
standard deviation.
28
SDM = (7.0 - 14.0)/5 = -7/5 = -1.4
2.
Using Table 9.1 on page 158 as a guide, how would you
characterize in words the size of the difference in standard
deviation units calculated in the prior problem.
This difference would likely be characterized as very large.
3.
Students who attend one or more help sessions earn an average
score of 92 on a statistics test. Students who do not attend any
sessions earn an average score of 87. What is the SDM?
Trick question.
deviation
4.
Cannot calculate the SDM without the standard
Presume the standard deviation within groups for the prior
problem is 8 points. What is the SDM?
Divide the difference in means by the standard deviation.
SDM = (92 - 87)/8 = 5/8 = .625
5.
Where the SDM is 1.00, and assuming equal standard deviations and
normal distributions, about what percentage of cases in the group
with the higher mean are located above the mean of the group with
the lower mean? See if, using your knowledge of the normal
curve, you can reason out in your head. (You may want to look at
Figure 9.2 at top of page 161.)
The mean of the lower group is located 1 standard deviation below
that of the higher group. Thirty-four percent of cases in the
higher group are located between its mean and the mean of the
lower group (68% within 1 standard deviation divided by 2 = 34%).
Further (by definition) 50% of cases in higher group are located
above its mean.
34% + 50% = 84% of cases in group with the higher mean are
located above the mean of the group with the lower mean.
6.
You observe the following information regarding the number of
police contacts for youth participating in two different
intervention programs.
Group
Mean
SD
n
Group 1
Group 2
10.0
7.0
6.0
3.5
25
25
29
Is the SDM an appropriate measure for this data?
Why or why not?
For the SDM to be an appropriate measure, the standard deviations
of the groups should be approximately equal. The text suggests
that the larger should be no more than one-third larger than the
smaller.
(6.0 - 3.5)/3.5 = 2.5  3.5 = 0.71  100 = 71% larger
The SDM is not an appropriate measure as the larger standard
deviation is 71% larger than the smaller.
7.
Is the SDM an appropriate measure for the following data?
what is the estimated SDM? If so, comment on the size of
association.
Group
Mean
SD
n
Group 1
Group 2
10.0
7.0
8.0
6.0
25
25
If so
For the measure to be appropriate, the larger standard deviation
should not exceed the smaller by more than about 33%. Assuming
that the measure is appropriate, one must calculate an average
standard deviation via Formula 9.2 on page 163.
(8 - 6)/6 = 2/6 = 1/3 = 33% -> just barely within 33% guideline
Hence, we may calculate the SDM. The first step is to calculate
the average standard deviation within the groups:
(8 + 6)/2 = 14  2 = 7
Now use the average standard deviation to estimate the SDM
SDM = (10.0 - 7.0)/7 = 3/7 = 0.42
Via Table 9.1 on page 158, the difference is approximately medium
in size.
8.
Estimate the SDM for the following data and comment on the size
of the observed difference/relationship:
Group
Mean
SD
n
Group 1
Group 2
20.0
30.0
12.0
15.0
25
25
30
First, check to see if the 33% guideline is met. Next, calculate
an average SD. Next, carry out formula 9.2 on page 163.
(15 - 12)/12 = 3/12 = .25  100 = 25%: The guideline is met.
The mean standard deviation within the groups is:
(15 + 12)/2 = 27/2 = 13.5
SDM = (20.0 - 30.0)/13.5 = 10.0/13.5 = 0.74
Most would regard the size of the difference as large.
9.
An instrument measuring reading skills is administered to first
graders who have been exposed to two different reading curricula.
Calculate the SDM and comment on the size of association.
Group
Mean
SD
n
Curriculum 1
Curriculum 2
10.0
17.0
8.0
10.0
25
40
After checking to see if the guideline is met, this problem
differs from prior ones as sample sizes are unequal. Hence,
formula 9.3 on page 163 will be used to estimate the standard
deviation within groups.
(8-10)/10 = 2/10 = .20 = 20% -> the guideline is met
Via formula 9.3 on page 163 the estimated swg is:
[(258) + (4010)]/(25+40) = (200+400)/65 = 600/65 = 9.23
The estimated SDM is:
(10 - 17)/(9.23) = -7/(9.23) = -0.76
In the context of social science, this difference is reasonably
large.
10.
Where eta equals 0.16, what does eta squared equal? In this
situation, what is the percentage of explained/shared variation?
Would you characterize the size of the observed association, as
trivial, quite weak, or strong?
To calculate eta squared, one squares eta. One, then multiplies
by 100 to determine the percentage of explained variation. Eta
squared can be a misleading measure, as it “sounds” smaller than
the actual size of association would be judged to be.
31
First calculate eta squared: .16  .16 = .026
Hence, 2.6% of the variance is explained.
As a rough gage for interpreting size of association, consult
Table 8.3 on page 137. Via this table, the association is weak,
but not trivial. (Note that in using Table 8.3, one uses the
value of eta (.16) not that of eta squared (.026).)
11.
The Kendall’s tau-b that measures the association between support
for public welfare programs (1 = very low ... 5 = very high) and
liberal versus conservative political views (1 = very
conservative ... 5 = very liberal) is .45. Is there a
relationship between support for welfare programs and (liberal)
political views? If so, does the relationship have direction? If
so, is the direction positive or negative? Comment on the
strength of the association.
Only a tau-b of 0.00 would convey the (complete) absence of
directional relationship. Tau-b measures directional
association. Values of tau-b greater than 0.00 convey positive
relationship. Table 8.3 on page 137, for the correlation
coefficient, can provide a very rough measure of the size of
association indicated by tau-b.
Yes, there is a relationship. Yes, it does have direction. The
relationship is positive (as support increases, so also does
liberalism). Via table 8.3, the relationship is reasonably
strong.
32
Chapters 10 and 11 (combined)
Here are some data on type of foster placement for a hypothetical
state. Placement with kin versus other placement (i.e., non-kin) for
minority and nonminority children is presented.
Number of
children placed
Number placed in
kinship home
% placed in
kinship home
Minority
children
1000
700
70%
Nonminority
children
1000
440
44%
Total
2000
1140
57%
1.
Use the just presented table on foster placement to respond to
the following questions/statements: 1) Is there an association
between minority (versus nonminority) status and type of
placement (kinship home versus other)? 2) If there is an
association, describe it in your own words. 3) If there is an
association, how would you characterize its size? 4) Can you
identify any potential confounding variables, ones that might be
affecting, for instance, the size of the association?
(Clarification on reading this table: If, for instance, 70% were
placed in kinship home, you may assume that 30% were placed in
other homes).
I will number the thoughts that I have according to the four subquestions.
1.
If percentages differ, the variables are related.
2.
Section 6.4.3 in Chapter 6 on page 104 shows some
different ways to describe association. I will pick
one that follows the guidelines for describing
association.
3.
Table 6.4 on page 109 provides a starting point for
describing the size of association represented by a
given difference in percentages.
4.
Just try to put on my thinking cap on this one.
Here are the “answers”:
1.
Yes, the variables are related.
33
2.
To describe the relationship, here are a couple of
possibilities:
Minority children are more likely than
nonminority children to be placed in kinship
homes.
Nonminority children are more likely than
minority children to be placed in nonkinship
(other) homes.
2.
3.
The difference in percentages is 36% (70 - 44 = 36).
This is a large difference; the association is a
strong one.
4.
Perhaps minority children, as a group, are placed at a
younger age than are nonminority children. If younger
children tend to be placed with kin, then age at
placement may be a (partial) explanation for the
relationship. What about geographic area,
particularly city environment versus other environment
(suburbs, rural, etc.)? These are also possible
confounding variables. You might think of others.
The following table presents the just-discussed data, controlling
for environment, city versus other:
Children who live in city
Children in other
environments
Number
of
children
placed
Number
placed
in
kinship
home
% placed
in
kinship
home
Number
of
children
placed
Number
placed
in
kinship
home
% placed
in
kinship
home
Minority
children
800
600
75%
200
100
50%
Nonminority
children
400
240
60%
600
200
33%
Total
1200
840
70%
800
300
38%
Respond to the following questions:
1)
In each subgroup (city environment and other environment)
is there an association between minority status and kinship
placement?
34
2)
Is the strength of association in each subgroup reasonably
similar or quite different?
3)
(Given that the strength of association in each subgroup is
similar): Is the strength of association when controlling
for environment – that is, the strength of association
within the subgroups -- stronger than, about the same as,
or weaker than that prior to control?
4)
Do you think that the relationship between minority status
and kinship placement is entirely due to type of
environment? Is it partially due to this?
My thoughts on each of the four just-posed questions:
1)
Just need to look to see whether percentages differ.
2)
Just compare the difference in percentages in the two
subgroups.
3)
Compare the (average) D% in the subgroups to that prior to
control.
4)
This probes ideas presented in Chapter 11. Basic logic that
would guide response would be: If there is still a
relationship within each subgroup following control, the
initial relationship cannot be due entirely to the
controlled for variable. Further, if the relationship
weakens when the variable is controlled for, then the
controlled for variable is indeed exerting some effect on
the size of the initial relationship.
“Answers” to the questions:
1)
Yes, in both subgroups, minority children are more likely
to be placed in kinship homes then are nonminority
children.
2)
For city: D% = 75% - 60% = 15%
For other: D% = 50% - 33% = 17%
Clearly the size of association within the subgroups (15%
and 17%) is similar.
3)
The average difference in the subgroups is: (15%+17%)/2 =
16%
The D% prior to control for environment was: 70% - 44% =
36%
35
Hence, following control, the size of association is
considerably smaller. Stated differently, the size of
association within the subgroups is smaller than the
association observed prior to control.
4)
3.
Type of environment is partially responsible as the
association weakens when it is controlled for. On the other
hand, as the association persists (does not disappear)
within the subgroups following control, it is not fully
responsible.
Regarding the just-presented table in Question 2, would you be
willing to conclude that the relationship between minority status
and kinship placement is causal? Why or why not?
Section 10.6, beginning on page 179, presents discussion on the
role of random assignment and difficulties in drawing causal
conclusions in its absence.
Most importantly, other variables not controlled for in the table
may be generating the relationship between minority status and
placement. For instance, age of child could be a factor if: 1)
minority children tend to be placed when younger and 2) younger
children are more likely to be placed in kinship care. Even if
the relationship persisted when one controlled for several
variables, s/he should be cautious in drawing a causal
conclusion. (There could still be other variables that are not
controlled for that have neither been thought of and/or
measured.) On the other hand, minority communities have
considerable experience with and expertise in raising children in
relative (kin) homes. So there are theoretical reasons to suggest
that minority kin step forward more often into the parenting
role. But the bottom line is that one should be cautious
regarding the attribution of causality.
4.
Suggest a causal model for the relationship.
Some of the following discussion perhaps takes you a bit beyond
the ideas presented in the Chapter 10 but, hopefully, is
instructive:
The fact that a relationship continues to exist even when type of
environment is controlled for opens up the possibility that
something about minority status (greater experience with raising
relatives outside of nuclear family?) does indeed contribute to
(cause) increased use of kin. We cannot draw a definitive
conclusion on this based on the current research design
(basically a survey rather than an experiment) because a myriad
of possible variables are not controlled for.
36
In causal models, the absence of direct effect is signaled by the
disappearance of an association following control (see Section
11.5.2, pages 198-200). As the relationship between minority
status and kinship placement does not disappear with control for
environment, we will presume that minority status does indeed
have a direct effect on the likelihood of kinship placement. (In
other words, we will presume a direct effect even though we
recognize that we have not controlled for all possible variables
and that it is conceivable that control for the “right” variable
could make the effect disappear.)
As has already been demonstrated, the presented table allows us
to examine the association of minority status to kinship
placement, controlling for city versus other environment. The
table also allows us to examine the association of type of
environment (city versus other) to kinship placement, controlling
for minority status. For instance, among minority children, 75%
of those in the city versus 50% of those in other environments
are placed in kinship homes, a difference in percentages of 25%
(D% = 75% - 50% = 25%). For non-minority children, this
difference in percentages is: D = 60%  33% = 27%. Hence, even
when controlling for minority status, type of environment and
kinship placement are associated. As such, it makes sense to
think of city environment as having a direct effect on kinship
placement.
Finally: Minority status and city environment are associated.
This association can be discerned by observing that 67% of
children placed in city environments (800/1200 = 67%) versus only
25% (200/800 = 25%) of those placed in other environments are
minority children: (D% = 67%  25% = 42%). One could view
minority status as causing increased likelihood of being in a
city environment. And, perhaps, one could also view city
environment as causing increased likelihood of minority status. I
think, however, that it makes most sense to think of minority
status and city environment simply as being associated. Where one
is not sure of or chooses not to speculate on the cause of an
association, that association is depicted via a curved arrow (see
for instance, Figure 10.7 on page 187).
In sum: 1) minority status may have a direct effect on kinship
placement, 2) city environment may have a direct effect on
kinship placement, and 3) minority status and city environment
are associated. So the best model is probably similar to that in
Table 10.8 on page 188.
37
Minority status

|


Kinship placement

City environment
(In case the above model does not print/appear correctly, a
curved arrow connects minority status and city environment and
straight arrows lead from each of these to kinship placement.)
4.
Which pattern from page 196 -- association persists at about the
same strength, association weakens, association disappears,
association varies by group -- is best illustrated by the table
presented in Question 2 above?
Examine the results.
“Association weakens” is the unequivocal choice. The association
clearly weakens, ruling out “association persists”. Yet, it
clearly does not disappear, ruling out “association disappears”.
The association is of similar size in the two subgroups (D% = 15%
versus D% = 17%), so this rules out “association varies by
group.”
5.
Does the Table in Question 2 above demonstrate an interaction
effect?
When there is an interaction, the pattern of association differs
according to the value of a third variable. The fourth pattern
“association varies by group” is an example of an interaction
effect.
Size of association is essentially the same (15% versus 17%) in
the two subgroups. In other words, size of association varies
hardly at all according to type of environment, city versus
other. Thus, there is no interaction effect.
Chapter 12
1.
A social work professor administers an instrument measuring
career motivations to the 30 students in her class. She reports
on their responses in the school’s newsletter. Is this sample of
students a random sample? Is she carrying out inferential or
descriptive statistics? Are the motivations of students in this
sample likely to be representative of the motivations of students
at the college/university?
38
Random samples are selected via formal methods of chance. If one
simply reports data for the sample (with no intent to draw
conclusions beyond the sample), this is descriptive statistics.
Though one does not know assuredly that a nonrandom sample is
biased, the more important point is that one does not know that
it is not.
Is a nonrandom sample. As there is no mention made that the
professor is attempting to use the data to draw any conclusions
beyond the sample, s/he is using descriptive statistics. One
suspects bias in the sample with respect to college population
for at least some career motivations. For instance, social work
students may be less motivated by money than the “typical”
student.
2.
A health researcher in a community is studying asthma in
children. Via random methods, he selects a sample of families in
the community. Where a family is selected, he includes all
children in that family in the study. With respect to children,
is the independence of selection criteria met?
For the independence of selection assumption to be met, the
selection of each case must have no bearing (effect) on whether
another case is selected.
The independence of selection assumption is not met (the children
and television example on page 216 is similar to this one).
3.
A research study examines a given group work therapeutic
intervention. Group members who receive the intervention interact
extremely closely with each other. Is the independence of
observations assumption met? What do you suggest doing?
As Section 12.5 on pages 219-220 discusses, it can be very
difficult to know whether, strictly speaking, this assumption is
met in situations where study participants interact considerably,
and thus the progress of one participant(s) could affect that of
another(s). Further, even where the assumption is violated, the
effect of this violation can be difficult to assess. And,
further, there may be no easy remedy.
To respond: We don’t know assuredly that the assumption has been
met. Pragmatically, one would likely assume that the assumption
was met and go ahead and carry out the appropriate statistical
procedures.
4.
A large state agency wants to determine the functioning level of
clients receiving services at its mental health clinics and,
thus, administers a scale to clients at one particular clinic.
39
The mean score at this clinic is 25 with a standard deviation of
10. Can one appropriately conclude that the mean score of
clients in the state is 25 and that the standard deviation is 10?
To make valid statistical inferences to a population, one’s
sample must have been randomly selected from that population.
No, as the sample is not a random one, the conclusion is not
appropriate.
5.
Presume for the prior problem that a random sample of 25 clients
from across the state was selected. (Such clients would be
selected from among those at all clinics across the state rather
than from just one.) Can one appropriately conclude that the mean
score of clients in the state is 25 and that the standard
deviation is 10? Is 25 the best estimate that one can make of
the mean score? Is 10 the best estimate of the standard
deviation? Is the estimate of the mean unbiased? Is the estimate
of the standard deviation unbiased?
(For this question, the reasoning behind and the actual answer
are combined into one response which follows:
As we are estimating the population parameters, we can’t
appropriately conclude the population mean is exactly 25 or that
the standard deviation is exactly 10. But given that the sample
is a random one, these are our best estimates. The estimate of
the mean is unbiased. The estimate of the standard deviation is
biased but the degree of bias is negligible. (That bias is so
small that we may ignore it.)
6.
The mean household income in a large city is $45,000 with a
standard deviation of $20,000 dollars. The distribution of
income is positively skewed. Suppose that one took an infinite
number of random samples of size 100 and for each one recorded
its mean.
a.
b.
c.
d.
e.
What is the special name of the distribution that one would
build by the above process?
What would be the mean of the distribution?
What would be the standard deviation of the distribution?
What would be the shape of the distribution?
What percentage of cases would be located within one
standard deviation of the distribution’s mean?
The central limit theorem as presented in Section 12.8 (pages
223-5) guides reasoning. The mean will be the population mean.
The standard deviation will be that in the population divided by
the square root of sample size. Given that N > 100, the
40
distribution’s shape will be nearly normal. The percentage of
cases within one standard deviation is common knowledge by now.
a) the sampling distribution of the mean
b) $45,000
c) $20,000  100 = $20,000  10 = $2,000
Note that in this document the square root symbol (√) is always
interpreted as spanning the number(s) that follows it. For
instance, √100 conveys the square root of 100.
d) close to normal
e) about 68%
7.
All facts the same as in prior problem except that sample size is
400 rather than 100. Respond to the same questions.
a)
b)
c)
d)
e)
8.
the sampling distribution of the mean
$45,000
$20,000  400 = $20,000  20 = $1,000
close to normal
about 68%
Continuing to think about the prior problem. If the mean of a
given sample is $44,000, what is the sampling error?
As presented on page 223, to determine sampling error, subtract
the population parameter from the sample statistic.
Sampling error = $44,000 - $45,000 = -$1,000
9.
(T or F) In actual research studies, the precise amount of
sampling error can be calculated.
False. In actual research, one knows only the value of the
sample statistic. As one has taken only a (random) sample, the
exact value of the population parameter is unknown. One does
know, however, that as samples get larger and larger, the likely
amount of sampling error tends to decrease. Even though one
knows this, s/he does not know the precise sampling error in any
particular study.
Chapter 13
41
1.
There are 100 students in a social work class. The mean age of
these 100 students is 25.0 years with a standard deviation of 6.0
years. What is the 95% confidence interval for the mean age of
students in the class? Trick question: Think about whether a
confidence interval is needed here, or, on the other hand,
whether you already know more than the confidence interval would
tell you.
Where one has access to the full population, there is no need for
a confidence interval.
We already know the mean age exactly. Hence, there is no need to
compute a confidence interval. Indeed, computing one would be
illogical, nonsensical, and wrong.
2.
Consider again the 100 students in the social work class in the
prior problem whose mean age is 25.0 years with a standard
deviation of 6.0 years. Presume that they attend a large state
university with, say, 60,000 students. What is the confidence
interval for the mean age of students at the University? (Think
carefully before responding.)
For a confidence interval to be accurate, the sample used to
compute it must be a random sample from the population to which
the confidence interval applies.
The social work students in the class are not a random sample of
students at their University. Hence, it is inappropriate to
calculate a confidence interval.
3.
In a given state, close to 100,000 persons receive a given
welfare benefit. The mean age of 100 such persons in one city in
the state is 22.0 with a standard deviation of 8.0 years. What
is the 95% confidence interval for the state? (Think.)
Confidence intervals are only valid when the sample is randomly
selected from the population for which the interval is formed.
As the persons all come from one city, they are not a random
sample of persons from across the state. Hence, a confidence
interval may not be formed.
4.
Exact same situation as presented in prior problem, except
presume that the sample is a random one drawn from the 100,000
persons in the state who receive the benefit. What is the 95%
confidence interval for the state?
As sample is a random one, one may calculate the confidence
interval. To do so use steps shown on page 237.
42
S
X
= 8/100 = 8/10 = 0.80
95%CI = 22.0  (1.96  0.80) = 22.0  1.57 = 20.43 to 23.57
5.
Calculate a 99% confidence interval for the same data, but this
time presume that the distribution is positively skewed.
So long as sample size is 100 or greater, the shape of the
sampling distribution will be nearly normal. This is the case
even when the distribution from which samples are selected is
distinctly nonnormal. Hence, the fact that the distribution is
positively skewed does not prevent us from calculating the
confidence interval. We simply need to use the formula for the
99% interval as presented on pages 238-9. (The estimate of the
standard error of the mean was calculated in the prior problem.)
99%CI = 22.0  (2.58  0.80) = 22.0  2.06 = 19.94 to 24.06
6.
144 participants in a welfare-to-work program are randomly
selected from a much larger number who participated in the
program. The mean income in the 12-month period following the
termination of the program for these 144 participants is $19,500
with a standard deviation of $8,100. What are:
a.
b.
c.
d.
e.
f.
g.
The shape of the sampling distribution of the mean?
To determine the CI, would the researcher actually
construct a sampling distribution of the mean by drawing
many random samples?
The estimate of the standard error of the mean?
The 95%CI?
The 99%CI?
Which confidence interval is wider?
Which confidence is better, the 95% or the 99%? (Think
carefully.)
Logic for the answers follows:
a.
b.
c.
d.
e.
f.
Wherever N > 100, we know the shape of the sampling
distribution of the mean is nearly normal (we know this
regardless of the shape of the distribution from which
samples are selected).
This distribution is theoretical and is estimated via
drawing only one random sample (the study sample).
Simply divide the standard deviation of the sample by the
square root of the sample size
Carry out the formula on pages 236-7
Carry out the formula on pages 238-9
We don’t even need to do the calculations to answer this
question – other things being equal (i.e. standard
43
g.
deviation and sample size), the 99%CI is always wider than
the 95% CI
Neither is “better” – the 95%CI is narrower (more precise)
but we have less confidence (95%) that it includes the
population mean. The 99% is wider (less precise) but we
have greater confidence (99%) that it includes the mean
Here are the answers:
a.
b.
c.
d.
e.
f.
g.
7.
Nearly normal
No.
$8,100 / (144) = $8,100/12 = $675
$19,000  1.96($675) = $19,000  $1,323 = $17,677 to
$20,323
$19,000  2.58($675) = $19,000  $1,742 = $17,258 to
$20,742
99% is wider
neither is better (though the 95%CI is certainly much more
common)
A random sample of 324 residents in a large city responds to a
poll on social services. Fifty-eight percent support plans to
build a new community facility to serve those with serious mental
illness and 42% oppose these plans.
a.
b.
c.
d.
e.
f.
g.
Is the sample size sufficient for calculating a confidence
interval for the proportion who support the facility?
What is the standard error of the proportion?
What is the 95%CI of the proportion?
Is the proportion in the city population (the population
from which the random sample was selected) located within
the range specified by the confidence interval?
How much confidence can we have that the true proportion
(unknown) in the city population is within the range
specified by the 95%CI?
What is the 99% confidence interval of the proportion?
How much confidence can we have that the proportion in the
population is within the range specified by the 99%CI?
Here are the answers:
a.
Yes, via Table 13.1 on page 241, where the sample
proportion is .58 the minimum needed sample size is 13.
(324 greatly exceeds 13).
b.
S
p
 (.58)(1  .58) / 324  (.58)(.42) / 324  .24 / 324  .0074  .027
44
c.
d.
e.
f.
g.
8.
95%CI = .58  1.96(.027) = .58  .05 = .53 to .63
No way to know for sure, as we have only selected a sample
for study
95% (as 95% of 95% CI’s include the population mean, we
have 95% confidence that any given one does so)
99%CI = .58  2.58(.027) = .58  .07 = .51 to .65
99%
About 100,000 clients participate in a public welfare program.
In a random sample of 289 clients, 72% hold jobs one year after
their discharge from the program.
a.
b.
c.
Is sample size sufficient for computing a confidence
interval?
What is the estimate of the standard error of the
proportion?
What is the 95%CI?
Answers follow:
a.
Where p = .72 via Table 13.1 on page 241, the minimum
sample size is 60, so the size guideline is met.
b.
S
c.
9.
p
 (.72)(1  .72) / 289  (.72)(.28) / 289  .20 / 289  .0070  .026
95%CI = .72  1.96(.026) = .72  .05 = .67 to .77
An agency serves 400 clients in a given year. It selects a
random sample of 200 clients and sends them a survey. 100 of
these clients respond. One question on the survey asked: “Would
you recommend this agency to a friend?” Eighty-three percent
replied yes.
a.
b.
c.
d.
Is sample size sufficient to calculate at 95%CI?
What percentage of the population was included in the study
sample? How does this affect whether one may carry out the
confidence interval formulae presented in the Chapter 13?
What is the response rate? Do you think the lower than
desired response rate introduces possible bias?
Should you compute a confidence interval?
Here are the answers:
a.
Where p = .83 the minimum recommended sample size is 180.
As the number who responded (N = 100) is less than this,
the formulae presented in the chapter are not sufficiently
45
b.
c.
accurate. (Footnote 3 on page 248 refers one to an
appropriate formula for this situation.)
50% of population was sampled (200/400 = 50%). This exceeds
the 10% guideline as presented at beginning of Section 13.4
on page 244. This being the case, the confidence interval
formulas presented in the Chapter do not yield accurate
results. See note 5 on page 248 for how to deal with this
situation.
100/200 = .50  100 = 50% response rate. Problem with a
low response rate is that one can never be sure of whether
response bias has been introduced and if so, of its
direction. A logical conjecture in this situation is that
those who respond have, on balance, more positive opinions
than those who do not. Hence, my best guess is that in the
full agency population of 400 clients, less than 83% would
make a positive recommendation to a friend.
Chapter 14
1.
What is the probability of selecting at random from a normal
distribution a case with a z score of 0.66 or higher?
For this problem, simply look in the third column of A.1 on page
448 which indicates the proportion of cases greater than the
given (positive) z score. This proportion is the probability.
The probability is .255.
2.
What is the probability of selecting at random from a normal
distribution a case with a z score of -0.66 or higher?
To find the proportion of cases greater than a negative z score,
one needs first to find the proportion of cases between that
score and the mean and then to add on .50 to reflect those cases
above the mean. This proportion is the probability.
The second column of Table A.1 on page 448 indicates that the
proportion of cases between the mean and a z score of -0.66 is
.245. Next: .245 + .500 = .745. Thus, the proportion of cases
with z scores greater than 0.66 is .745. Hence, p = .745.
3.
The appropriate statistical significance test is used to examine
which of two interventions is more effective at reducing
depression. (The test compares results for clients randomly
assigned to one intervention or the other. Presume also that
study participants were randomly selected from a very large
46
population.) The chosen statistical significance level is .05.
The probability resulting from the test is .29. As .29 > .05,
the null is accepted. May one:
a.
b.
c.
d.
Decide/conclude that the null is true.
Be sure that the null is true.
Be confident that the null is true.
If your response to b. and c. above is “No”, what may you
conclude?
The answers follow:
a.
b.
c.
d.
4.
Yes one decides/concludes that the null is true. Yet, in
making this decision, they recognize that they do not have
access to all of the cases in the population and, thus,
hold open the possibility that their decision could be
wrong (in other words, they recognize the possibility that
the null could be false).
No. Given that only a sample is studied, one can never be
100% sure that a null is true. (And it is also the case
that one can never be 100% sure that it is false.)
No. Even this language is too strong. One cannot be
confident that the null is true.
Basically, one may conclude that the null is not
inconsistent with study results. In other words, given a
true null, the study results are not unusual or unexpected.
The fact that the null is consistent with (not inconsistent
with) study results does not allow us to conclude that the
null is the only hypothesis that is consistent. As such, we
cannot be confident that it is true. One more comment: In
this example, we may also conclude that sampling error
alone is a plausible explanation for the study results.
(See first paragraph of Section 14.8 on page 257.)
A researcher finds extremely strong support in the literature
that Therapy A enhances self-esteem in young adults. She finds
also many studies indicating that Therapy B does not do so. She
decides to do a study comparing these two therapies. As such, she
randomly assigns participants to Therapy A or Therapy B and
conducts the appropriate statistical test on self-esteem scores.
a.
b.
Would you be inclined to use a directional or a
nondirectional hypothesis pair?
Formulate a directional pair. (Your pair should pertain to
mean self-esteem scores.)
Here are the answers:
47
a.
b.
5.
Though the choice of which type of pair to use is up to the
researcher, the compelling findings from prior studies
would recommend a directional pair.
The “trick” here would be to be sure that the direction of
expected study results is stated in the research hypothesis
(rather than in the null). The pair might read something
like:
Null:
The mean self-esteem score of those who receive
Therapy A is less than or equal to that of those
who receive Therapy B.
Research:
The mean self-esteem score of those who receive
Therapy A is greater than that of those who
receive Therapy B.
For each of the following, indicate whether one would accept or
reject the null given that the .05 statistical significance level
has been selected. (p stands for the probability of the study
sample result.)
a.
b.
c.
d.
e.
p
p
p
p
p
=
=
=
=
=
.008
.052
.0500000000000... (exactly .05)
.23
.02
Answers: a. reject;
e. reject
6.
c. reject;
d. accept;
Using the information presented in the prior problem, indicate
whether to accept or reject the null using the .01 statistical
significance level.
Answers: a. reject;
e. accept
7.
b. accept;
b. accept;
c. accept;
d. accept;
(T or F) Whenever one accepts (fails to reject) the null, the
study result is not statistically significant.
True. These two decisions always go together.
8.
(T or F) Whenever one rejects the null, the study result is
statistically significant.
True.
9.
These two decisions always go together.
(T or F) In deciding that the null is true, one is essentially
deciding that sampling error alone is a plausible explanation for
48
the difference between the study sample result and the condition
stated in the null.
True. One is concluding that the study sample result could
simply reflect sampling error. (If the sample result is due to
chance alone, there is no real difference in the population and,
thus, the null is true.)
Chapter 15
1.
(T or F) Where a test is robust to an assumption, violation of
that assumption greatly reduces the accuracy of that test.
False. The key idea behind robustness is that even when the
assumption is not met, the probability resulting from the test is
still highly accurate. Where a test is not robust to an
assumption, violation of that assumption can greatly reduce
accuracy. In such a situation, the test should not be used.
2.
(T or F) Where a test is not robust to an assumption, violation
of that assumption can greatly reduce the accuracy of that test.
True.
3.
(T or F) Where all of the assumptions of a test are met, that
test yields accurate results, that is, accurate probabilities
True. So, the gist is, one may carry out a given test when: 1)
its assumptions are met or 2) its assumptions are not met but the
test is robust to violated assumptions. But one should not carry
out a test where both of the following are true: 1) an assumption
is violated and 2) the test is not robust to that assumption.
4.
(T or F) Hypothesis statements pertain to the sample rather than
to the population from which the sample was randomly selected.
False.
5.
Hypothesis statements pertain to the population.
Presume that the mean of a given population is 10.0 points with a
standard deviation of 1.0 point and that that the population is
negatively skewed.. Suppose that one picks an infinite number of
random samples of a size 100. What is ...
a.
b.
c.
The mean of the sampling distribution
The standard deviation of the sampling distribution
The shape of the sampling distribution
49
d.
e.
f.
g.
The percentage of samples with means greater than 10.1
The percentage of samples with means that are either > 10.1
or < 9.9
The probability of selecting at random a sample with a mean
> 10.1
The probability of selecting at random a sample with a mean
> 10.1 or < 9.9
Answers follow:
a.
b.
c.
d.
e.
f.
g.
6.
10.0
.1 (1.0  100 = 1.0  10 = 0.1)
nearly normal (the key point here is that the sampling
distribution has a normal shape even though the population
from which samples were selected has a non-normal shape)
16% (this is the percentage of cases in a normal
distribution that are more than one standard deviation
above the mean; a score of 10.1 is 1 standard deviation
above the mean)
a score of 9.9 is one standard deviation below the mean; so
this question is asking for the percentage of cases that
are either more than one standard deviation above the mean
or more than one standard deviation below (the combined
percentage); this percentage is 32%
if 16% of samples have means greater than 10.1 then the
probability of randomly selecting such a sample is .16
.32 (just double the probability from the prior question)
Presume that the probability resulting from a given large sample
test equals .08. Presume also that the test is two-tailed, that
the .05 significance level is used, and that test assumptions are
met.
a.
b.
c.
d.
e.
f.
g.
What percentage of means in the sampling distribution
differ from the mean stated in the null by more than does
the study sample mean?
What percentage of sample means in the sampling
distribution are more extreme than the study sample mean?
What is the probability of selecting at random from the
sampling distribution, a sample with a mean that differs
more from the mean stated in the null by more than does the
study sample mean?
What is the probability that the study sample result is due
to chance alone?
Given a true null, what is the probability of obtaining the
study sample result or an even more extreme result?
Is the study sample result likely to be due simply to
chance?
Is chance alone a plausible explanation for the study
sample result?
50
h.
i.
Should the null be rejected?
Is the result statistically significant?
Answers follow:
a.
b.
c.
d.
e.
f.
g.
h.
i.
6.
8% (see top of p. 275 for discussion)
8% (see top of p. 275 for discussion)
.08 (see top of p. 275 for discussion)
.08
.08
The results of significance testing indicate the
probability that the study sample result is due to chance
alone. As this probability is only .08 (8 times in 100),
we would not conclude that it is likely that the result is
due to chance alone. On the other hand, via the criterion
of the .05 significance level the result (p = .08) is not
sufficiently unlikely (sufficiently rare) for us to
conclude with 95% confidence (see page 257) that chance
alone is not the explanation. In a nutshell, though chance
alone is a reasonably unlikely explanation (8 times in
100), it is not sufficiently unlikely (needs to be 5 times
in 100), for us to rule out chance with sufficient (95%)
confidence to reject the null.
Yes, even if chance alone is a somewhat unlikely
explanation, it is a plausible explanation
No, the probability given by the test (.08) exceeds that
associated with the significance level (.05) and so the
null is accepted (we fail to reject the null)
No. Whenever one fails to reject the null, the result is
not statistically significant.
(T or F) When the probability that chance alone could lead to a
result that is as extreme as or more extreme than the study
sample result is less than alpha, one rejects the null.
True. (As you should know, alpha is the probability
associated with the significance level.) When the
probability that the result is due to chance alone is
less than alpha, one rejects the null.
7.
In a large state, the mean family income of persons who have
graduated from its public welfare programs is $14,000. A random
sample of 225 persons takes part in an intensive program. The
mean family income in this sample is $14,900 with a standard
deviation of $4,500. The null hypothesis is that the mean of the
population from which the study sample was drawn equals $14,000.
The research hypothesis is that it does not equal $14,000. Based
on the just-presented information, respond to the questions 23a
to 23l on page 293 of the text.
51
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
8.
Nondirectional (as equal/not equal logic rather than
greater than/less than logic is used.)
The mean is simply the value stated in the null: $14,000.
To estimate the standard deviation, divide the sample
standard deviation by the square root of the sample
size:$4,500  225 = $4,500  15 = $300 (see p. 280 for an
example of calculation). The shape is extremely close to
normal.
z = ($14,900 - $14,000)/$300 = $900/$300 = 3.00
Let’s reword this question to ... a mean as high as or even
higher than”. The study sample mean has a z score of 3.00.
Via the normal distribution table on page 448-9, the
proportion of cases with z scores > 3.00 is .0013. So, the
probability is .0013.
To solve this, simply multiply the proportion from the
prior problem by 2: 2  .0013 = .0026
As the hypothesis pair is nondirectional, the statistical
test is two-tailed.
This is the proportion of samples with means that are as
extreme as or more extreme than the study sample mean. This
proportion is .0026 (as the test is two-tailed, cases in
both directions were considered)
.0026 is < .05
Where alpha equals .05, the rejection area in each tail
includes 2.5% of cases
Yes. [In a two-tailed test where alpha equals .05, the
lower rejection area consists of cases where z  1.96 and
the upper area consists of cases where z  1.96. In our
example, z = 3.00.]
We reject the null and accept the research hypothesis.
.0026 [The probability that result is due to chance alone
is always the same as – indeed is the same thing as - the
probability of obtaining the result given a true null.]
(T or F) When a result is not statistically significant, the
researcher (correctly) reasons that it may be due to chance
alone.
True. This is the nuts and bolts meaning of a result that is not
statistically significant. (In this case, one accepts the null.)
9.
(T or F) When a result is statistically significant, the
researcher (correctly) reasons that it is unlikely to be due to
chance alone.
True. This is the nuts and bolts meaning of a result that is
statistically significant. (In this case, one rejects the null.)
52
10.
At which level of significance, .05 or .01, does one have greater
confidence that the null is indeed false? (In responding presume
that the null is rejected.)
One has 99% confidence that the null is false when the .01 level
is used, but only 95% confidence when the .05 level is used. So,
one has greater confidence at the .01 level.
11.
Where the research hypothesis states “greater than,” the
rejection region is in the ________ tail. Where it states “less
than,” it is in the _______ tail.
Greater than -> upper; less than -> lower
12.
Indicate the appropriate decision regarding the null (accept or
reject) for each large sample test of X
� result :
a.
b.
c.
d.
e.
f.
g.
h.
Two-tailed test, alpha = .05,
Two-tailed test, alpha = .01,
One-tailed test, alpha = .05,
greater than, z = 2.22
One-tailed test, alpha = .01,
greater than, z = 2.22
One-tailed test, alpha = .05,
greater than, z = 2.22
One-tailed test, alpha = .05,
less than, z = -2.22
Two-tailed test, alpha = .05,
One-tailed test, alpha = .05,
hypothesis states less than
z = 2.22
z = 2.22
research hypothesis states
research hypothesis states
research hypothesis states
research hypothesis states
z = -1.88
z = -1.88, research
Answers follow:
a.
b.
c.
d.
e.
f.
g.
h.
reject as absolute value of obtained z  1.96
accept as absolute value of obtained z is < 2.58
reject as obtained z  1.645
accept as obtained z < 2.33
accept as obtained z < 1.645 (result in this example is in
opposite direction to that expected; opposite to that
stated in the research hypothesis)
reject as obtained z  1.645 (as research hypothesis states
less than, rejection region is in the negative tail)
accept as absolute value of obtained z is < 1.96
reject as absolute value of obtained z is  1.645 (Here is
an example of where a one-tailed test rejects the null but
a two-tailed one (see prior example) does not.)
53
Chapter 16
13.
(T or F) As sample size increases, other things being equal,
statistical power increases.
True. As sample size gets larger, the width of the sampling
distribution “shrinks” and, thus, smaller differences from the
null result in rejection. In essence, it gets easier to reject
the null and power increases.
14.
(T or F) The greater the likely amount of sampling error, the
greater the statistical power.
False. Absolutely false. When a result differs from the
condition stated in the null, one never knows assuredly whether
this difference is 1) simply due to sampling error or 2) also
reflects a real difference (i.e., that the null is false). The
less the likely amount of sampling error, the greater the ability
of the statistical test to rule it out as the sole explanation
for the sample result. (Viewed differently, the less the likely
amount of sampling error, the greater the test’s ability to
detect real differences.) Where sampling error can be ruled out
with sufficient confidence (at the .05 level, with 95%
confidence), one may reject the null.
15.
Where the risk of a type II error is .25, what is the statistical
power.
The risk of type II error is symbolized by  (beta).
As presented on page 297: Power = 1 - 
Power = 1.00 - .25 = .75
16.
Ho:  = 50, s = 20,  = .05, two-tailed test. For each of the
following: 1) estimate the standard error of the mean and 2)
indicate (approximately) the range of sample results that result
in rejection of the null at each of the following sample sizes.
a. n = 100
b. n = 400
c. n = 1600
d. n = 6400
In case you need help in “decoding” symbols,  (pronounced “mew”)
conveys the hypothesized sample mean; s is the standard deviation
in the study sample; and  is alpha, the probability associated
with the significance level. For simplicity, we will assume that
a difference of two standard deviations (rather than 1.96)
provides sufficient accuracy. Basically, any result that is more
54
than two standard deviations (two times the estimated standard
error of the mean) less than or greater than  will result in
rejection. So to answer the problem(s), one needs to: 1)
estimate the standard error (s divided by the square root of the
sample size), 2) multiply the estimated standard error by 2, 3)
add and subtract from the value stated in the null ( = 50), and
4) write out the approximate range of values that result in
rejection.
Here follow the answers:
a.
b.
c.
d.
17.
S
S
S
S
X
X
X
X
= 20/100 = 20/10 = 2; 2  2 = 4;
 46 or  54
= 20/400 = 20/20 = 1; 1  2 = 2;
 48 or  52
= 20/1600 = 20/40 = 0.5; .5  2 = 1;
 49 or  51
= 20/6400 = 20/80 = 0.25; .252 = .5;  49.5 or  50.5
Consider results for the prior problem as you respond to the
following questions.
a.
b.
As sample size increases, does the size of the difference
from the condition stated in the null that is necessary for
rejection increase or decrease?
(Presuming that the null is false), as sample increases
does power increase or decrease?
Some general comments follow.
As sample size increases, the standard error of the mean gets
increasing smaller, which tells us that the sampling distribution
is getting progressively narrower (see page 300 for visual
presentation of this concept). As sample size gets progressively
larger, the likely (expected) amount of sampling error is getting
smaller. (This is illustrated by the progressively “shrinking”
sampling distributions on page 300.). The less the expected
sampling error, the easier the job of the statistical test, which
is to see the degree to which it can rule out sampling error as a
plausible explanation for the difference between the condition
stated in the null and the sample result.
Answers follow:
a.
b.
18.
The necessary size of the difference decreases.
Power increases.
Suppose that you want to know which of two interventions A or B
is more effective in helping mothers of high risk new-born babies
keep their post-natal medical appointments. Suppose further that
55
you have 10 such mothers on your caseload. Presume that you
randomly assign 5 to receive each intervention. Comment on:
a.
The degree to which confounding variables can be ruled out
as explanation for observed differences that you may see.
b.
The likely statistical power.
c.
Presume that the null is accepted. This being the case,
should one conclude that the two interventions are equally
effective?
My comments follow:
a.
The fact of random assignment rules out confounding
variables. There is no reason to be concerned that
systematic differences between the two groups would affect
your results. So, if you would see a statistically
significant result you could have reasonably good
confidence that the explanation for that result is the
treatment intervention (greater effectiveness of one
intervention than the other) rather than some confounding
variable (one group had greater initial motivation to seek
medical care than the other, etc.)
b.
The problem here is poor statistical power. (One can’t be
sure the power is poor as, if, for instance, one
intervention was anticipated to work much, much, much
better than the other, the small sample size here might be
sufficient for adequate power (see Section 16.5.4 on page
305). But presuming that one was not expecting a huge
difference in effectiveness, power would be poor.) Given
the small sample size, chance fluctuations (same thing as
sampling error) are likely to be quite large in this
situation. Even if the study result reveals a large
difference from the condition stated in the null, this
difference will, most likely, not be sufficient to rule out
sampling error as a plausible explanation. Another way to
think about this is to reason that just by the luck of the
draw (luck of the random assignment process), the
characteristics of the two groups may differ considerably.
For instance, with only five in each group, it is not
unlikely that those mothers (randomly) assigned to one
group may be a much more motivated group of mothers than
those assigned to the other.
c.
Here is where “fail to reject” so much more clearly
expresses the conclusion that you should draw than does
“accept”. Your conclusion should be that there is
insufficient evidence to conclude that one intervention
works better than the other. You appropriately conclude
that the null (no difference) provides a plausible
explanation for the study result. In “accepting” the null,
56
you are concluding that sampling error alone is a
sufficient explanation for the difference between the study
sample result and the condition stated in the null (no
difference). The null is the simplest explanation, and,
so, if it is sufficiently consistent with study results, we
accept it. However, the fact that the null is consistent,
does not affirm that it is true. Another hypothesis may
also be consistent. In the current example, power was low
and we never had a realistic opportunity to reject the
null. Viewed differently, we never had a realistic
opportunity to consider whether the research hypothesis
might be true. (See Section 14.10, pages 260-1.)
19.
(T or F) Presuming that the study sample result is in the
expected direction, the size of the difference that is necessary
to reject the null is smaller for a one-tailed test than for a
two-tailed one.
True. With a one-tailed test, assuming the result is in the
expected direction, a smaller difference will result in
rejection. (And, thus, power is greater.)
20.
A school of social work wants to see whether there is a
statistically significant difference between the mean salaries of
its female(n = 80) and male (n = 8) graduates. Comment on the
likely statistical power.
Via Table 16.1 on page 308, one is tempted to respond
“fair/moderate” and this is not a bad answer. However, there is
very little variability in gender (more than 90% of graduates are
female). This low variability reduces power (see last sentence
in Section 16.7 on page 308), so “poor” or “low” are better
responses.
21.
You read the results of a very large experiment involving 10,000
women and 10,000 men that tells you that “the risk for men of
having “Health Problem A” exceeds that for women by a
statistically significant amount.” Assuming that this is the
only information reported, why should one not conclude that there
is (necessarily) a large difference in risk between men and
women? (If it is a statistically significant difference, it
follows that it is a large difference, doesn’t it?)
Where sample size is very large, even trivial-sized differences
may be statistically significant. This is so because with large
sample sizes, expected sampling error is very small, so even tiny
differences between the sample result and the condition stated in
the null (in this example, equal risk for men and women) may be
sufficient to reject the null. This being the case, you may not,
in this situation, conclude that the difference in risk is large.
57
The difference may be large, it may be medium, it may be small,
or it may be trivial. From the provided information, you don’t
know which of these is the case. (Some related examples are
presented on page 303.)
22.
When one uses the .05 significance level, does p = .04 convey a
statistically significant result? Should one reject the null?
Yes, the result is statistically significant. Yes, one should
reject the null. Statistical significance and rejecting the null
go hand in hand; they are one and the same. When the result is
statistically significant, you always reject the null.
23.
Pragmatically speaking, may statistical significance testing be
carried out in the absence of random sampling?
Most researchers say yes. This is the case because whether
chance is a possible explanation for the study sample result is
often an important question even when the sample is nonrandom. In
essence, one presumes random sampling (as this is a basic
assumption of, in essence, all statistical tests), and then
carries out the significance test.
24.
(T or F) (Presuming that the null is false), where power equals
.60, the probability that a correct decision to reject the null
will be made is .60.
True. Power is the probability that a false null will be
rejected.
25.
(T or F) Where power equals .30, the significance test has an
excellent opportunity to reject the null.
False. In this situation (given that the null is false), the
test has only a 30% chance of resulting in the correct decision
to reject.
26.
(T or F) The statistical power in a given situation is .50. Via
the standards of social science research, this is an acceptable
level of power.
False. By tradition, social science’s standard for acceptable
power is .80.
27.
(T or F) When a result is statistically significant, one accepts
(fails to reject) the null.
False. When a result is statistically significant, one rejects
the null. These two decisions/events go hand-in-hand and are, in
essence, the same thing.
58
28.
(T or F) When a result is statistically significant, the
difference between the condition stated in the null and the study
sample result is likely to be due to chance alone.
False. Absolutely false. When result is statistically
significant, this difference is unlikely to be due to chance
alone. It likely reflects a real difference, something more than
chance. Stated differently, where a result is statistically
significant, sampling error alone is an unlikely explanation for
the sample result. As such, one concludes that (in the
population from which the study sample was randomly selected) the
null is false, that is, one rejects the null.
Chapter 17
1.
(T or F) When the shape of the distribution of the population is
strongly skewed, one should have a sample size of at least 100 in
order to conduct a highly accurate one-sample t test.
False. In such a situation a sample size of 30 suffices (see
Table 17.1 on page 324).
2.
(T or F) As sample size increases, the shape of the t
distribution and that of the normal distribution become
increasingly different.
False. As sample size increases, the shape of the t distribution
increasingly resembles that of the normal distribution.
3.
The shape of the distribution of scores in a study sample has a
very strong (extreme) positive skew (N = 11). In this situation,
should one calculate a confidence interval? Why or why not?
No. One uses the shape of the study sample distribution to
estimate that of the population from which the sample was
(randomly) selected. Where the shape of the population is skewed
to an extreme degree, Table 17.1 guidelines recommend a minimum
sample size of 60. (Given extreme skew), where the sample size is
less than 60, the shape of the sampling distribution does not
closely approximate a t distribution. This being the case,
confidence intervals that make use of the t distribution are not
accurate.
4.
The mean score of a random sample of 25 clients on a depression
inventory is 50.0 points with a standard deviation of 10.0
59
points. The shape of the study sample’s distribution is a modest
negative skew.
a.
b.
c.
d.
e.
f.
g.
h.
Is sample size sufficient to calculate a confidence
interval?
What is the estimated standard error of the mean (sX�)
How many degrees of freedom are there?
What is the value of t in Table A.2 on page 450?
What is the 95%CI?
Interpret in your own words the “meaning” of the justcomputed 95% confidence interval.
What is the 99% confidence interval?
Interpret the just-computed 99% confidence interval.
Answers follow:
a.
b.
c.
d.
e.
f.
g.
h.
5.
Yes, via Table 17.1 on page 324 a sample of 8 would have
been sufficient.
= 10.0/25 = 10/5 = 2.00
X
For confidence intervals for sample means using the t
distribution: df = N - 1, so df = 25 - 1 = 24
Tracing down the 95%CI column to df = 24 yields a t of
2.064
95%CI = 50  (2.064)(2.00) = 50  4.128 = 45.872 <> 54.128
We can be 95% confident that the mean depression score in
the population from which the study sample was selected is
between 45.872 and 54.128.
99%CI = 50  (2.797)(2.00) = 50  5.594 = 44.406 <> 55.594
We can be 99% confident that the mean depression score in
the population from which the study sample was selected is
between 44.406 and 55.594.
S
The mean behavior problems score on a given scale of 64 children
who were adopted at or near in infancy is 80.0 points with a
standard deviation of 24.0 points. The mean score of a large
representative sample of “typical” children is 75.0 points. Your
task is to determine whether the difference between the mean
score of the sample of adopted children differs to a
statistically significant degree from that of the representative
sample of “typical” children. Your hypothesis pair should be
directional and the research hypothesis should state a higher
mean for the adoptees. Use the .05 significance level. Presume
that the study sample has a moderate negative skew. Use the justprovided information to respond to questions a. through m. for
question 21 on page 335.
a.
Null:
the mean of the population from which the sample
of adopted children was randomly selected is less
than or equal to 75.0.
60
Research:
the mean of the population from which the sample
of adopted children was randomly selected is
greater than 75.0.
Or to state the same in shorter format:
Null:
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
the mean for the adopted children is less than or
equal to 75.0
Research: the mean for the adopted children is greater than
75.0
The one-sample t test should be used. You would not use
the large sample test from Chapter 15 as the sample size is
less than 100.
Yes, as 64 is greater than the recommended minimum of 15
presented in Table 17.1 on page 324.
You were instructed to use a directional pair.
Whenever a directional hypothesis is stated, the
appropriate test is a one-tailed one.
df = N - 1 = 64 - 1 = 63
Via Table A.2 on page 450: Locating the column for a onetailed test where  = .05, the critical value of t for 60
degrees of freedom is 1.671 (the table does not list
degrees of freedom for 63 degrees; as such, one references
the next lower degrees of freedom, which is 60). Note that
in this example the research hypothesis stated “greater
than”; had it stated “less than” the critical t would have
been -1.671 (see decision rules on pages 324-5 in Section
17.6.5).
= 24.0/64 = 24.0/8 = 3.00
X
Carrying out formula 17.2 from page 319 and as illustrated
on page 330: t = (80.0 - 75.0)/3.0 = 5.0/3.0 = 1.60
In making decision, we reference the decision rule for a
directional hypothesis where the research hypothesis states
“greater than” as presented on page 328. In our example,
the obtained t of 1.60 is less than the critical t of 1.67.
Hence, via the decision rule, the null is accepted (we fail
to reject the null).
Accepting the null, does not allow us to conclude that it
is likely that the null is true. When we reject the null,
we can be confident that the null is false (at the .05
level, we are 95% confident that the null is false). But
when we accept it, the best that we can do is conclude that
it is plausible that the null is true. In the present
example, the t test just missed rejecting the null, so we
can be almost 95% confident that the null is false. But
via the rules of significance testing being almost 95%
confident is not sufficient for rejection;, so, thus, we
accept the null. (See Section 14.10 in Chapter 14 on pages
260-1 for more on this.)
S
61
Chapter 18
1.
Suppose that you have a first population with the following
characteristics:  = 10.0,  = 10.0 and a second where  = 10.0,
 = 7.0. What is the mean, standard deviation, and shape of the
sampling distribution of the difference between means if N = 100
for the first population and N = 49 for the second? Presume that
both population distributions have moderate positive skews.
Comments: Section 18.2.2 on pages 337-9 presents the logic for
responding here. The question is asking for the characteristics
that would result from: 1) selecting a sample of size 100 from
the first population, one of size 49 from the second, 2)
calculating the mean of each sample, 3) subtracting the mean of
the second from that of the first, and 4) plotting this result.
One would repeat steps 1 through 4 for an infinite number of
pairs of samples.
Answers: The mean of the sampling distribution is obtained by
subtracting 2 from 1: 10 - 10 = 0. Formula 18.1 on page 338 is
used to determine the standard deviation of the sampling
distribution (note that variances rather than standard deviations
are used in the formula
100 49

 1  1  2  1.41
100 49
The shape of the sampling distribution is very close to normal.
2.
Twenty-five married couples participate in a study to determine
whether wives or husbands report greater marital satisfaction.
Consider wives to be Sample 1 and husbands to be Sample 2.
Respond to the following questions:
a.
b.
c.
Are the samples independent or dependent?
Are the scores in Sample 1 independent from those in Sample
2
Which test, the dependent samples t test or the independent
samples t test should be used in this situation?
Answers follow:
a.
Wives and husbands “go together” as pairs, so the samples
are dependent
62
b.
c.
3.
No, the scores in the two samples would be presumed to be
related. Stated differently, one would expect that scores
of wives and their husbands are associated. More
specifically, one would expect a positive correlation
between these scores. (For instance, where wives have high
satisfaction one would expect (at least to some degree) the
same for husbands. Where wives are unsatisfied one expects
(at least to some degree) the same for their husbands.)
Given that the samples are dependent, the dependent samples
t test should be used.
Sally’s score is 30.0 in Sample 1 and 40.0 in Sample 2.
her difference score?
What is
To determine a case’s difference score, subtract the mean of the
second sample from that of the first.
Sally’s difference score is:
4.
30.0 - 40.0 = -10.0
A researcher prepares to conduct a dependent samples t test. The
mean of the first sample is 25.0. That of the second sample is
21.0. What is the mean of the difference scores.
As discussed in Section 18.6.3 on page 352, the mean of the
difference scores is equal to the difference between the sample
means.
The mean of the difference scores is: 25.0 - 21.0 = 4.0
5.
(T or F) The equal and unequal variances formulae for the
independent samples t test always yield identical values of t.
False. They typically yield different values. They always yield
the same value only when sample sizes are equal.
6.
(T or F) When sample sizes are equal, degrees of freedom for
equal and unequal variances formulae are always equal.
False. Though the value of t will be identical if sample sizes
are equal, the degrees of freedom will typically differ. Hence,
even when sample sizes are equal, it is important to know which
formula should be used, so that the appropriate degrees of
freedom can be referenced.
7.
At a given point in time, a child guidance clinic serves 72
children. Thirty-six children participate in an intensive play
therapy intervention. The children who participate were selected
by their therapists. The other 36 children do not participate
(presumably because their therapists did not think that play
therapy would be beneficial for them). Following the conclusion
63
of the play therapy workshop, mothers of children in both groups
fill out a scale that is designed to measure their children’s
overall mood. The higher the score, the better the mood. For
children who received play therapy, the mean score on the scale
is 68.0 points with a standard deviation of 12.0 points. For
children who do not receive the intervention, the mean score is
50.0 points with a standard deviation of 12.0 points. Scores in
both samples have a moderate positive skew. Presume a
nondirectional hypothesis pair. Refer to Question 19 on pages
355-6 and respond to a. through w. [Note: Question g, should be
changed to read “the assignment process was not random.”]
a.
b.
c.
d.
e.
f.
g.
The two samples are independent, so the independent samples
t test should be used. (The samples are independent because
the two groups of children were not linked or paired in any
way.)
Null:
Means are equal in the populations from which the
samples were randomly selected.
Yes. The guideline given for the independent samples t test
is that both samples should meet the guidelines presented
in Table 17.1 on page 324. Where the degree of skew is
moderate, Table 17.1 recommends a minimum sample size of
15. As both samples meet this guideline, sample size is
adequate.
In a sense, the 50 clients are neither a random or a
nonrandom sample. They are not a sample at all, but
instead represent the full population of children at the
clinic. They are certainly not a random sample. But as
discussed in Section 16.8 on pages 308-10, we will assume
the sample to be random and, by so doing, meet the
assumption of the statistical test that this is the case.
Clearly, the population is abstract/ hypothetical/
imaginary. (We have presumed the population so that a
significance test may be conducted.
True.
We altered Question G so that it states “the assignment
process was not random”. Now to respond: Given that the
assignment process was nonrandom, confounding variables
make it difficult to reach a causal conclusion. If we find
that the difference between means (Mean = 68.0 in therapy
group vs. 50.0 for other group) is statistically
significant, we can be confident (though not certain) that
it is not due to chance (sampling error) alone. However, we
will not know regarding whether the difference is due to:
1) the positive impact of the play therapy or 2)
differences between the play therapy kids and the other
kids that existed prior to the intervention. (These would
be confounding variables. For instance, maybe the kids in
the play therapy group had closer relationships with their
parents at the time that the study began than did kids in
64
h.
i.
j.
the other group and this greater closeness, not the effects
of play therapy, explains their higher score on the mood
scale.) So, if the difference is statistically
significant, we can rule out chance alone as a plausible
cause but cannot conclude whether the real factor causing
the difference is play therapy (the intervention) or a
preexisting difference (a confounding variable(s)). Had
the study used random assignment and obtained the same
result, we could have: 1) ruled out chance alone as a
plausible cause and 2) concluded that the real factor
causing the difference was indeed the intervention rather
than a confounding variable. (The fact of random assignment
would have eliminated systematic bias due to confounding
variables.)
SDM = (68.0 - 50.0)/12.0 = 1.5
Given that all of three conditions described on page 342 –
small sample sizes, unequal sample sizes, greatly differing
variances – are not met the present situation is not an
exception to the general rule that one uses the equal
variance formula when the equality of variance test does
not reject the null. So, use the equal variances formula.
The formula to estimate this is:
144 144

 288 / 36  8  2.83
36 36
Note that even though the equal variance formula is appropriate,
we are taking advantage of fact that the unequal variance formula
is computationally simpler, and using it (this because the two
formulas yield the same value of t when sample sizes are equal).
Note also that variances rather than standard deviations are
used; to determine the variance simply square the standard
deviation: 122 = 144)
k.
l.
m.
n.
o.
p.
t = (68.0 - 50.0)/2.83 = 6.36
We have established that we are using the equal variance
formula, so degrees of freedom for this formula apply
df = 36 + 36 - 2 = 70 (this formula in Section 18.3.4 on
page 343
Consulting Table A.2 on page 450: df = 70 is not listed, so
we reference the next lower listed df which is 60. Tracing
down the two-tailed column where  (alpha) = .05 yields a t
of 2.000. As this is a two-tailed test, there are two
critical values, 2.000 and -2.000
Values of t  -2.000 or  2.000 result in rejection
The absolute value of the obtained t, 6.36, is greater than
2.000, so the null is rejected and the research hypothesis
is accepted.
65
q.
r.
s.
t.
u.
v.
w.
8.
Rejection of the null signals that chance alone is an
unlikely explanation.
True. Where one rejects using the .05 level, s/he has (at
least) 95% confidence that the null is false. (See the last
paragraph of Section 14.8 on page 259.)
True (Even though in reality there is no such population,
if we assume that such a population exists, the statement
is true.)
As there is no such population, this conclusion is useful
for carrying out the hypothesis testing model but doesn’t
really have real-world applicability.
True – rejecting the null rules out sampling error as a
likely explanation
Yes, more so than for question t. above. This conclusion
pertains to the study sample which is comprised of real
people; so, it does have real-world applicability
Neither True or False. The key point is that due to the
absence of random assignment we don’t know whether the
treatment intervention or a confounding variable(s) is the
explanation for the difference between the means of the two
samples. We just don’t know.
The mean high school GPA of 16 first-born siblings (Sample 1) is
3.25. The mean GPA of the (16) second-born siblings (Sample 2)
in these same families is 3.00. The standard deviation of
difference scores is 0.64. The difference scores have a moderate
positive skew. Presume a directional hypothesis stating that
first-borns have higher GPA’s. Set alpha equal to .05. State the
null and research hypotheses and after having done so, respond to
a. to m. for Question 36 on page 357.
Null: The mean difference score is less than or equal to zero.
Research: The mean difference score is greater than zero.
Alternatively, one could state:
Null:
Research:
The mean GPA of first-borns is less than or equal to
that of second-borns.
The mean GPA of first-borns is greater than that of
second-borns.
(See Section 18.6.1 on page 349 and Section 18.6.2 on page 351
for discussion.)
a.
The cases of the two samples form linked pairs, each pair
consisting of a first-born and a second-born from the same
family. As the cases form linked pairs, the samples are
dependent (paired), and, hence, the dependent samples t
test is the appropriate test. (Section 18.5 on pages 3489
discusses dependent samples.)
66
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
m.
9.
For the dependent samples t test, the shape of the
distribution of difference scores is examined in order to
see whether the sample size guideline is met. Sample size
refers to the number of pairs, which in this case is 16.
Via the guidelines in Table 17.1 on page 324, the guideline
is met, the number of pairs being 16 and the recommended
minimum for a moderate skew being 15.
3.25 - 3.00 = 0.25
The mean of the difference scores is always equal to the
difference between the sample means, so this is also 0.25
16
0.64  16 = 0.64  4 = 0.16
t = 0.25  0.16 = 1.56
df = N - 1 = 16 - 1 = 15
As the hypothesis pair is directional, the appropriate t
test is one-tailed with a rejection region in the upper
tail. (The rejection is in the upper tail because the
research hypothesis states “greater than”)
As the research hypothesis states “greater than,” we will
reject the null if the obtained t is greater than or equal
to the t in Table A.2 on page 450. The obtained t of 1.56
is less than the t in the table (for a one-tailed test with
15 degrees of freedom and alpha equal to .05, this t, the
critical t, is 1.753). (The applicable decision rule is the
rule listed at the bottom of page 328.)
Fail to reject the null. Reject the research.
We do not want to conclude that it is likely that the
difference in means is due to chance alone, but instead
that it is plausible (not unlikely) that this is the case.
We cannot be 95% confident that the difference is not
simply due to chance alone.
Suppose that an independent samples t test with 18 degrees of
freedom yields an obtained t of 2.32. Presume a two-tailed test
at the .05 level. What is the critical value(s) of t? What
decision should one make regarding the null?
In Table A.2 on page 451, for 18 degrees of freedom, a two-tailed
test, and alpha equal to .05, the listed t is 2.101. As presented
in Section 17.6.5 on page 328, the decision rule for a two-tailed
test calls for rejection when the absolute value of the obtained
t equals or exceeds the listed t. Therefore, obtained t’s  the
negative of the listed t or  the listed t result in rejection.
Hence, in this example, the critical values of t are 2.101 and
2.101. The absolute value of the obtained t (|2.32| = 2.32),
exceeds the listed t (2.101). Therefore, the null is rejected.
67
Chapter 19
1.
A mother surmises that her young son wanders into her bedroom to
sleep on 50% of nights. She wants to decrease how often this
occurs. Working with a counselor, she develops a “demonstration”
plan designed to decrease this behavior. In the next month, the
son wanders into the mother’s room on 7 of 30 (23.3%) days. Your
task is to determine whether the percentage of times that the son
wandered in during the month following intervention represents a
statistically significant decrease from 50%. Your hypothesis
should be directional with the research hypothesis stating that
the percentage following intervention is less than 50%. Set alpha
equal to .05. To respond, refer to questions a. to i. in question
3 on page 372, but use information appropriate to the current
problem. For instance, in this problem, p = .233 and  = .50. (p
is the proportion in the sample and  is that in the population
(the hypothesized proportion in the null)). For question b. state
a directional hypothesis pair rather than a nondirectional one.
Prior to doing the problem, read the following comment:
Where data are gathered sequentially over time – in this example,
the 30 observations are gathered over 30 consecutive days - the
independence of observations assumption sometimes does not hold.
This (most often) occurs when observations close together in time
resemble one another more than do those spaced further apart.
For instance, presume that all of the seven days where the child
wandered into the bedroom occurred very close together in time –
say, for instance, within the first 10 days of the month. This
presumed to be the case, whether a child walks in on a given
night can be used to predict whether they will do so on the next
night. For instance, if a child walks in on a given night, we
could predict that they would do so on the next, this because all
of the incidents of walking in are clustered together closely in
time. Similarly, if s/he did not walk in on a given night, we
could predict that s/he would not walk in on the next, this
because incidents of “not walking in” are also clustered together
(in our example, in the second half of the month.) Given the
pattern described, “nearby” (close together in time) observations
are not independent (unassociated) but instead are associated or
dependent.
Such a pattern of dependency, that is dependency
according to time, is termed serial dependency. Where serial
dependency is present the independence of observations assumption
is not met and the statistical tests presented in this text
should not be used. Where observations are serially dependent,
specialized forms of time-series analysis are carried out. (See
the Bloom, Fisher, and Orme reference listed in text on page 507
for time-series methods.) We are assuming that the data for this
problem are not serially dependent.
68
a.
b.
c.
d.
e.
f.
g.
2.
One-sample test of p
Null:
The proportion of days wandering in is greater
than or equal to .50.
Research: The proportion of days wandering in is less than
.50.
As presented in first paragraph of Section 10.6.1 on page
367, one uses  which is .50.
Where  = .50, the recommended minimum sample size is 10.
The study sample size is 30 which is sufficient.
As demonstrated on page 362:
p = [(.5)(.5)/50] = (.25/50) = (.005) = .071
z = (.233 - .50)/.071 = -.267/.071 = -3.76
Reject the null; accept the alternative/research, these
decisions made because the obtained z (-3.76) is less than
the critical value of z (-1.645) stated in the decision
rule. The decision rule where alpha equals .05 and the
research hypothesis states “less than” is stated near to
the bottom of page 290). (As stated in Section 19.2.2 near
the bottom of page 361, the decision rules for the onesample test of p are the same as those for the large sample
test of the mean.)
A state agency holds a workshop for adoptive families who adopted
children with special needs. Of the 40 families who decide to
attend, 10 are from urban communities, 20 are from suburban ones,
and 10 are from rural ones. According to the statewide registry
of all families who adopted children with special needs, 35% of
these families live in urban communities, 35% live in suburban
ones, and 30% live in rural ones. Conduct a statistical
significance test to test whether the actual proportions
attending the workshop differ significantly from the statewide
proportions. Set alpha to .05. As you conduct the test, respond
to the following questions. Answers are listed as you proceed:
a.
b.
c.
d.
What is the hypothesis pair?
Null:
The proportions attending are .35 urban, .35
suburban, and .30 rural.
Research: At least one proportion differs from the
corresponding proportion stated in the null.
Is the study sample actually a random sample?
No. The workshop attendees decided on their own to attend.
Given that the sample is not random, may a statistical test
be conducted?
Pragmatically speaking, yes (see Section 16.8 beginning on
page 308).
What are the observed proportions?
Urban: .25(10/40); Suburban: .50(20/40); Rural:
.25(10/40)
69
e.
f.
g.
h.
i.
What are the expected proportions?
Urban and suburban: .35; rural: .30
Are observed proportions and expected proportions used in
the preferred calculation formula (19.4 on page 368)?
No.
What are the observed frequencies?
Suburban: 20;
Rural and urban:10
What are the expected frequencies?
To determine these, multiply the expected proportion by the
sample size. Do so for each group:
Urban and suburban: 40  .35 = 14;
Rural: 40  .30 = 12
2
What is the value of  ? Best to make a grid similar to
table 19.1 on page 369
Category
fo
fe
fo - fe
(fo  fe)2
(fo - fe)2/fe
urban
10
14
-4
16
16/14 = 1.143
suburban
20
14
6
36
36/14 = 2.571
rural
10
12
-2
4
4/12 = 0.333
 = ()2 = 4.05
j.
k.
l.
m.
n.
3.
How many degrees of freedom?
df = 3 - 1 = 2
What is the critical value of 2?
Via Table A.3 on page 451, 5.99
What is your decision regarding the null? The research?
Fail to reject the null. Reject the research. These
decisions made because the obtained 2 of 4.05 is less than
the critical value of 2 (5.99).
Is the result statistically significant?
No.
Is it plausible that the differences between the
proportions in the study sample and those in the null are
simply due to chance?
Yes.
For each of the following chi-square test results indicate first
the critical value of 2 at the .05 level and second the decision
that should be made regarding the null (at the .05 level).
a.
b.
c.
d.
e.
2
2
2
2
2
=
=
=
=
=
5.20
22.65, df = 14
7.35, df = 2
7.35, df = 5
11.85, df = 5
70
Answers follow:
a.
b.
c.
d.
e.
Is a “trick” question. One can’t know whether this value
of 2 is significant without knowing the degrees of freedom.
22.68, accept the null
5.99, reject the null
11.07, accept the null
11.07, reject the null
Chapter 20
1.
One hundred freshmen students are identified as having an
increased risk of failing to complete their first year of
college. Fifty of these students are randomly assigned to a
mentoring program while the other 50 (also randomly assigned)
receive no special services. Forty-three of the 50 mentored
students (86%) in comparison to 32 (68%) of the nonmentored
students successfully complete their freshmen year. Respond to
a. to o. for Question 6 on page 387. Note that o. has been
reworded below.
a.
b.
c.
Null:
Participation in the mentoring program and
successful completion of freshmen year are
unassociated.
Research: Participation in the mentoring program and
successful completion of freshmen year are
associated.
Null:
The percentage of mentored students who
successfully complete their freshmen year is
equal to the percentage of nonmentored students
who do so.
Research: The percentages of mentored and nonmentored
students who successfully complete their freshmen
years are not equal.
We will do best to create a table of expected frequencies
similar to Table 20.2 on page 380. Before doing so: Note
that the numbers not completing their freshmen year
(dropping out) can be determined by subtraction: Mentored
students: 50  43 = 7; nonmentored: 50  32 = 18. Note
further that the 75 in right margin was derived by summing
the successful students in the two groups: 43 + 32 = 75 and
that the 25 in the right margin was derived by summing the
unsuccessful students: 7 + 18 = 25. The actual calculation
of expected frequencies for the cells uses Formula 20.4 at
the bottom of page 379.
71
Mentored
Not mentored
Successful
(5075)/100 = 37.5
(5075)/100 = 37.5
75
Dropped out
(5025)/100 = 12.5
(5025)/100 = 12.5
25
50
50
100
d.
e.
f.
g.
h.
i.
Section 20.3 on page 371 lists requirements. For 22
table, average expected frequency should be at least 6.00
and minimum expected frequency should be at least 1.00.
These guidelines are easily met. (The average frequency is
25 (100 students  4 cells = 25. The minimum expected
frequency is 12.5; both cells in the “dropped out” row have
expected frequencies of 12.5)
[Note: sum to 1.00 within categories of mentoring, that is
down the columns.]
Among mentored students, the proportion who are
successful is .86 (43/50 = .86); the proportion who
drop out is .14 [(5043)/50 = 7/50 = .14)].
Among nonmentored: the proportion successful equals
32/50 = .64; the proportion dropping out equals
(5032)/50 = 18/50 = .36.
Associated, as the percentages who succeed differs in the
two groups.
[Note: sum to 100 within categories of mentoring, that is
down the columns.]
Formula 20.2 on page 379 is used here. For both
mentored and nonmentored, the expected proportion of
successful students is: 75/100 = .75 where 75 is the
row margin total and 100 is the total sample size.
For both groups the expected proportion of
nonsuccessful students is: 25/100 = .25.
No, they differ (compare the proportions in e. and g.
above).
The best way to compute 2 will be to build a grid similar
to Table 20.4 on page 382. (Hopefully, you see how to
build observed frequencies for those who dropped out via
subtraction as is demonstrated in calculating observed
proportions in c. and e. above.)
72
Cell
fo
fe
fo - fe
(fo - fe)2
(fo - fe)2/fe
mentor/
success
43
37.5
5.5
30.25
30.25/37.5= 0.81
mentor/
fail
7
12.5
-5.5
30.25
30.25/12.5= 2.42
nonmentor/
sucess
32
37.5
5.5
30.25
30.25/37.5= 0.81
nonmentor/
fail
18
12.5
-5.5
30.25
30.25/12.5= 2.42
 = 2 = 6.42
j.
k.
l.
m.
n.
o.
df for the chi-square test of independence = (number of
rows minus 1)  (number of columns minus 1):
df = (2  1)(2 - 1) = 1  1 = 1
Via table A.3 on page 451, the critical value is 3.84
2 = 6.42 (1, N = 100) = 6.42, p < .01)
As the obtained 2 (6.42) is greater than the critical value
of 2 in Table A.3 (3.84) the decision is to reject the null
and accept the research hypothesis.
Yes. Whenever one rejects the null at the .05 level they
have 95% confidence that: 1) the null is false and 2) the
observed difference is not due solely to chance. (If you
have 95% confidence that the null is false then, by
definition, you have 95% confidence that study result is
not due to chance alone.)
Question o. is rephrased to: Given that the null is
rejected, should you conclude that the mentor program
caused greater success?
Given that the null is rejected, we conclude (with 95%
confidence) that the study result is not due to chance
alone. As we can be (95%) confident that the result is not
due to chance, we can be (95%) confident that it is due to
something real. So the question becomes what is the
something real that is causing the study result (that being
higher success rates for those in the mentor program).
(Though it perhaps oversimplifies), the fact that students
were randomly assigned to mentor vs. nonmentor allows us to
conclude that the something real is not a confounding
variable on which the two groups differ – for instance,
something like greater initial motivation to succeed in the
mentoring group - but instead is the mentoring program. In
short, we may conclude that the mentoring program causes
greater success. We don’t know exactly what facet of the
mentoring program is responsible – perhaps it is something
so simple as charismatic leaders and the program would
73
minimal positive impact with other leaders - but we are
confident that something(s) connected with mentoring
explains the study result. (For discussion of random
assignment see Section 1.13.2 on pages 13-4, Section 10.6
on pages 179-81, Section 15.13 on pages 282-3, and the last
paragraph of Section 18.4.2 on pages 346-7.)
2.
A given contingency table has 5 rows and 3 columns. How many
degrees of freedom are there for a 2 test of independence?
Carry out Formula 20.1 on page 378: (5-1)(3-1) = (4)(2) = 8
3.
(T or F) Presuming the same number of degrees of freedom, the
critical value of 2 where  = .01 is greater than that where  =
.05.
True. One may examine Table A.4 on page 453 to see that for any
given degrees of freedom, the critical value is always higher
where  = .01 than where  = .05. This should make intuitive
sense. As presented in Sections 15.14 and 15.15 on pages 283-6,
choice of the .01 level makes it more difficult to reject the
null. Hence, to reject at the .01 level, a higher value of 2 is
needed.
4.
(T or F) The chi-square test of independence is the most common
statistical test for examining differences between group means.
False. It is the most common test for examining differences
between proportions.
5.
A 2 test of independence yields an obtained 2 of 12.25 with 10
degrees of freedom. At the .01 significance level, what is the
critical value of 2? What decision should one make regarding
the null?
Via Table A.3 on page 451, the critical value is 23.21. As the
obtained 2 is less than this, the null is accepted.
6.
(T or F) Where a 2 test of independence is statistically
significant, one concludes that the association between variables
in the contingency table is likely due to chance alone.
False. The statistically significant finding rules out chance as
a likely cause of the association. So, presumably the
association is real, that is, the variables indeed are associated
in the population from which the study sample was randomly
selected.
74
Chapter 21
1.
To give you a feel for hand-calculation of ANOVA, follow the
steps presented on page 397 to calculate ANOVA for the data that
follow. (Note that sample sizes here are too small for one to be
confident that ANOVA would yield accurate probabilities.
Probabilities resulting from ANOVA where sample sizes are this
small are accurate only when the shape of populations from which
samples have been (randomly) selected is very close to normal.
Given the small sample size here, it is difficult to use the
shape of the distribution in the samples to estimate that in
populations. See discussion in Section 17.6.2 on page 326. So,
one would be hesitant to use ANOVA in this situation. The
situation is provided to give practice at hand calculation.)
Let’s presume that the scores presented here represent selfesteem scores in three different treatment groups:
Group 1
7,8,9
Group 2
Group 3
3,4,5
4,6,8
Calculation steps follow:
1) The grand mean is: 7+8+9+3+4+5+4+6+8=54, 54/9 = 6
2) The group means are:
Group 1: 7+8+9 = 24, 24/3 = 8
Group 2: 3+4+5 = 12, 12/3 = 4
Group 3: 4+6+8 = 18, 18/3 = 6
3) the SSw is:
Group 1: (7-8)2 + (8-8)2 + (9-8)2 = 1 + 0 + 1 = 2
Group 2: (3-4)2 + (4-4)2 + (5-4)2 = 1 + 0 + 1 = 2
Group 3: (4-6)2 + (6-6)2 + (8-6)2 = 4 + 0 + 4 = 8
Summing: SSw = 2 + 2 + 8 = 12
4) MSw = SSw/(N - J) = 12/(9 - 3) = 12/6 = 2.0
5) SSb = 3(8-6)2 + 3(4-6)2 + 3(6-6)2 = 3(4) + 3(0) + 3(4) = 24
6) MSb = SSb/(J - 1) = 24/(3 - 1) = 24/2 = 12.0
7) F = MSb/MSw = 12.0/2.0 = 6.0
2.
Given that F for the prior problem is 6.0, what is the critical F
and what decision should be reached regarding the null? Presume
that alpha equals .05.
75
To determine the critical F, one needs to know the degrees of
freedom for both the MSb and the MSw. Degrees of freedom for the
MSb = J - 1 = 3 - 1 = 2. Degrees of freedom for the MSw = N  J
= 9  3 = 6. Via Table A.4 on page 452, the critical value for an
F distribution with 2 degrees of freedom for the MSb and 6 for
the MSw where alpha equals .05 is 5.14. As the obtained F
exceeds the critical F, the null is rejected.
3.
Indicate the degrees of freedom first for the MSb and second for
the MSw for each of the following situations.
a.
b.
c.
N = 64, J = 8
N = 25, J = 4
N = 90, J = 6
For each of these, one simply carries out the degrees of freedom
formulae: MSb = J - 1 and
MSw = N  J.
a. 7 and 56,
4.
b. 3 and 21,
c. 5 and 84
Using Table A.4 on page 451, indicate the critical value of F
with alpha set to .05 for the three situations presented in the
prior problem.
As you do these problems, note that, as stated in Section 21.4.4
on page 396, where the exact number of degrees of freedom is not
listed: use the closest larger number for the MSb and use the
closest smaller number for the MSw.
a.
b.
c.
5.
2.20 (as 56 degrees was not listed for the MSw, 50 was
used)
3.07
2.22 (as 90 degrees was not listed for the MSw, 75 was
used)
(T or F) Where the obtained F is greater than or equal to the
critical F, one rejects the null.
True. (And where obtained F is less than critical F, one accepts
the null.)
6.
In a given situation, the MSw equals 5.00 and the MSb equals
22.00. What is the value of F?
To compute F divide the MSb by the MSw: 22.0/5.0 = 4.4
7.
Is the F ratio computed in the prior problem statistically
significant at the .05 level?
76
Trick question. Can’t know this unless we know the degrees of
freedom, and these were not given.
8.
Presume that degrees of freedom for the MSb were 2 and those for
the MSw were 12. Given the F ratio from problem 6 (F = 4.4),
what decision should be made regarding the null? (Use the .05
level.)
Given these degrees of freedom, the critical value of F is 3.89.
As the obtained F exceeds the critical F, the null is rejected.
9.
In a given situation, the SSw equals 100 and the degrees of
freedom for the MSw equals 20. What is the value of the MSw?
As indicated by Formula 21.3 on page 395, to compute the MSw,
divide the sums of squares within by the degrees of freedom for
the MSw: MSw = 100/20 = 5.0
10.
In a given situation, the SSb equals 40 and the degrees of
freedom for the MSb equals 10. What is the value of the MSb?
As indicated by Formula 21.5 on page 396, to compute the MSb,
divide the sums of squares between by the degrees of freedom for
the MSb: MSb = 40/10 = 4.0
11.
In a given situation the value of F is 0.70. Even without the
degrees of freedom or knowledge of the significance level, how
can one know that one should accept the null?
Values of F less than 1.00 indicate even less sampling error than
one would expect (on average) just from chance alone. Where F is
less than 1.00, the null will always be accepted.
12.
(T or F) When the null is true, one knows assuredly that the
statistical test will result in a (correct )decision to accept
the null.
False. Most of the time – 95% of the time at the .05 level and
99% at the .01 – the statement is true. However, occasionally
the study sample result (by the luck of the draw) will be
sufficiently extreme to result in the (incorrect) decision to
reject the null. This will happen in 5% of (random) samples at
the .05 level and 1% of (random) samples at the .01 level.
13.
Assuming that the null is true, what is the probability that the
(appropriate) statistical test will result in an (incorrect)
decision to reject the null, at the .01 level? At the .05 level?
77
At the .01 level, the probability of an incorrect decision to
reject the null – that is, the probability of a Type I error – is
.01. At the .05 level, this probability is .05.
14.
(T or F) As the number of significance tests that are conducted
increases, so also does the likelihood of making a Type I error
on at least one test.
True. So, when many many tests are reported in an article, be
somewhat skeptical of study results. There may well be one or
two instances where a statistically significant finding (and
therefore a decision to reject the null) represents Type I error.
Chapter 22
1.
(T or F) Other things being equal, as sample size increases so
also does the statistical power of a significance test of r.
True. (This is true of all statistical tests, not just the test
of r.)
2.
(T or F) One may formulate either a directional or a
nondirectional hypothesis test for a significance test of r.
True.
3.
(T or F) Where the hypothesis pair is directional the test of r
is two-tailed.
False.
4.
It is one-tailed.
Given a directional hypothesis where the research hypothesis
states “less than” and sample size is 52. State the hypothesis
pair. What values of r result in rejection of the null at the
.05 level? At the .01 level?
Null:
In the population from which the study sample was
randomly selected,   0.00.
Research:
In the population from which the study sample was
randomly selected,  < 0.00.
(Hopefully you recall that  (rho, pronounced “row”) is the
symbol for the correlation coefficient in a population.)
To see what values result in rejection, reference Table A.5 on
page 454. Where the hypothesis is directional, the test is one-
78
tailed. Where the research hypothesis states less than, the
rejection region is in the lower tail. The applicable decision
rule is the last rule listed in Section 22.2.6 on page 410. The
value of r in the table is .231. Applying the decision rule, at
the .05 level, one would reject the null for all values of r 
.231. At the .01 level one would reject the null for all values
 .322.
5.
In the correlation matrix on page 412. What is the correlation
between family income and the parents’ perception of the impact
of the adoption on the family? Is the correlation negative or
positive? Is the correlation statistically significant (.01
level, two-tailed test)? Should the null be rejected? Is the
relationship strong or weak? Interpret/describe the correlation
in your own words.
r = .170. Negative. Yes, it is statistically significant.
Yes, the null should be rejected. Via Table 8.3 on page 137, the
relationship between income and impact is reasonably weak. To
interpret: As income level increases, the impact of adoption on
the family, as reported by the parent, decreases (becomes more
negative). (Sections 8.3.1, 8.3.2, and 8.3.3 on pages 130-3
provide guidance on interpreting/describing correlation.)
6.
(T or F) The one-sample test of z is a parametric test.
True.
7.
(T or F) A nondirectional test of Spearman’s r examines whether
the correlation between two rank orderings differs significantly
from 0.00.
True.
8.
(T or F) Tests of the significance of taub and tauc are typically
conducted with categorical variables that are at the nominal
level of measurement.
False. These tests test for the possible significance of
directional association between categorical variables. For a
variable to have a directional association, its level of
measurement must be at least ordinal. So, these tests are used
with ordinal-level variables, not nominal-level ones.
9.
(T or F) The Mann-Whitney U test is a parametric test.
False.
10.
It is nonparametric.
A social work professor asks 24 students which class they enjoy
more, research methods or practice. Eighteen state practice, two
79
state research, and four state that they enjoy each the same.
What test would be a good one to use to see whether the
difference in preferences is statistically significant.
The sign test (see Section 22.8 on page 416).
11.
What test is regarded as the nonparametric alternative to the
independent samples t test?
The Mann-Whitney U test. (Presuming that the dependent variable
is measured at the interval/ratio level), this test is preferable
to an independent samples t test when both of the following are
true: 1) sample size is very small and 2) the degree of skew is
very strong or extreme. (See page 415.) In this situation the
independent samples t test may yield inaccurate probabilities.
(Of course, the Mann-Whitney test is often preferred over the t
test when one’s level of measurement is ordinal.)
12.
What test is a good nonparametric alternative to ANOVA?
Kruskal-Wallis test. (Presuming that the dependent variable is at
the interval/ratio), this test is preferred to ANOVA when both of
the following conditions hold: 1) sample sizes are small and 2)
skewness is extreme. In such a situation, ANOVA may yield
inaccurate probabilities. (See page 415.) (Of course, the
Kruskal-Wallis test is also often preferable to ANOVA with
ordinal-level data.)
13.
Where one has information on the size of differences as well as
on the direction, is the sign test or the Wilcoxon signed ranks
test preferred?
Most would prefer the Wilcoxon test in this situation (see
Section 22.8 pages 416-7)
14.
(T or F) In multiple regression, the ’s (beta’s) are
unstandardized coefficients and, thus, convey change in the units
in which variables were originally measured rather than in
standardized units.
False. ’s are standardized and convey change in terms of
standardized units (that is, in terms of z scores). B’s are
unstandardized and convey change in terms of the original units
of measure (in terms of raw scores). (See page 419 in Section
22.11)
15.
To respond to the next series of questions refer to Table 22.2 on
page 420. (Answers are given after each question.)
80
a.
As parent’s mean educational level increases by one (1.00)
unit (in its original unit of measure), what is the
predicted change in parent/child relationship score (in its
original unit of measure)?
The B coefficient provides this information. It is .0530.
This conveys that as educational level increases by 1.00
point, predicted relationship score decreases by .0530
points (by about 1/20th of a point).
b.
As parents’ mean educational level increases by 1.00
standard deviation, what is the predicted change in
standard deviations for parent/child relationship score?
The  coefficient provides this information. It is -.087.
Thus, as education level increases by 1.00 standard
deviation the parent-child relationship score is predicted
to decrease by .087 standard deviations.
c.
Controlling for other predictors in the regression
equation, about how strong is the relationship between
parent educational level and parent-child relationship
score? Is the association (controlling for other
predictors) a directional one? If so, what direction?
Strength of relationship controlling for the other
predictors is conveyed by the  coefficient of -.087. Via
the descriptors for r in Table 8.3 on page 137, the
association is a weak one. The association is directional.
The negative sign conveys that the direction of association
is negative. As education level increases (controlling for
other predictors), predicted relationship score decreases.
d.
How can it be that the association between education level
and relationship score is significant (p = .011, see last
column of Table 22.2) when this relationship is weak?
Though the sample size was inadvertently omitted in Table
22.2, it is very large, about 700. When sample size is
large even weak associations may achieve statistical
significance. (See page 303 for discussion on this point
and also Table 16.1 on page 308)
e.
As parental income increases by 1.00 standard deviation,
what is the predicted change in standard deviations for
parent-child relationship score?
The  of -.122 conveys that the predicted score decreases by
.122 standard deviations.
81
Chapter 23
1.
The answers given here expand on questions a. to m. in Question
22 on page 443. Answers to these questions are presented on page
497 of the text.
a.
See page 497.
b.
The total sample size is 400. Referencing Table 16.1 on
page 308, power is “good/fairly high”. Note that the
variability of rehospitalization is somewhat restricted
(low)– only 15% in one group and 25% in the other require
hospitalization – but this restriction is not sufficient to
substantially reduce power. If only, say, 5% in one group
and 10% in the other had been rehospitalized, power would
have been reduced somewhat. The fact that group sizes were
equal (200 in each) enhances power to some degree. (See
Section 16.5.5 on pages 305-6 for discussion of the effects
of variability on power.)
c.
Yes, at the .05 level; the “p < .05" symbol in the question
description conveys this information. See second paragraph
on page 311 for discussion of p symbol.
d.
Less than or equal to .05. Where the result is
statistically significant, it is not likely to be due to
chance alone.
e.
Less than or equal to .05 (Given a true null, the
probability of obtaining this or a more extreme result is
quite low; given a true null, a result this extreme would
be rare, unexpected, uncommon, and “atypical” – it would be
a result in a tail of the sampling distribution.)
f.
Can be at least 95% confident. See last paragraph of
Section 14.8 on page 259, last paragraph of Section 14.10
on page 261, and Table 16.2 on page 313, third grouping
from the bottom of the table.)
g.
Statistical significance and rejection of the null go hand
in hand; one rejects the null.
h.
Here one would calculate the difference in percentages and
then assess size making use of Table 6.4 on page 109: D% =
25%  15% = 10%. Via Table 6.4 the relationship is
small/weak.
i.
The fact that the difference in rehospitalization rates
achieves statistical significance rules out chance as a
plausible explanation. However, no mention is made of
random assignment. (Presumably clients went to center A
because they lived in its catchment area or to B because
they lived in its area.) In the absence of random
assignment, one should be suspicious that some differing
characteristic(s) of clients served in the two centers
82
j.
k.
l.
m.
rather than differential effectiveness of the two treatment
approaches (halfway house versus own residence) is the
explanation for the study result. (This is the same as
saying that one should be suspicious that a confounding
variable is the cause of the result.) So, one would not
want to conclude that relationship between type of
treatment and rehospitalization is (necessarily) a causal
one. Random assignment would have greatly strengthened the
ability to draw a causal conclusion. See Sections 10.5 and
10.6 on pages 178-81 for discussion of random assignment
and related issues.
No, as the study sample was not randomly sampled from any
larger (real world) population. See Section 16.8 on pages
308-10.
See answer on page 497; see discussion in Section 23.2.2 on
page 431.
See page 497.
See page 497 and Section 23.3 on pages 433-5.
83