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Basic Quantitative Methods
in the Social Sciences
(AKA Intro Stats)
02-250-01
Lecture 7
In Review…
• Sampling distribution = The distribution of a
statistic over repeated sampling from a
specified population.
• Standard error = The standard deviation of a
sampling distribution (tells us how much
variability we will get over repeated sampling)
• If we know the shape and parameters (e.g.,
mean and standard deviation) of the sampling
distribution of a statistic, we can derive the
position of a particular statistic in the overall
distribution.
More Review…
• John got a 76% on the last midterm in
statistics. The class mean was 65%, and the
standard deviation was 6.
• Can we determine the position of John’s score
in the class distribution? Yes! We can calculate
a Z-score.
• SO: We know the position of John’s score (i.e.,
z-score), the probability of this score occurring
in the class (which we get from the z-score),
and the amount of sampling error.
And More Review…
• What if we know that our class has a mean on
the midterm of 68% with a Standard Deviation of
7, and we want to know if the first 3 rows did
better than the rest of class…
• Can we consider the first 3 rows as a sample of
the class (population) and do a Z-test? YES! Why?
We know the POPULATION PARAMETERS (mean
and standard deviation – so the modified Z
formula will work).
More Review….
• Crucial to understand: We can do this because
we KNOW the standard deviation of the
population.
• What if we want to know how our class mean
(that is, 65%) compares with introductory
statistics courses across the country.
• Can we calculate the Z-score of our class mean
to find out it’s position?
More Review…
• NO! Why not? Because we do not know
the standard deviation of the sampling
distribution of the mean (i.e., our class is
no longer the population. Now our class
is a sample in a larger population of
statistics classes)
Last Review Slide… 
• Central Limit Theorem: Given a population with
mean  and standard deviation , the sampling
distribution of the mean (the distribution of
sample means) will have a mean equal to  and
a standard deviation equal to  =  /N.
The distribution will approach the normal
distribution as N (sample size) increases.
Introduction to t-Distributions
• The reality is, we rarely know the population
standard deviation ( ).

• The t-distributions are a “family of theoretical
distributions” that can be used when:
 We are dealing with interval or ratio data
 Our data is normally distributed
 The population standard deviation ( ) is unknown.
t-Distributions continued..
• Review (ok, I lied, this is
the last review slide):
• A normal distribution is a
population of z-scores
where z is defined as:
• Note: Here we know the
population’s S.D ()
Z
X 

t-Distributions continued..
• A t-distribution is a
population of t-scores
where t is defined as:
• X-bar is mean of
random sample
 Do you see the
standard error of the
mean in the formula?
t
X 
s
n
t-Distributions continued….
• t-distributions are
similar to the normal
distribution in that they
are unimodal and
symmetrical. They have
a mean of 0, negative
values below the mean,
and positive values
above the mean.
t-distributions and degrees of
freedom
• Because the definition of t involves a term
obtained from a sample (that is the estimated
standard error of the mean), which in turn
involves the degrees of freedom associated
with the sample, there is a different t
distribution for every degrees of freedom
(sample size).
• The t-distribution can be found in Table E.6
(p.444 in Howell). You will notice an extra
column not in the normal curve table (df).
t-distributions and degrees of
freedom (continued)
• Here we see a representation of a set of t distributions.
• Note that if the df is large t and z are the same, and they
depart as the df gets smaller.
Basic Properties of t-Curves
• Property 1: The total area under a t-curve is equal to 1.
• Property 2: A t-curve extends indefinitely in both
directions, approaching, but never touching the horizontal
axis as it does so.
• Property 3: A t-curve is symmetrical about 0.
• Property 4: As the number of degrees of freedom
becomes larger, t-curves look increasingly like the standard
normal curve.
• Property 5: Every t-score has a certain probability of
occurrence in a specific t-distribution. As such, the values
of t which enclose or cut-off given proportions of the
appropriate t-distribution can be calculated.
z table vs. t table
• z table:
 Gives the area above
and below each
specified value of z.
• t table:
 A different t distribution
is defined for each
possible number of
degrees of freedom.
 Gives values of t that
cut off particular critical
areas, for example, the
.05 and .01 levels of
significance.
Values of t
Finding the t-value having area
0.05 to it’s right
Intro to Confidence Intervals
• Recall: Although random sampling has no
inherent bias, we cannot expect any
given sample to perfectly represent its
population. Why? Sampling error!
• SO: The sample mean will almost always
be a different value than the population
mean.
A Confidence Interval Is:
• A score interval calculated by a
procedure with a specified probability of
producing an interval containing the
parameter (i.e., from the population).
• Can we make a statement about how
confident we are that a sample mean is
close to the (unknown) population mean?
Example:
• 25 people around Windsor are approached at
random and asked to rate how good a job Jean
Chretien is doing as Prime Minister, on a scale of
1 (he stinks) to 20 (he’s great). The mean rating
was 8, and the standard deviation was 7.558.
How confident are we that this mean of 8 is
close to the mean of the overall population of
Ontario? SO:
X
= 8, s = 7.558, n = 25
• The sample mean (8) is deemed to be the mean of
a distribution which conforms to the t distribution
at df = n-1.
• By choosing t values which enclose a specified
proportion of that t distribution, we can construct
an interval of plausible values of .
Don’t worry, it’s not as complicated as
that last sentence seemed
• If we choose critical values for t at 0.05
confidence level, there is a 0.95
probability that the score interval we
generate will contain .
• This score interval is termed the 95%
confidence interval (95% C.I.)
Confidence Limits on Mean
• Sample mean (8) is a point estimate
• We want an interval estimate
 Probability that interval computed this way
includes  = 0.95
CI.95  X  t.05 s X
How do we get the t value?
• n = 25, so df = n-1 = 24
• Look at the t distribution table for df =
24 at 0.05 level of confidence (for two
tails).
• The critical value for t = 2.064
SX =
s
= 7.558 / 5 = 1.5116
n
t.05 S X = 2.064 (1.5116) = 3.12
X
 t.05 S X
= 8  3.12, so 4.88 to 11.12
SO: We can be 95% confident that the
interval 4.88 to 11.12 inclusive contains the
population mean rating on Jean Chretien.
That is to say….
• If we took 100 samples (25 people in
each sample) from the same population,
95% of the samples would produce a
mean between 4.88 and 11.12.
Our example illustrated…
t
What if we wanted to be 99%
confident?
• t.01 S X = 2.797 (1.5116) = 4.23
SO: 8  4.23 = 3.77 to 12.23
SO: If we took 100 samples of 25 people, 99 of the
samples would produce a mean between 3.77
and 12.23.
Things to Remember…
• Other things being equal, increasing the
confidence level (say from 95% to 99%)
increases the size of the confidence
interval. Why?
• Because less certainty (confidence) is
associated with greater precision (smaller
interval).
Things to Remember
(continued)
• Other things being equal, an increase in
the size of the sample standard deviation
increases the size of the confidence
interval. Why?
• More variable data indicate more
sampling error which in turn means less
certainty can be attached to the accuracy
of a particular estimate.
Things to Remember
(continued)
• Other things being equal, an increase in
the size of the sample decreases the size
of the confidence interval. Why?
• Because larger samples provide more
stable (less variable) estimates which in
turn means that on average, sampling
error is less and greater certainty can be
attached to the accuracy of an estimate.
One more example:
• 16 University of Windsor students were
polled regarding how much they pay for
rent each month, producing a mean of
$500.00 a month, with a standard
deviation of $60.00. Compute 95%
confidence limits for the population mean
of University of Windsor students.
Here we go…
• t.05 (with df =15) = 2.131
•
t.05 S X = 2.131 (60/4) =
• t.05 S X = 2.131 (15) = 31.97
SO: 500  31.97 = 468.03 to 531.97
SO: If we took 100 samples of 16 people, 95 of the samples
would produce a mean of $468.03 to $531.97 per month.
One sample t-tests: Rationale
•
Sometimes we know the population mean () of
a variable, and we wish to determine whether
the mean of a sample X differs significantly
from the population mean.
Assumptions:
1. Normal population or large sample.
2. The population’s standard deviation is not
known.
Why t and not z?: Review
• Gosset noted that when we use the
sample’s standard deviation instead of
the population’s (which we do not know),
the distribution changes as a function of
sample size. If n is large, it is very close
to the normal distribution. But smaller
sample sizes lead to skewed
distributions, which would give us too
many “significant” results.
• To compensate, we compare our t-value
with it’s own distribution.
• Say we know that the average cell phone user
uses 3000 minutes of cellular air time each year
() .
• Dr. Z hypothesizes that business executives
spend more time on their cell phone each year
than does the average cell phone user. She
interviews a sample of 20 business executives,
and finds that they use on average 3500
minutes of cellular air time each year, with a
standard deviation 300 minutes. Did this
sample of business executives use significantly
more cellular air time than the average cell
phone user? Test at the .01 level of
significance.
Hypothesis testing with the
one sample t-test
• We can test the null hypothesis:
• H0: The mean number of cell phone minutes
used per year by business executives does not
differ from the mean of the average cell phone
user.
Ho : X    3000
t
SX
X 
Let’s try it!
SX
s

We know , we know
what else do we need?
X
Luckily, we know that
s = 300. If we didn’t, we
would need to calculate it
from the raw data.
n
= 300 / 20 = 300 /
4.4721
= 67.0826
Calculating t…
t
X 
SX
= 3500 – 3000
67.0826
= 7.453 = tobt.
One tailed or two tailed?
• The null hypothesis is
H o : X   and the
alternative hypothesis is one of the following:
• Ha:  ≠  : 2-tailed test.
• Ha:  <  : right tailed test
• Ha:  >  : left tailed test
Do we want to use a two-tailed, a left tailed,
or a right-tailed test in this example?
Since our hypothesis is that business
executives use more cellular air time……
We’ll use a one-tailed test (in this case, a
right-tailed test)
Is it significant? P values revisited
p-value for a t-test if the test is (a) two-tailed,
(b) left-tailed, (c) right-tailed:
Refer to the t-table…
• Remember, df = n –1 = 19.
• As mentioned, we’ll use a one-tailed test.
• If we set our level of significance at .01,
the critical t-value is 2.539 (this is called
tcrit).
• tobt (7.454) > than tcrit (2.539).
Therefore: We reject the H0.
• Can we state our conclusion in words?
The size of t and the Decision
about H0 are affected by…..
• The actual obtained difference
( X  )
• The magnitude of sample variance
• The sample size
• The significance level (.05? .01?)
• Whether the test is 1 or a 2 tailed
t
X 
SX
Underlying Assumptions
• Most statistical tests make certain
presumptions regarding the data and the
distributions of whatever parameters are at
issue – These are known as the underlying
assumptions of the test.
• If the underlying assumptions are violated,
the validity of the test may be compromised,
such that for instance, the probability of a
Type I error might be higher than the alpha
level of the test.
Underlying Assumptions
• The one-sample t-test • The one-sample t-test
assumes that the
assumes that the raw
dependent variable is
data were a random
normally distributed in
sample – that is, the
the population.
raw scores must be
independent of each • This assumption can be
other.
(and usually is) violated
to some degree – the t• This assumption must test is a “robust” test, it
not be violated, or the tolerates some violation
t-test is worthless.
of the normality
assumption.
A second Approach: Confidence
Intervals Re-visited
• An alternative to the one-sample t-test is
to calculate confidence intervals for the
sample mean. If the population’s mean
falls outside the sample mean’s
confidence interval, H0 is rejected.
• It’s logical: A 95% Confidence Interval
suggests that 95% of samples will
produce means within the interval. If the
population mean falls outside the
interval, it is significantly different than
the sample mean.
Let’s look at our example:
CI.99  X  t.01s X
3500  2.861 (67.0826) =
3500  191.9233 = 3308.08 to 3691.92
So: Since  = 3000 is outside the interval, we
reject H0, as 99% of samples of business
executives will use an average of 3308.08 to
3691.92 minutes of cellular air time per year.
• Another Example: A random sample of 65
freshman college students was selected to
participate in a new look-say teaching program
designed to increase reading speed in French.
The final exam consisted of a French passage
that the students translated. The time required
for each student to complete the translation was
recorded. The sample statistics were x = 302
and s = 56 sec. According to department
records, the mean for students in conventional
classes was 320 sec. Let alpha () = .05.
• Hypothesis: the new program will increase
reading speed.
• The steps:
Step 1 - State the statistical hypothesis.
• Ho - The students in the special program
will take the same amount of time to
complete the task as students in
conventional classes.
• Ha - The students in the special program
will take less than 320 seconds on the
reading task (they will be faster than
students in the conventional classes; a
one tailed test).
Step 2 - State the test statistic.
•Since we do not know the population
standard deviation, we will use the t-test.
Step 3 - Specify the sample size.
•Our Sample size for this experiment is 65.
We will use the t-distribution for 65-1 or 64
df.
Step 4 - Specify the significance level.
•We will set alpha () = .05.
Do the experiment, calculate t, and make a
decision.
• Student’s t distribution - Example
 1. Calculate/state the mean of the sample - 302
 2. Calculate/state the standard deviation of the sample - 56
 3. Calculate t
t
X 
302  320
t
 2.591
s
56
n
65
From the table we learn that to be significant, the
ABSOLUTE VALUE OF t must be equal to or
greater than the table value of 1.671 in order to reject
the null hypothesis. (Why the absolute value!?)
Thus, we reject the null hypothesis.
And using the CI approach…
CI.95  X  t.05 s X
302  1.984 (6.9459) =
302  13.7807 = 288.22 to 315.78
So: Since  = 320 is outside the interval, we
reject H0, as 95% of samples will produce
means between 288.22 and 315.78
Some Examples
• A researcher is concerned that police
officers are not in good physical shape
because they eat too many doughnuts.
He hypothesizes that police officers eat
significantly more doughnuts than do the
population. If the average Windsorite
eats 30 doughnuts every year (), and a
sample of 15 police officers produce a
mean of 26 doughnuts a year with a
standard deviation of 10, what
conclusions can be drawn at the .05 level
of significance?
Get to work!
• Ho:?
• Ha:?
• One tail or two tail?
• What test do we use?
• Formula?
• Decision re: the Ho?
• Conclusion?
• The average Canadian has an IQ of 100
with a standard deviation of 15. Dr. F
hypothesizes that chronic Ecstasy users
would lose intelligence over time. She
samples 30 chronic Ecstasy users and
gives them IQ tests. This sample
produces a mean IQ of 94 with a
standard deviation of 13.
• What conclusions can be drawn at the
.01 level of significance?
• HINT: Watch out for extraneous
information!
Get to work!
• Ho:?
• Ha:?
• One tail or two tail?
• What test do we use?
• Formula?
• Decision re: the Ho?
• Conclusion?
• The average movie-goer spends $8.00 on food
each time he or she goes to the movies. George
Lucas thinks he can help theatres sell more
food by putting subliminal messages in Star
Wars that show Yoda with a bag of popcorn
stating “May the Corn be with you”. A theatre
manager evaluates George Lucas’ claim by
keeping track of how much Star Wars viewers
spend on food. His sample of 120 Star Wars
viewers generates a mean of $8.75 (spent on
food while viewing Star Wars) with a standard
deviation of $0.84. Is George’s subliminal ad
campaign effective?
• What conclusions can be drawn at the .05 level
of significance?
Get to work!
• Ho:?
• Ha:?
• One tail or two tail?
• What test do we use?
• Formula?
• Decision re: the Ho?
• Conclusion?
• The average University of Windsor
student spends a total of 25 hours
studying for final exams each semester.
The Dean is concerned about students
who live in Residence – she believes they
may be partying too much at the
expense of studying. She samples 40
students who live in Res, and find that
they study on average 18 hours for final
exams, with a standard deviation of 4.3
hours. Does the Dean have reason to be
concerned?
• Test at the .01 level of significance.
Get to work!
• Ho:?
• Ha:?
• One tail or two tail?
• What test do we use?
• Formula?
• Decision re: the Ho?
• Conclusion?
Last one
• Now solve the last problem using the
confidence interval approach. Create
95% and 99% confidence limits.
• Hint: Watch your tails!
And Now the Answers
• The following slides contain the answers
to these last few problems. No peeking
until you try the problems and calculate
them through to the end!
Problem #1
• A researcher is concerned that police
officers are not in good physical shape
because they eat too many doughnuts.
He hypothesizes that police officers eat
significantly more doughnuts than do the
population. If the average Windsorite
eats 30 doughnuts every year (), and a
sample of 15 police officers produce a
mean of 26 doughnuts a year with a
standard deviation of 10, what
conclusions can be drawn at the .05 level
of significance?
Problem #1 - Solution
H o : X    30
Ha : X  
• One tail or two tail? 1 tailed
• What test do we use? t-test – we don’t have the population standard
deviation
tobs 
X 
26  30
t 
 1.549
s
10
n
15
• n=15, df=14 tcrit.05 = 1.761
• To reject Ho, tobs would have to be > +1.761. Therefore, we retain the
Ho, police officers do not eat significantly more than do the general
population.
• Problem #2:
• The average Canadian has an IQ of 100 with
a standard deviation of 15. Dr. F hypothesizes
that chronic Ecstasy users would lose
intelligence over time. She samples 30
chronic Ecstasy users and gives them IQ
tests. This sample produces a mean IQ of 94
with a standard deviation of 13.
• What conclusions can be drawn at the .01
level of significance?
• HINT: Watch out for extraneous information!
Problem #2 - Solution
H o : X    100
Ha : X  
• One tail or two tail? 1 tailed
• What test do we use? z-test – we have the population standard
deviation
Z obs
X 
94  100

t 
 2.1909

15
n
30
• 1-tailed test at .01 alpha, so Zcrit.01 = -2.33
• Zobs > Zcrit so we retain the Ho, ecstasy users do not differ in
intelligence from the average Canadian.
• Problem #3:
• The average movie-goer spends $8.00 on food
each time he or she goes to the movies. George
Lucas thinks he can help theatres sell more
food by putting subliminal messages in Star
Wars that show Yoda with a bag of popcorn
stating “May the Corn be with you”. A theatre
manager evaluates George Lucas’ claim by
keeping track of how much Star Wars viewers
spend on food. His sample of 120 Star Wars
viewers generates a mean of $8.75 (spent on
food while viewing Star Wars) with a standard
deviation of $0.84. Is George’s subliminal ad
campaign effective?
• What conclusions can be drawn at the .05 level
of significance?
Problem #3 - Solution
H o : X    8.00
Ha : X  
• One tail or two tail? 1 tailed
• What test do we use? t-test – we don’t have the population standard
deviation
tobs 
X 
8.75  8.00
t 
 9.778
s
0.84
n
120
• n=120, df=119 tcrit.05 = 1.645 (note: use the last row in the table)
• tobs > tcrit so we reject the Ho, the subliminal messages seem to work,
Starwars viewers seem to spend more on food than do other
moviegoers.
• Problem #4:
• The average University of Windsor student
spends a total of 25 hours studying for final
exams each semester. The Dean is concerned
about students who live in Residence – she
believes they may be partying too much at the
expense of studying. She samples 40 students
who live in Res, and find that they study on
average 18 hours for final exams, with a
standard deviation of 4.3 hours. Does the Dean
have reason to be concerned?
• Test at the .01 level of significance.
Problem #4 - Solution
H o : X    25
Ha : X  
• One tail or two tail? 1 tailed at .01
• What test do we use? t-test – we don’t have the population standard
deviation
tobs 
X 
s
n
t 
18  25
7
7


 10.2956  10.296
4.3
4.3
0.6799
6.3246
40
• n=40, df=39 tcrit.01 = -2.423 (note: use df=40 since there is no row for
39 and 40 is the one right after)
• tobs < tcrit so we reject the Ho, the dean’s concern does hold up –
students in res do seem to study less than other students.
• Problem #4b:
• Now solve the last problem using the confidence
interval approach. Create 95% and 99%
confidence limits.
• Hint: Watch your tails!
Problem #4b Solution
CI.95  X  t.05 s X
Remember to use 2-tails
18  2.021(0.6799) = 18  1.3741 = 16.6259 – 19.3741
= 16.63 - 19.37
So: Since  = 25 is outside the interval, we reject H0.
CI.99  X  t.01s X
Remember to use 2 tails!
18  2.704(0.6799) = 18  1.8384 = 16.1616 – 19.8384
= 16.16 – 19.84
So: Since  = 25 is outside the interval, we reject H0.