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1.
A new teacher is analyzing whether or not there is an association between scores earned by students on
their first exam in the course and the course grade earned by students at the end of the term. Exams are scored
using a 100 point scale (0 to 100 points) and course grades use a 100% scale (0% to 100%). There are 35
students in the course.
HaL Find the equation of the regression line y` = a + b x. The mean and standard deviation for the
Course Grade variable is 0.766 and 0.123 The mean and standard deviation for the Exam 1 Score
variable is 83.943 and 11.295 The correlation is 0.7845 Be very sensitive to roundoff errors.
sy
We'll use the formula, y` = a + b x, with b = r ÅÅÅÅ
ÅÅ and a = êêy - b êê
x.
sx
0.123
b = .7845 H ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ L º .008543
11.295
a = 0.766 - .008543 H83.943L º .04887
So the regression line is given by y` = .04887 + .008543 x.
HbL Predict the course grade for a student who scores a 91 on their first exam. Use the equation of the
regression line found in (b).
y` = .04887 + .008543 H91L º .826
2.
A Gallup Poll asked "Do you think the United States needs to: change its current strategy in Iraq, keep
its current strategy, but change its tactics, or keep both its current strategy and tactics in Iraq? ... Nearly 6 in
10 Americans (59%) say the United States should change its strategy in Iraq." Gallup's report said "Results
for this panel study are based on telephone interviews with 1,001 national adults, aged 18 and older, conducted Oct. 23-26, 2006." Furthermore, the article included the statements "For results based on the total
sample of national adults, one can say with 95% confidence that the maximum margin of sampling error is ±3
percentage points. In addition to sampling error, question wording and practical difficulties in conducting
surveys can introduce error or bias into the findings of public opinion polls."
HaL
Describe in as much detail as possible given the excerpts from the article above the population.
The population are adults in the United States aged 18 and older.
HbL
How many members of the sample felt that the United States should change its strategy in Iraq?
59% which is roughly 590 people.
HcL If the confidence level was changed from 95% to 99% would the maximum margin of sampling
error be (1) less than 3%, (2) greater than 3%, or (3) remain at 3%?
greater than 3%
HdL Suppose that there are 240 million adults living in the US. How many members of the population
described by the article would you estimate feel that the United States should change its strategy in Iraq?
59% which is roughly 141,600,000 people
3.
Choose an adult aged 25 years or over at random. The probability is 0.151 that the person chosen did
not complete high school, 0.322 that the person has a high school diploma but no further education, and 0.289
that the person has at least a bachelor's degree.
HaL What must be the probability that a randomly chosen adult has some education beyond high school
but does not have a bachelor's degree?
1 - .151 - .322 - .289 = .238
HaL What must be the probability that a randomly chosen adult has some education beyond high school
but does not have a bachelor's degree?
1 - .151 - .322 - .289 = .238
HbL What is the probability that a randomly chosen young adult has at least a high school education?
1 - .151 = .849
4.
The distribution of the weights of all bags of M&M candies measured in grams is NH48.5, 1.3L:
44.6 45.9 47.2 48.5 49.8 51.1 52.4
HaL If a single bag of M&M's was selected at random, what is the probability that the weight of the
bag would be between 48 grams and 48.5 grams?
We need to first find the z values for 48 and 48.5.
The z value for 48.5 is 0, since this is the mean.
48-48.5
For 48, the z value is z = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅ º -.38.
1.3
The probability of getting a bag that weighs less than 48.5 grams is .5.
From table A, the probability of getting a bag that weighs less than 48 grams is .352.
So the probability of getting a bag between 48 and 48.5 grams is .5 - .352 = .148.
HbL If a SRS of 40 bags of M&M's was selected and the average weight of the 40 bags was computed,
what is the probability that the average weight for the 40 bags would be between 48 grams and 48.5
grams?
1.3
The sampling distribution for the average weight of a sample of 40 bags is NI48.5, ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ M = NH48.5, .21L.
è!!!!!!
40
The z value for 48.5 is 0.
48-48.5
The z value for 48 is z = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅ º -2.38.
.21
The probability of getting a sample with average weight less than 48.5 grams is .5.
From table A, the probability of getting a sample with average weight less than 48 is .0087.
So the probability of getting a sample with average between 48 and 48.5 grams is .5 - .0087 = .4913.
5.
Suppose a population is composed of uniformly distributed values ranging from 0 to 90 ( m = 45 and
s = 25.98 ). The population distribution is shown below.
Uniformly Distributed Population
0
20
40
60
80
HaL What is the probability that a single value selected from the population would be greater than 40
and less than 45?
45-40
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅ º .056
90
HbL What is the probability that a SRS of size 100 taken from this population would have a sample
mean greater than 40 and less than 45?
Since the sample size is relatively large, the sampling distribution for the mean will be roughly normal
25.98
with distribution NI45, ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅ M = NH45, 2.598L.
è!!!!!!!!
100
The z value for 45 is 0.
40-45
The z value for 40 is z = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅ º -1.92.
2.598
The probability of getting a sample with mean less than 45 is .5.
From table A, the probability of getting a sample with mean less than 40 is .0274.
So the probability of getting a sample with mean between 40 and 45 is .5 - .0274 = 0.4726.
6.
To estimate the mean height m of adult women in your city, you will measure an SRS of adult women.
You know from government data that heights of adult women are approximately Normal with standard deviation 2.3 inches. How large an SRS do you need to collect so that the standard deviation of êê
x is 0.25?
2.3
.25 = ÅÅÅÅ
ÅÅÅÅ!ÅÅ
è!!!
è!!!
2.3
n = ÅÅÅÅ
ÅÅÅÅ
.25
è!!!
n = 9.2
n = 84.64
n
So we need an SRS of size 85.
7.
To estimate the mean weight of all bags of M&M candies you collect a SRS of 75 bags and find the
average weight of the 100 bags to be 48.95 grams. Give a 96% confidence interval for the mean weight of all
bags of M&M's. Assume s = 1.4 for the weights of all bags of M&M's.
From table C, we get that z* = 2.054 for a 96% confidence interval. So our confidence interval comes from
the
following.
1.4
48.95 ± 2.054 I ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ M
è!!!!!!
75
48.95 ± .332
or from 49.282 to 48.618
8.
You would be satisfied to estimate the mean weight of all bags of M&M's to within ± 0.1 grams with
95% confidence. How many bags of M&M candies must be in your SRS? Assume s = 1.4 for the weights of
all bags of M&M's.
From table C, we get that z* = 1.96 for a 95% confidence interval. Using this with the formula for margin of
error gives the following.
1.4
.1 = 1.96 I ÅÅÅÅ
ÅÅÅÅ!Å M
è!!!
.1
ÅÅÅÅ
ÅÅÅÅÅÅ
1.96
=
1.4
ÅÅÅÅ
ÅÅÅÅ!Å
è!!!
n
1.4
.051 = ÅÅÅÅ
ÅÅÅÅ!Å
è!!!
n
è!!!
1.4
n = ÅÅÅÅ
ÅÅÅÅÅÅ
.051
è!!!
n = 27.44
n = 752.9536
n
So we need an SRS of size 753.
all bags of M&M's.
From table C, we get that z* = 1.96 for a 95% confidence interval. Using this with the formula for margin of
error gives the following.
1.4
.1 = 1.96 I ÅÅÅÅ
ÅÅÅÅ!Å M
è!!!
.1
ÅÅÅÅ
ÅÅÅÅÅÅ
1.96
=
1.4
ÅÅÅÅ
ÅÅÅÅ!Å
è!!!
n
n
1.4
.051 = ÅÅÅÅ
ÅÅÅÅ!Å
è!!!
è!!!
1.4
n = ÅÅÅÅ
ÅÅÅÅÅÅ
.051
è!!!
n = 27.44
n = 752.9536
n
So we need an SRS of size 753.
9.
The mean height of American women is about 64 inches, with standard deviation about 2.7 inches. The
mean height of American men is about 69 inches, with standard deviation about 2.8 inches. If the correlation
between the heights of husbands and wives is about r = 0.5, find the equation of the regression line that will
predict the height of the wife when given the height of the husband.
sy
We'll use the formula, y` = a + b x, with b = r ÅÅÅÅ
ÅÅ and a = êê
y - b êêx .
sx
2.7
b = .5 H ÅÅÅÅ
ÅÅÅÅ L º .48
2.8
a = 64 - .48 H69L º 30.73
So the regression line is given by y` = 30.73 + .48 x.
10. Use your answer from the previous problem to predict the height of the wife of a man who is 72 inches
tall. Would you consider this estimate to be an accurate one?
y` = 30.73 + .48 H72L = 65.29 inches
This can't be considered an accurate estimate because the correlation between the variables is only .5, which
suggests that a linear model is not a good choice for the data.
11. A study of children in elementary school shows a high positive correlation between the heights and
scores on a test of reading comprehension. Johnny, an exceptionally bright and tall fifth grader concludes that
he must have a high reading comprehension. Explain to Johnny what is wrong with his conclusion.
Age is a lurking variable in this situation. Older children tend to be taller and thus tend to have higher reading
comprehension.
12. The following is a list of the 10 most popular boys names. Susy wants to name her newborn son one of
these names, but can't decide which one to use. Use line 108 of table B to find a SRS of size 3 of the names in
the list, so that Susy will only have three names to chose from.
AIDAN ETHAN
CADEN JADEN
CALEB DYLAN
JACOB CONNOR
LOGAN HAYDEN
First we'll number the names using 0 - 9.
0 AIDAN
5 ETHAN
1 CADEN
6 JADEN
2 CALEB
7 DYLAN
3 JACOB
8 CONNOR
4 LOGAN
9 HAYDEN
Using line 108 of table B, the first three digits are 609, so we'll choose names 6, 0, and 9 from the list.
These names are Jaden, Aidan, and Hayden.
Note: Depending on how you number the names, you may get a different answer. As long as you follow this
basic procedure, however, the list can be considered random.
0 AIDAN
5 ETHAN
1 CADEN
6 JADEN
2 CALEB
7 DYLAN
3 JACOB
8 CONNOR
4 LOGAN
9 HAYDEN
Using line 108 of table B, the first three digits are 609, so we'll choose names 6, 0, and 9 from the list.
These names are Jaden, Aidan, and Hayden.
Note: Depending on how you number the names, you may get a different answer. As long as you follow this
basic procedure, however, the list can be considered random.
13. A political scientist wants to know how college students feel about the Social Security program. The
scientist obtains a list of all the students at WWCC and mails a questionnaire to 200 of the students selected at
random. Based upon 98 returned surveys the scientist concludes that college students are dissatisfied with the
Social Security program. What are the population and sample of this study?
Population: WWCC students
Sample: The 98 students who responded
14. Give an explanation for what is wrong with the conclusion of the study described in the previous problem.
Only the students with strong feelings are likely to respond, so we can't consider the sample to be a SRS.
15.
Read the description for each of the following studies, then answer the question.
Study 1: A publishing company wants to measure the effectiveness of one of its textbooks. The company
tests a group of students for comprehension and then divides the students into two groups. One group uses the
book and one doesn't. The company then retests the students and compares the results.
Study 2: A political scientist wants to know how college students feel about the Social Security program.
The scientist obtains a list of all the students at WWCC and mails a questionnaire to 200 of the students
selected at random. Based upon 98 returned surveys the scientist concludes that college students are dissatisfied with the Social Security program.
For each study, determine whether it is an observational study or an experiment and give reasons for your
answers.
Study 1: This is an experiment. The use of the new textbook is the treatment.
Study 2: This is an observational study. There is no treatment imposed on the individuals.
16. In a bag full of marbles there are three red marbles, five blue marbles, and seven yellow marbles. Give
the probability model for the act of drawing one marble at random from the bag (i.e. what is the sample space
and probability of each event).
Sample Space: red, blue, yellow
3
5
7
Probabilities: PHredL = ÅÅÅÅ
ÅÅ = ÅÅÅÅ15 , PHblueL = ÅÅÅÅ
ÅÅ = ÅÅÅÅ13 , PHyellowL = ÅÅÅÅ
ÅÅ
15
15
15
17. What is the probability of drawing either a red marble or a blue marble? Explain your answer using the
appropriate probability rule.
8
PHred or blueL = PHredL + PHblueL = ÅÅÅÅ15 + ÅÅÅÅ13 = ÅÅÅÅ
ÅÅ
15
18. Voter registration records in a small town in Washington show that 52% of all voters are Republican.
On Mayberry Ave., 50 people are polled and it is found that 49% of them are Republican. Which of these
numbers is a parameter and which is a statistic? Explain your answers.
52% is a parameter because it is the actual percent of voters who are Republican.
49% is a statistic because it is an estimate from a small sample of the percent of Republicans on Mayberry
Ave.
18. Voter registration records in a small town in Washington show that 52% of all voters are Republican.
On Mayberry Ave., 50 people are polled and it is found that 49% of them are Republican. Which of these
numbers is a parameter and which is a statistic? Explain your answers.
52% is a parameter because it is the actual percent of voters who are Republican.
49% is a statistic because it is an estimate from a small sample of the percent of Republicans on Mayberry
Ave.
19. The weight of the eggs produced by a certain breed of hen is Normally distributed with mean 65g and
standard deviation 5g. Think of cartons of such eggs as SRSs of size 12 from the population of all eggs.
What is the approximate distribution of the mean weight from each carton of eggs?
5
NI65, ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ M º NH65, 1.44L
è!!!!!!
12
20. For the situation described in the previous problem, what is the probability that the weight of a carton
falls between 750g and 825g?
If the carton weighs 750g, then the average weight per egg is 62.5g.
62.5-65
The z-value for this weight is z1 = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅ º -1.74.
1.44
PHz § -1.74L = .0409 (from table A)
If the carton weighs 825g, then the average weight per egg is 68.75g.
68.75-65
The z-value for this weight is z2 = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ º 2.6.
1.44
PHz § 2.6L = .9953 (from table A)
From this we get PH750 § êê
x § 825L = .9953 - .0409 = .9544
21. A class survey in a large class for first-year college students asked, "About how many minutes do you
study on a typical weeknight?" The mean response of the 200 students was 135 minutes. Suppose we know
that the study time of all first year students at this university follows a Normal distribution with standard
deviation s = 60 minutes. Find a 99% confidence interval for the mean study time of all first-year students.
The approximate distribution for the mean study time of students from a SRS of size 200 is
60
NI135, ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅ M º NH135, 4.24L.
è!!!!!!!!
200
For a 99% confidence interval z* = 2.576 (from table C).
We'll
use
these
to
get
the
following
confidence
interval.
135 ± 2.576 H4.24L
135 ± 11.39
Or between 147 and 123 minutes
22. How large a sample is needed to cut the margin of error for the previous problem to
6?
2.576 H60L
n = I ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅ M º 663.58
6
So we need a sample of 664.
2
23. If the distribution of the study time of all first year students at this university didn't follow a Normal
distribution but still had standard deviation s = 60, could you still find the same confidence interval asked for
above? Give a reason for your answer.
Yes, since the sample size is relatively large we can apply the Central Limit Theorem and assume that the
sampling distribution is approximately normal.
24. A manufacturer of small appliances wants to estimate retail sales of its hand mixers by gathering information from a sample of retail stores. This month an SRS of 75 stores finds that these stores sold an average of
24 mixers, with standard deviation 11. Give a 95% confidence interval for the mean number of mixers sold
by all stores.
11
The distribution of êê
x = average number of mixers is approximately NI24, ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ M º NH24, 1.27L.
è!!!!!!
75
For a 95% confidence interval t* = 2 (from table C, with df = 60 because 74 isn't on the table).
24. A manufacturer of small appliances wants to estimate retail sales of its hand mixers by gathering information from a sample of retail stores. This month an SRS of 75 stores finds that these stores sold an average of
24 mixers, with standard deviation 11. Give a 95% confidence interval for the mean number of mixers sold
by all stores.
11
The distribution of êê
x = average number of mixers is approximately NI24, ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ M º NH24, 1.27L.
è!!!!!!
75
For a 95% confidence interval t* = 2 (from table C, with df = 60 because 74 isn't on the table).
This
gives
the
following
confidence
interval.
24 ± 2 H1.27L
24 ± 2.54
or between 26.54 and 21.46 mixers