Practice Exam 3 Answers
1. A simple random sample of commuting UF students was taken to determine the
average amount of money spent on gas per week. The 34 students surveyed gave an
average expenditure of $15.80 with a standard deviation of $4.25. Does it appear that
the population average gas expenditure per week is greater than $17.50?
(i) Is this a means problem or a proportion problem?
Means
(ii) Are the assumptions met?
SRS – yes, because it says so
n> 30 or X is Normal  n = 34 Met.
(iii)State the null and the alternative hypothesis.
Ho: μ = 17.50
Ha: μ > 17.50
(iv) Calculate the test statistic.
Entered into the calculator. (t-test)
μ0 = 17.50
n = 34
s= 4.25
x  15.8
Test Statistic = -2.33
(v) Find the p-value and interpret.
p-value = 0.9870
With a p-value equal to 0.9870, we have no statistically significant evidence that the
population mean amount spent on gas is greater than 17.50 dollars.
2. A national study estimated that in 2005, college students spent an average of 2 hours
per day online. An SRS of 150 UF students produces an average of 1.8 hours. The
standard deviation is 0 .4. Do UF students spend less time than average online?
(i) Is this a means problem or a proportion problem?
Means
(ii) Are the assumptions met?
SRS – yes, because it says so
n> 30 or X is Normal  n = 150 Met.
(iii)State the null and the alternative hypothesis.
Ho: μ = 2
Ha: μ < 2
(iv) Calculate the test statistic.
Entered into the calculator. (t test)
μ0 = 2
n =150
s= 0.4
x  1.8
Test Statistic = -6.12
(v) Find the p-value and interpret.
p-value = almost zero
With a p-value equal to almost zero, we have very strong evidence to show that the
population mean time spent online is less than 2 hours.
3. A company tries to fill their canned drinks to 12 fl. oz. They control this by taking a
SRS of 6 cans every 30 minutes. The data is below. Is the level of fill in the drinks
significantly different from what it should be?
11.2
11.3
11.4 11.4 10.9 10.7
(i) Is this a means problem or a proportion problem?
Means
(ii) Are the assumptions met?
SRS – yes, because it says so
n> 30 or X is Normal  n = 6 make a plot to check for outliers
(iii)State the null and the alternative hypothesis.
Ho: μ = 12
Ha: μ ≠ 12
(iv) Calculate the test statistic.
Entered into the calculator. (t test)
μ0 = 12
n =6
s= 0.288
x  11.15
Test Statistic = -7.23
(v) Find the p-value and interpret.
p-value = 0.0007912 = almost zero
With a p-value of almost zero, we have very strong evidence that the population
mean amount of fill in the canned drinks is significantly different from 12.
4. Is the percentage of college students looking for a job higher than the percentage
looking for a job in the general population? A survey reported that at any given time,
19% of Americans are looking for a job. A survey of 200 college students was conducted
and 60 reported looking for a job.
(i) Is this a means problem or a proportion problem?
proportion
(ii) Are the assumptions met?
SRS – does not say
npo ≥ 15 and n(1-po) ≥ 15
po= 0.19 n = 200
npo = 0.19*200 = 38 Met.
n(1-po) = 200(1-.19) = 162 Met.
(iii)State the null and the alternative hypothesis.
Ho: p = 0.19
Ha: p > 0.19
(iv) Calculate the test statistic.
Values entered into Calculator. (t test)
Po= 0.19
n = 200
x = 60
Test Statistic equals 3.96
(v) Find the p-value and interpret.
p-value equals almost zero.
With a very small p-value, we have very strong evidence to show that the population
proportion of students looking for a job is greater than 0.19 as long as this was a
simple random sample.
5. The GSS asked participants whether they agree with the statement that a marriage
without children isn’t fully complete. 623 agreed or strongly agreed with the statement,
while 731 did not. Is the proportion that agrees or strongly agrees different from 83%?
n = 623+731 = 1354
(i) Is this a means problem or a proportion problem?
Proportion
(ii) Are the assumptions met?
SRS – yes GSS
npo ≥ 15 and n(1-po) ≥ 15
po= 0.83 n = 623+731 = 1354
npo = 0.83*1354 = 1123.82 Met.
n(1-po) = 1354*(1-.83) = 230.18 Met.
(iii)State the null and the alternative hypothesis.
Ho: p = 0.83
Ha: p ≠ 0.83
(iv) Calculate the test statistic.
Values entered into Calculator. (1 Prop Z Test)
Po= 0.83
n = 1354
x = 623
Test Statistic equals -36.24
(v) Find the p-value and interpret.
p-value = “almost” zero
With a p-value of almost zero, we have very strong evidence that the population
proportion who things that a marriage is incomplete without children is different
from 0.83.
6. Is the percentage of people who call themselves very religious less than 25%? The
1998 GSS reported that 268 of 1427 people considered themselves very religious. Is
there evidence to show that the percentage is significantly less than 25%?
(i) Is this a means problem or a proportion problem?
Proportion
(ii) Are the assumptions met?
SRS yes because GSS
npo ≥ 15 and n(1-po) ≥ 15
po= 0.25 n= 1427
npo = 0.25*1427 = 356.75 Met.
n(1-po) = 1427*(1-.25) = 1070.25 Met.
(iii)State the null and the alternative hypothesis.
Ho: p = 0.25
Ha: p < 0.25
(iv) Calculate the test statistic.
Values entered into Calculator. (1 Prop Z Test)
Po= 0.25
n = 1427
x = 268
Test Statistic equals -5.43
(v) Find the p-value and interpret.
p-value = almost zero
With a p-value of almost zero, we have strong evidence that the population
proportion of people that consider themselves very religious is less than 0.25.
7. The GSS asked 1594 people if they had been to an amateur or professional sports
event within the past twelve months. 855 people reported that they had. Is there
evidence to show that the proportion of people who have been to a sports event in the
past twelve months is different from 0.50?
(i) Is this a means problem or a proportion problem?
Population proportion
(ii) Are the assumptions met?
SRS – yes because GSS
npo ≥ 15 and n(1-po) ≥ 15
po= 0.50 n= 1594
npo = 0. 5*1594 = 797 Met.
n(1-po) = 1594*(1-.5) = 797 Met.
(iii)State the null and the alternative hypothesis.
Ho: p = 0.5
Ha: p ≠ 0.5
(iv) Calculate the test statistic.
Values entered into Calculator. (1 Prop Z test)
Po= 0.5
n = 1594
x = 855
Test Statistic equals 2.91
(v) Find the p-value and interpret.
p-value = 0.0037
With a p-value equal to 0.0037, we have very strong evidence that the population
proportion of those that have attended a sporting event is significantly different
from 0.50.
8.An inspector inspects large truckloads of potatoes to determine the proportion in the
shipment with major defects prior to using the potatoes to make chips. If there is clear
evidence that this proportion is less than 0.10, she will accept the shipment. To do so, she
selects a simple random sample of 200 potatoes from the more than 3000 potatoes on the
truck. Only 8 of the potatoes sampled are found to have major defects. Does she accept
the shipment?
(i) Is this a means problem or a proportion problem?
Proportion
(ii) Are the assumptions met?
SRS – yes b/c it says random
npo ≥ 15 and n(1-po) ≥ 15
po= 0.10 n= 200
npo = 0. 1*200 = 20 Met.
n(1-po) =200*(1-.1) = 180 Met.
(iii)State the null and the alternative hypothesis.
Ho: p = 0.1
Ha: p < 0.1
(iv) Calculate the test statistic.
Values entered into Calculator. (1 Prop Z Test)
Po= 0.1
n =200
x=8
Test Statistic equals -2.83
(v) Find the p-value and interpret.
With a p-value equal to 0.0023, we have very strong evidence that the population
proportion of defective potatoes is less than 0.1.
9. A newsletter recently reported that 90% of adults drink milk. A regional farmers’
organization is planning a new marketing campaign across its tri-country area. They
randomly poll 600 people in the area. In this sample, 525 people said that they drink
milk. Do these data provide strong evidence that the 90% figure is not accurate for this
region?
(i) Is this a means problem or a proportion problem?
proportion
(ii) Are the assumptions met?
npo ≥ 15 and n(1-po) ≥ 15
po= 0.90 n= 600
npo = 0. 9*600 = 540 Met.
n(1-po) =600*(1-.9) = 60 Met.
(iii)State the null and the alternative hypothesis.
Ho: p = 0.9
Ha: p ≠ 0.9
(iv) Calculate the test statistic.
Values entered into Calculator. (1 prop Z test)
Po= 0.9
n =600
x =525
Test Statistic equals -2.04
(v) Find the p-value and interpret.
With a p-value equal to 0.0412, we have strong evidence that the population
proportion of adults that drink milk is significantly different from 0.90.
For questions 10-19, answer the questions for each problem.
10. A high school biology teacher thinks that a new hands-on method of teaching will
improve student scores on an end of the year test. She teaches 10 students the old way,
and she teaches 7 students the new hands-on way. Are the test scores for the students that
learned the new way higher than those that learned the old way?
New Method
Old Method
new
old
N
7
10
Mean
87.7
83.50
88 78
65 86
StDev
11.6
9.58
95 96 100
92 95 84
67 90
77 96
83 75 82
SE Mean
4.4
3.0
Difference = mu (new) - mu (old)
Estimate for difference: 4.21429
95% CI for difference:
(-7.48721, 15.91578)
Boxplot of new, old
100
Data
90
80
70
60
new
old
a.) What type of test is this: two independent proportions, two independent means or
means of dependent samples?
Two independent means
b.) What are the assumptions of the confidence interval? Are they satisfied?
SRS -- doesn’t say
Data from a Normal Distribution – yes, b/c no outliers
c.) What is the 95% confidence interval? ___(-7.48, 15.91)____________
(If you had done this in the calculator, it would have been a 2Sample T Interval)
d.) What is your conclusion? We are 95% confident that the population mean of the
end of the year test for the new method is between 7.48less to 15.91 more as long
as it was a random sample.
e.) What is the short conclusion? No significant difference.
11. Each of a random sample of ten college freshman takes a mathematics aptitude
test both before and after undergoing an intensive training course designed to
improve such test scores. Then, results of the students’ scores before and after the
course are listed below. Using the below results and the computer output below to
answer the following questions.
Student
Before
After
1
60
70
2
73
80
3
42
40
4
88
94
5
66
79
6
77
86
7
90
93
8
63
71
9
55
70
10
96
97
Paired T-Test and CI: before, after
Paired T for before - after
before
after
Difference
N
10
10
10
Mean
71.00
78.00
-7.00
StDev
17.07
16.77
5.25
SE Mean
5.40
5.30
1.66
95% CI for mean difference: (-10.76, -3.24)
T-Test of mean difference = 0 (vs not = 0): T-Value = -4.22
P-Value = 0.002
a.) What type of test is this: two independent proportions, two independent means or
means from dependent samples?
Means from dependent samples
b.) What are the assumptions of the confidence interval? Are they satisfied?
SRS  met, says random sample
N greater than or equal to 30  No, says n=10
Population of differences could be normally distributed.
c.) What is the 95% confidence interval? (-10.76, -3.24)_______________
d.) What is the short conclusion? Before scores are lower
e.) What is the full interpretation?
We are 95% confident that the population mean mathematics aptitude score for
before the intensive course is between 10.76 less to 3.24 less than after the course.
12. It is widely believed that those ages 16-24 feel differently than those 25 or older
about the legalization of marijuana. A national survey, which randomly selected from all
across the United States, found the following results:
Yes-Legalize
503
525
Ages 16-24
Age >25
No-Don’t Legalize
1021
1263
Total
1524
1788
Does it appear that the younger group feels differently from the older group about the
legalization of marijuana?
Sample
1
2
X
503
525
N
1524
1788
Sample p
0.330052
0.293624
Difference = p (1) - p (2)
Estimate for difference: 0.0364283
95% CI for difference: (0.00475862, 0.0680980)
a.) What type of test is this: two independent proportions, two independent means or
means of dependent samples?
Two independent proportions
b.) What are the assumptions of the confidence interval? Are they satisfied?
SRS – yes says randomly selected.
At least 15 successes and 15 failures in both groups.
In group 1: 503 successes and 1021 failures
In group 2: 525 successes and 1263 failures
c.) What is the 95% confidence interval? ___(0.0047, 0.068)____________
d.) What is your conclusion?
We are 95% confident that the population proportion of those that are for legalizing
marijuana in the 16-23 years of age group is between 0.0047 more to 0.068 more
than those over age 25.
(To do this using a calculator, use 2-samplePropZInterval.)
13. In 2010, the General Social Survey included a question that asked males and females
if they thought that Antarctic penguins were threaten. The possible responses were “A
great deal” and “not at all”.
Males (group 1)
Females (group 2)
A great deal
121
213
Not at all
30
17
Total
151
230
Make a 95% confidence interval for the difference in the population proportion between
men and women that feel that penguins in Antarctica are threatened a great deal.
Test and CI for Two Proportions
Sample
1
2
X
121
213
N
151
230
Sample p
0.801325
0.926087
Difference = p (1) - p (2)
Estimate for difference: -0.124762
95% CI for difference: (-0.196828, -0.0526971)
Test for difference = 0 (vs not = 0): Z = -3.39
P-Value = 0.001
Fisher's exact test: P-Value = 0.000
a.) What type of test is this: two independent proportions, two independent means or
means of dependent samples?
Two independent proportions
a) What are the assumptions of the confidence interval? Are they satisfied?
SRS – yes, says GSS
At least 15 successes and 15 failures in both groups.
In group 1: 121 successes and 30 failures
In group 2: 213 successes and 17 failures
b) What is the 95% confidence interval? ______(-0.197, -0.053)_______________
c) What is your conclusion?
We are 95% confident that the population proportion of those that feel that
Antarctic penguins are threatened a great deal for group 1 (males) is between 0.197
less to 0.053 less than group 2 (females).
d) What is your quick interpretation? More females than males feel that Antarctic
penguins are threatened.
(To do this using a calculator, use 2-samplePropZInterval.)
14. A health professional believes that a new diet will increase average energy levels.
Eight randomly selected women are given a survey to measure their average energy
levels before they begin the diet. After 3 months, they are again given the survey.
Higher numbers relate to higher energy levels. The results are as follows:
Before
After
15
18
Before
After
Difference
N
8
8
8
22
23
25
23
18
22
24
25
Mean
20.5000
21.6250
-1.12500
StDev
4.7509
3.6228
2.41646
SE Mean
1.6797
1.2809
0.85435
14
18
19
17
27
27
95% CI for mean difference: (-3.14521, 0.89521)
Histogram of Differences
(with Ho and 95% t-confidence interval for the mean)
2.0
Frequency
1.5
1.0
0.5
0.0
_
X
Ho
-4
-3
-2
-1
Differences
0
1
2
a) What type of test is this: two independent proportions, two independent means or
means of dependent samples?
Paired Means
b.) What are the assumptions of the confidence interval? Are they satisfied?
SRS – yes stated
n> 30 (no) or X is from a Normal Distribution ( no outliers so o.k. )
c.) What is the 95% confidence interval? _(-3.14, 0.895)_____________
d.) What is your conclusion? We are 95% confident that the population mean energy
level before the diet before the diet was between 3.14less to 0.895 more than after
the diet.
e.) What is the quick interpretation? No statistically significant difference
15. Do UF students from out of state spend significantly less time at their parents’ house
(homes) than those from in state? A SRS found that out-of-state students spend an
average of 3.22 days per semester (std. dev. = 2.49) at their parents’ house (homes)
while in-state students spend an average of 5.89 days per semester (std. dev.=4.23) at
their parents’ house (homes).
Out of state
Instate
N
9
9
Mean
3.22
5.89
StDev
2.49
4.23
SE Mean
0.83
1.4
Difference = mu (Out of state) - mu (Instate)
Estimate for difference: -2.66667
95% CI for difference: (-6.22877, 0.89544)
T-Test of difference = 0 (vs not =): T-Value = -1.63 P-Value = 0.129 DF = 12
T-Test of difference = 0 (vs <): T-Value = -1.63 P-Value = 0.064 DF = 12
T-Test of difference = 0 (vs >): T-Value = -1.63 P-Value = 0.936 DF = 12
Boxplot of Out of state, Instate
14
12
Data
10
8
6
4
2
0
Out of state
Instate
a.) What type of test is this: two independent proportions, two independent means or
means of dependent samples?
Two independent means
b.) Are the assumptions met?
SRS – yes stated.
n> 30 for both samples – NO n= 9 for both samples.
Population is normal – Could be because there are no outliers
c.) What is the null hypothesis? _Ho: μ1-μ2=0______________
d.) What is the alternative hypothesis? __ Ha: μ1-μ2<0___________
e.) Look at the output below. What is the corresponding p-value for this hypothesis?
__p-value=0.064_________
f.) Write a conclusion.
With a p-value equal to 0.064, we have some evidence that the population mean
amount of time spent at home is at less for out of states students then for in state
students.
(This can be done on the calculator, using the 2-samplet test. )
16. A high school biology teacher thinks that a new hands-on method of teaching will
improve student scores on an end of the year test. She teaches 10 students the old way,
and she teaches 7 students the new hands-on way. Are the test scores for the students that
learned the new way higher than those that learned the old way?
New Method (1) 88 78 95 96 100 67 90
77 96 83 75 82
Old Method (2) 65 86 92 95 84
new
old
N
7
10
Mean
87.7
83.50
StDev
11.6
9.58
SE Mean
4.4
3.0
Difference = mu (new) - mu (old)
Estimate for difference: 4.21429
T-Test of difference = 0 (vs <): T-Value = 0.79
P-Value = 0.778
T-Test of difference = 0 (vs not =): T-Value = 0.79
T-Test of difference = 0 (vs >): T-Value = 0.79
DF = 11
P-Value = 0.445
P-Value = 0.222
DF = 11
DF = 11
a.) What type of test is this: two independent proportions, two independent means or
means of dependent samples?
Two independent means
b.) What is the null hypothesis? ___Ho: μ1-μ2=0____________
c.) What is the alternative hypothesis? ___ Ha: μ1-μ2>0__________
d.) Look at the output below. What is the corresponding p-value for this hypothesis?
___0.222________
e.) Write a conclusion.
With a p-value equal to 0.22, we have no statistically significant evidence that the
population mean end of the year test score with the new old method is greater than
with the old method.
(This can be done on the calculator, using the 2-samplet test. )
17. It is widely believed that those ages 16-24 feel differently than those 25 or older
about the legalization of marijuana. A national survey found the results below. Does it
appear that the younger group feels differently from the older group about the
legalization of marijuana?
Yes-Legalize
No-Don’t Legalize Total
503
1021
1524
Ages 16-24
525
1263
1788
Age >25
Sample
1
2
X
503
525
N
1524
1788
Sample p
0.330052
0.293624
Difference = p (1) - p (2)
Estimate for difference: 0.0364283
Test for difference = 0 (vs not = 0):
Z = 2.25
P-Value = 0.024
Test for difference = 0 (vs < 0):
Z = 2.25
P-Value = 0.988
Test for difference = 0 (vs > 0):
Z = 2.25
P-Value = 0.012
a.) What type of test is this: two independent proportions, two independent means or
means of dependent samples?
Two independent proportions
b.) What is the null hypothesis? ___Ho: p1-p2 = 0 ____________
c.) What is the alternative hypothesis? ___ Ho: p1-p2 ≠ 0 __________
d.) Look at the output. What is the corresponding p-value for this hypothesis?
__0.024_________
e.) Write a conclusion.
With a p-value equal to 0.024, we have strong evidence that the population
proportion of those that believe that marijuana should be legalized is different from
those 16-24 years old and those over 25.
(This can be done on a calculator using the 2 sample prop test.)
18. Each of a random sample of ten college freshman takes a mathematics aptitude
test both before and after undergoing an intensive training course designed to
improve such test scores. Then, results of the students’ scores before and after the
course are listed below. Using the below results and the computer output below to
answer the following questions. Is there evidence to show that the intensive training
course helped?
Student
Before
After
1
60
70
2
73
80
3
42
40
4
88
94
5
66
79
6
77
86
7
90
93
8
63
71
9
55
70
10
96
97
Paired T-Test and CI: before, after
Paired T for before - after
before
after
Difference
95% CI
T-Test
T-Test
T-Test
N
10
10
10
Mean
71.00
78.00
-7.00
StDev
17.07
16.77
5.25
for mean difference:
of mean difference =
of mean difference =
of mean difference =
SE Mean
5.40
5.30
1.66
(-10.76, -3.24)
0 (vs not = 0): T-Value = -4.22 P-Value = 0.002
0 (vs < 0): T-Value = -4.22 P-Value = 0.001
0 (vs > 0): T-Value = -4.22 P-Value = 0.999
a.) What type of test is this: two independent proportions, two independent means or
means of dependent samples?
Mean of dependent samples
a) What is the null hypothesis? ___ Ho:μd = 0 ____________
b) What is the alternative hypothesis? __ Ha: μd < 0___________
c) Look at the output below. What is the corresponding p-value for this hypothesis?
__0.001_________
d) Write a conclusion.
With a p-value equal to 0.001, there is very strong evidence that the population
mean test score before intensive training course is less than the population mean
after the training
(This can be done in the calculator, by adding another row in the table. Then find
before minus after. Put these differences into L1 in the calculator. Use the t –
interval to get a confidence interval. )
19. In 2010, the General Social Survey included a question that asked males and females
if they thought that Antarctic penguins were threaten. The possible responses were “A
great deal” and “not at all”.
Males (group 1)
Females (group 2)
A great deal
121
213
Not at all
30
17
Total
151
230
Is there evidence to show that the population proportion of those that think that penguins
are threatened a great deal is bigger for females than males?
Test and CI for Two Proportions
Sample
1
2
X
121
213
N
151
230
Sample p
0.801325
0.926087
Difference = p (1) - p (2)
Estimate for difference: -0.124762
95% CI for difference: (-0.196828, -0.0526971)
Test for difference = 0 (vs not = 0): Z = -3.39 P-Value = 0.001
Test for difference = 0 (vs < 0): Z = -3.39 P-Value = 0.000
Test for difference = 0 (vs > 0): Z = -3.39 P-Value = 1.000
a.) What type of test is this: two independent proportions, two independent means or
means of dependent samples?
Two independent proportions
a) What is the null hypothesis? ___ Ho: p1-p2 = 0 ____________
b) What is the alternative hypothesis? ___ Ho: p1-p2< 0 __________
c) Look at the output below. What is the corresponding p-value for this hypothesis?
_0.000__________
d) Write a conclusion.
With a p-value equal to 0.000, we have very strong evidence that the population
proportion that think that ANtartic pengins are threatened a great deal for women
is higher than for men.
(To do this in the calculator, use the 2 sample prop Z interval.)
20. A health professional believes that a new diet will increase average energy levels.
Eight women are given a survey to measure their average energy levels before they begin
the diet. After 3 months, they are again given the survey. Higher numbers relate to
higher energy levels. The results are as follows:
22
25
18
24
14
19
27
Before 15
18
23
23
22
25
18
17
27
After
Before
After
Difference
N
8
8
8
Mean
20.5000
21.6250
-1.12500
StDev
4.7509
3.6228
2.41646
SE Mean
1.6797
1.2809
0.85435
T-Test of mean difference = 0 (vs not = 0): T-Value = -1.32
P-Value = 0.229
T-Test of mean difference = 0 (vs < 0): T-Value = -1.32
P-Value = 0.115
T-Test of mean difference = 0 (vs > 0): T-Value = -1.32
P-Value = 0.885
a.) What type of test is this: two independent proportions, two independent means or
means of dependent samples?
Mean of dependent samples
b.) What is the null hypothesis? _Ho: μd = 0 ______________
c.) What is the alternative hypothesis? ___ Ho: μd < 0 __________
d.) Look at the output below. What is the corresponding p-value for this hypothesis?
__0.115_________
e.) Write a conclusion.
With a p-value equal to 0.115, we have no statistically significant evidence to show
that the population mean energy level is different before or after the diet.
(This can be done in the calculator, by adding another row in the table. Then find
before minus after. Put these differences into L1 in the calculator. Use the t –
interval to get a confidence interval. )
21. Do UF students from out of state spend significantly less time at their parents’ house
(homes) than those from in state? A survey found that out-of-state students spend an
average of 3.22 days per semester (std. dev. = 2.49) at their parents’ house (homes)
while in-state students spend an average of 5.89 days per semester (std. dev.=4.23) at
their parents’ house (homes).
Out of state
Instate
N
9
9
Mean
3.22
5.89
StDev
2.49
4.23
SE Mean
0.83
1.4
Difference = mu (Out of state) - mu (Instate)
Estimate for difference: -2.66667
T-Test of difference = 0 (vs not =): T-Value = -1.63
P-Value = 0.129
DF = 12
T-Test of difference = 0 (vs <): T-Value = -1.63
P-Value = 0.064
DF = 12
T-Test of difference = 0 (vs >): T-Value = -1.63
P-Value = 0.936
DF = 12
a) What type of test is this: two independent proportions, two independent means or
paired means?
Two independent means
b) What is the null hypothesis? ____ Ho: μ1-μ2=0___________
c) What is the alternative hypothesis? ___ Ha: μ1-μ2<0__________
d) Look at the output below. What is the corresponding p-value for this hypothesis?
__0.064_________
e) Write a conclusion.
With a p-value equal to 0.064, we have some evidence that the population mean time
spent at the parent’s house is less for out of state students than in state students.
(This can be done on the calculator using the 2 sample t test.)
Short Answer
22. The cost of hiring an employee (excluding salary) can range from about $1,500 for a
secretary to more than $40,000 for a manager. To estimate its mean cost of hiring an
entry-level secretary, a large corporation randomly selected eight of the entry-level
secretaries it had hired during the last two years and determined the costs (in dollar)
involved in hiring each. The following data were obtained:
2,100 1,650 2250
2,035 2,245 1,980 1,700 2,190
a) State the assumptions for a confidence interval for the population mean and discuss if
the assumptions are met.
No outliers, so it could be
normal
1600
1700
1800
1900
2000
2100
2200
2300
SRS – yes because it says “Randomly Selected.
b) Make a 95% CI for µ and interpret this interval.
x = 2018.8
⁄
√
s = 233.0
n=8
df = 7
(1823.98, 2213.6)
We are 95% confident that the population mean cost of hiring an entry level
secretary is between 1823.98 and 2213.6 dollars.
c) Compute a 90% confidence interval for the population mean.
x = 2018.8
⁄
√
s = 233.0
n=8
df = 7
( 1862.7, 2174.9)
t = 1.895
We are 90% confident that the population mean cost of hiring an entry level
secretary is between 1862.70 and 2174.90 dollars.
d) Which interval is wider and why?
95% is a wider because a higher confidence requires a wider interval
23. A federal bank examiner is interested in estimating the mean outstanding principle
balance of all home mortgages foreclosed by the bank due to default by the borrower
during the last 3 years. A SRS of 12 foreclosed mortgages yielded the following data ( in
dollars):
95,982
81,422
39,888
46,836
66,899
62,331
105,812
55,545
56,635
72,123
69,110
a) State the assumptions for a confidence interval for the population mean price per
bushel of corn that October and discuss if the assumptions are met.. No outliers, so ok.
SRS –met because it states “SRS”.
30,000 40,000 50,000 60,000 70,000 80,000 90,000 100,000 110,000
b) Construct a 95% CI for μ. Interpret the interval.
x = 67,648.58
s = 19,171.28
⁄
√
n = 12
df = 11
(55,467.6, 79,829.5)
t = 2.201
We are 95% confident that the population mean outstanding
principal balance of all foreclosed mortgages is between 55,467.6 and 79829.5
dollars.
c) Construct a 99% confidence interval.
x = 67,648.58
s = 19,171.28
n = 12
df = 11
t = 3.106
⁄
√
(50459.13, 84838.03)
We are 99% confident that the population mean outstanding principal balance of all
foreclosed mortgages is between 50459.13 and 84838.03 dollars.
d) Which interval is wider and why?
99% wider
higher confidence intervals requires a wider interval
59,200
24. We want to know μ, the true average temperature in Florida on a certain day. We
take a SRS of 10 cities and get the following:
75 78 84 86 80 92 85 84 90 88
i) What are the assumptions that need to be met for constructing a confidence interval?

SRS – yes, its says it

Data comes from normal or n ≥ 30
n = 10, so not greater than 30
70
75
80
85
90
not outliers, so yes, could be normal
ii) Can we construct a 99% confidence interval for μ?
Yes, because it is a SRS and we have no outliers.
iii) If so, construct one. If not, explain why not.
x = 84.2
s = 5.31
n = 10
df = 9
t = 3.250
⁄
√
(78.7, 89.7)
We are 95% confident that the population mean average temperature in Florida is
between 78.7 and 89.7 degrees.
.
25. A survey asks 231 college students how many sexual partners they have had in their
lives. The sample mean was 4.641 and the sample standard deviation equals 6.33.
i) Is the parameter being estimated the population proportion or the population mean?
population mean
ii) Are the assumptions for a confidence interval for the population parameter met?

SRS – don’t know

Data comes from normal or n ≥ 30
231 ≥ 30
iii) Find the 95% confidence interval for the parameter.
x = 4.641
s =6.33
n = 231
⁄
√
t =1.96
(3.82, 5.46)
iv) Interpret the interval.
We are 95% confident that the population mean number of sexual partners is
between 3.82 and 5.46.
26. If you increase the sample size of a confidence interval, what happens to the width of
a confidence interval? ______it decreases________
27. A survey was conducted to University of Florida students after they returned from
Spring Break. Thirty-five students were asked how much money they spent during Spring
Break. A 95% confidence interval was created from the results. The interval was from
150 to 200 dollars.
For each of the following statements about the above confidence interval, determine if
they are correct or incorrect. If they are incorrect, state why.
1.) The 95% confidence interval for the average amount of money spent by all UF
students over Spring Break is between 150 and 200 dollars.
True
(same as population mean)
2.) 95% of all students spend between 150 and 200 dollars during Spring Break.
False –not about population mean
28. A 95% confidence interval for the population mean number of televisions per
American household is (1.15, 4.20). For each of the following statements about the
above confidence interval, determine if they are correct or incorrect. If they are incorrect,
state why.
a. The probability that  is between 1.15 and 4.20 is .95.
False- can’t use the term probability with a specific confidence interval
b. We are 95% confident that the true mean number of televisions per American
household is between 1.15 and 4.20.
True
c. 95% of all samples should have x-bars between 1.15 and 4.20.
False – not about population mean
d. 95% of all American households have between 1.15 and 4.20 televisions.
False – not about population mean
e. Of 100 intervals calculated the same way (95%), we expect 95 of them to capture
the population mean.
True
f. Of 100 intervals calculated the same way (95%), we expect 95 of them to capture
the sample mean.
False – all confidence intervals contain sample mean
29. The Harvard School of Public Health College Alcohol Study Survey surveys college
students in about 200 colleges in 1993, 1997, and 1999. They asked the students
demographic questions as well as questions about their drinking habits. They were
especially interested in the binging habits of college students. The survey defines an drink
as “12 oz bottle or can of beer, a 4 oz (120 mL ) glass of wine, a 12 oz. (360mL) bottle or
can of wine cooler, or a shot (1.25 oz or 37 mL) of liquor straight or in a mixed drink.”
Binge Drinking is considered drinking 5 drinks in a row for males and 4 drinks in a row
for females. This information is from the 2001 study in which a SRS of college students
was taken. Suppose that we want to find the population proportion of American male
college students that are binge drinkers. For 3925 males, 1908 were binge drinkers and
2017 were not binge drinkers.
a.) Are the assumptions met for the confidence interval for the population proportion of
male college students that binge drink?
̂
SRS – yes because it says SRS.
n ̂ ≥ 15
n(1- ̂ ) ≥ 15
n = 3925
̂ = 0.486
3925(0.486) = 1907.55
3925(1-0.486) = 2017.45
All assumptions are met
b.) Make a 95% CI for p, the proportion of male college students that binge drink and
interpret.
̂
√
̂
̂
√
(0.4704, 0.5016)
We are 95% confident that the population proportion of male college students that
binge drink is between 0.4704 and 0.5016.
Suppose that we want to determine the proportion of female American college students
that are binge drinkers. The same survey asked 6979 females. Out of 6979 females,
2854 were binge drinkers and 4125 were not binge drinkers.
c.) Are the assumptions met for the confidence interval for the population proportion of
female college students that binge drink?
̂
SRS – yes because it says SRS
n = 6979
n ̂ ≥ 15
̂ = 0.409
n(1- ̂ ) ≥ 15
All assumptions are met
6979(0.409) = 2854.41
6979(1-0.409) = 4124.5
d.) Make a 99% CI for p, the proportion of female college students that binge drink and
interpret.
√
(0.394, 0.424)
We are 99% confident that the population proportion of female college students that
binge drink is between 0.394 and 0.424.
30. A SRS of high school students was surveyed. Each student was asked about their
biological parents smoking habits and their own smoking habits. Suppose that we want to
estimate the population proportion of students that smoke given that at least one of the
parent’s smokes. Out of 4024 students, 816 students smoked and 3208 students did not
smoke.
a.) Are the assumptions met?
SRS – yes
n = 4024
n ̂ ≥ 15
̂ = 0.2027
n(1- ̂ ) ≥ 15
n = 816 + 3208 = 4024
4024(0.2027) = 815.66
4024(1-0.2027) = 3208.34
𝑝̂
All assumptions are met
b.) Make a 99% CI for p, the proportion of students that smoke given that at least one
of the parents smokes.
̂
√
̂
̂
√
(0.1862, 0.2191)
We are 99% confident that the population proportion of students that smoke given
that at least one parent smokes is between 0.1862 and 0.2191.
31. You decide to survey people and ask them if they intended to spend or save their tax
refund. You want to be 99% confident in your answer and you want to have a margin of
error of 0.02. What size sample do you need if . . . .
a.) You have no clue what proportion of people will spend their tax refund
̂
̂
b.) You think that the proportion will be close to 0.80.
̂
̂
32. You decide to survey college students and ask them how much they spend on
entertainment per month. You want to be 95% confident in the answer and you want to
be within 2 dollars of the population mean. What size sample do you need if . . .
a.) All you know is that the amount spent on entertainment is typically between 0
dollars and 50 dollars a month.
Estimate the standard deviation by: Range/6
Range = 50-0 = 50
S is estimated with 50/6=8.33
b.) Last year, the standard deviation was 10 dollars.