Practice Exam 4 - University of Floridammeece/Spring2023/Spring2012/practiceexam3... · Practice Exam 3 Answers 1. A simple random sample of commuting UF students was taken to determine

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

15.8x

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?


Means



n> 30 or X is Normal n = 150 Met.


Ho: μ = 2

Ha: μ < 2


Entered into the calculator. (t test)

μ0 = 2

n =150

s= 0.4

1.8x



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


Means



n> 30 or X is Normal n = 6 make a plot to check for outliers


Ho: μ = 12

Ha: μ ≠ 12


Entered into the calculator. (t test)

μ0 = 12

n =6

s= 0.288

11.15x



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.


proportion


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.


Ho: p = 0.19

Ha: p > 0.19


Values entered into Calculator. (t test)

Po= 0.19

n = 200

x = 60

Test Statistic equals 3.96


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


Proportion


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.


Ho: p = 0.83

Ha: p ≠ 0.83


Values entered into Calculator. (1 Prop Z Test)

Po= 0.83

n = 1354

x = 623

Test Statistic equals -36.24


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%?


Proportion


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.


Ho: p = 0.25

Ha: p < 0.25



Po= 0.25

n = 1427

x = 268



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?


Population proportion


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.


Ho: p = 0.5

Ha: p ≠ 0.5


Values entered into Calculator. (1 Prop Z test)

Po= 0.5

n = 1594

x = 855

Test Statistic equals 2.91


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?


Proportion


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.


Ho: p = 0.1

Ha: p < 0.1



Po= 0.1

n =200

x = 8




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?


proportion


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.


Ho: p = 0.9

Ha: p ≠ 0.9


Values entered into Calculator. (1 prop Z test)

Po= 0.9

n =600

x =525



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 88 78 95 96 100 67 90

Old Method 65 86 92 95 84 77 96 83 75 82

N Mean StDev SE Mean

new 7 87.7 11.6 4.4

old 10 83.50 9.58 3.0

Difference = mu (new) - mu (old)

Estimate for difference: 4.21429

95% CI for difference: (-7.48721, 15.91578)

Da

ta

oldnew

100

90

80

70

60

Boxplot of 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 1 2 3 4 5 6 7 8 9 10

Before 60 73 42 88 66 77 90 63 55 96

After 70 80 40 94 79 86 93 71 70 97

Paired T-Test and CI: before, after Paired T for before - after


before 10 71.00 17.07 5.40

after 10 78.00 16.77 5.30

Difference 10 -7.00 5.25 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


means from dependent samples?

Means from dependent samples


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 No-Don’t Legalize Total

Ages 16-24 503 1021 1524

Age >25 525 1263 1788

Does it appear that the younger group feels differently from the older group about the

legalization of marijuana? Sample X N Sample p

1 503 1524 0.330052

2 525 1788 0.293624

Difference = p (1) - p (2)


95% CI for difference: (0.00475862, 0.0680980)



Two independent proportions


SRS – yes says randomly selected.

At least 15 successes and 15 failures in both groups.

In group 1: 503 successes and 1021 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”.

A great deal Not at all Total

Males (group 1) 121 30 151

Females (group 2) 213 17 230

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 X N Sample p

1 121 151 0.801325

2 213 230 0.926087


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 are the assumptions of the confidence interval? Are they satisfied?

SRS – yes, says GSS

At least 15 successes and 15 failures in both groups.



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 15 22 25 18 24 14 19 27

After 18 23 23 22 25 18 17 27


Before 8 20.5000 4.7509 1.6797

After 8 21.6250 3.6228 1.2809

Difference 8 -1.12500 2.41646 0.85435

95% CI for mean difference: (-3.14521, 0.89521)

Differences

Fre

qu

en

cy

210-1-2-3-4

2.0

1.5

1.0

0.5

0.0X_

Ho

Histogram of Differences(with Ho and 95% t-confidence interval for the mean)

a) What type of test is this: two independent proportions, two independent means or


Paired Means


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 9 3.22 2.49 0.83

Instate 9 5.89 4.23 1.4

Difference = mu (Out of state) - mu (Instate)


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

Da

ta

InstateOut of state

14

12

10

8

6

4

2

0

Boxplot of Out of state, Instate




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

Old Method (2) 65 86 92 95 84 77 96 83 75 82


new 7 87.7 11.6 4.4

old 10 83.50 9.58 3.0

Difference = mu (new) - mu (old)


T-Test of difference = 0 (vs <): T-Value = 0.79 P-Value = 0.778 DF = 11

T-Test of difference = 0 (vs not =): T-Value = 0.79 P-Value = 0.445 DF = 11

T-Test of difference = 0 (vs >): T-Value = 0.79 P-Value = 0.222 DF = 11




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

Ages 16-24 503 1021 1524

Age >25 525 1263 1788 Sample X N Sample p

1 503 1524 0.330052

2 525 1788 0.293624



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




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_________


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 1 2 3 4 5 6 7 8 9 10

Before 60 73 42 88 66 77 90 63 55 96

After 70 80 40 94 79 86 93 71 70 97

Paired T-Test and CI: before, after Paired T for before - after


before 10 71.00 17.07 5.40

after 10 78.00 16.77 5.30

Difference 10 -7.00 5.25 1.66

95% CI for mean difference: (-10.76, -3.24)


T-Test of mean difference = 0 (vs < 0): T-Value = -4.22 P-Value = 0.001

T-Test of mean difference = 0 (vs > 0): T-Value = -4.22 P-Value = 0.999



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”.

A great deal Not at all Total

Males (group 1) 121 30 151

Females (group 2) 213 17 230

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 X N Sample p

1 121 151 0.801325

2 213 230 0.926087



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 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.


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:

Before 15 22 25 18 24 14 19 27

After 18 23 23 22 25 18 17 27


Before 8 20.5000 4.7509 1.6797

After 8 21.6250 3.6228 1.2809

Difference 8 -1.12500 2.41646 0.85435


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



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_________


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 9 3.22 2.49 0.83

Instate 9 5.89 4.23 1.4

Difference = mu (Out of state) - mu (Instate)


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?


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 69,110 59,200

62,331 105,812 55,545 56,635 72,123

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

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 70 75 80 85 90

n = 10, so not greater than 30 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 = 3925 n ≥ 15 3925(0.486) = 1907.55

= 0.486 n(1- ) ≥ 15 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 6979(0.409) = 2854.41

= 0.409 n(1- ) ≥ 15 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 4024(0.2027) = 815.66

= 0.2027 n(1- ) ≥ 15 4024(1-0.2027) = 3208.34


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.

��

n = 816 + 3208 = 4024

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.

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Practice Exam 4 - University of Floridammeece/Spring2023/Spring2012/practiceexam3... · Practice Exam 3 Answers 1. A simple random sample of commuting UF students was taken to determine