Test Bank Chapter 6 Inference For Means And Proportions

Statistics - Unlocking the Power of Data, 3e (Lock)

Chapter 6 Inference for Means and Proportions

6.1 Inference for a Proportion

Use the following to answer the questions below:

Consider taking samples of size 100 from a population with proportion 0.33.

1) Find the mean of the distribution of sample proportions.

A) 0.0033

B) 0.033

C) 0.33

D) 33

Diff: 2 Type: BI Var: 1

L.O.: 6.1.1

2) Find the standard error of the distribution of sample proportions.

A) 0.002211

B) 0.0033

C) 0.047

D) 0.33

Diff: 2 Type: BI Var: 1

L.O.: 6.1.1

3) Is the sample size large enough for the Central Limit Theorem to apply so that the sample proportions follow a normal distribution?

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 6.1.2

Use the following to answer the questions below:

Consider taking samples of size 25 from a population with proportion 0.65.

4) Find the mean of the distribution of sample proportions.

A) 0.026

B) 0.65

C) 0.13

D) 16.25

Diff: 2 Type: BI Var: 1

L.O.: 6.1.1

5) Find the standard error of the distribution of sample proportions.

A) 0.0954

B) 0.0091

C) 0.0455

D) 0.0191

Diff: 2 Type: BI Var: 1

L.O.: 6.1.1

6) Is the sample size large enough for the Central Limit Theorem to apply so that the sample proportions follow a normal distribution?

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 6.1.2

Use the following to answer the questions below:

Suppose that the makers of M&M's claim that 24% of their Milk Chocolate M&M's are blue.

7) Assume that Fun-Size bags of Milk Chocolate M&M's hold 20 candies. Find the standard error of the distribution of sample proportions of blue candies for Fun-Size bags (i.e., samples of size 20). Use four decimal places when reporting the standard error.

Diff: 2 Type: SA Var: 1

L.O.: 6.1.1

8) Assume that the bags of Milk Chocolate M&M's sold in vending machines have 55 candies. Find the standard error of the distribution of sample proportions of blue candies for vending machine bags (i.e., samples of size 55). Use four decimal places when reporting the standard error.

Diff: 2 Type: SA Var: 1

L.O.: 6.1.1

9) Assume that bags of Milk Chocolate M&M's labeled as "Medium" size contain 415 candies. Find the standard error of the distribution of sample proportions of blue candies for Medium bags (i.e., samples of size 415). Use four decimal places when reporting the standard error.

Diff: 2 Type: SA Var: 1

L.O.: 6.1.1

10) Would you expect using bags of Milk Chocolate M&M's labeled as "Large" size, which contain more candies than the "Medium" size bags, to result in a larger or smaller standard error?

A) Larger

B) Smaller

Diff: 2 Type: BI Var: 1

L.O.: 6.1.1

11) For which sample sizes (Fun-Size with 20, Vending Machine with 55, or Medium with 415) would the Central Limit Theorem apply?

A) Vending Machine and Medium size bags

B) Fun-Size bags

C) Medium size bags

D) Fun-size and Vending Machine size bags

Diff: 2 Type: BI Var: 1

L.O.: 6.1.2

12) Suppose you purchase a bag of Milk Chocolate M&M's from a vending machine and only 8 of your 55 candies are blue. Assuming that the sample proportions are normally distributed, what percent of vending machine bags (i.e., samples of size 55) will have a sample proportion smaller than 0.145? Use two decimal places when reporting your answer.

Diff: 2 Type: SA Var: 1

L.O.: 5.1.0;6.1.0

Use the following to answer the questions below:

Admissions records at a small university indicates that 6.7% of the students enrolled are international students.

13) Find the mean and standard error of the sample proportion of international students in random samples of size 50. Use four decimal places when reporting the standard error.

Diff: 2 Type: SA Var: 1

L.O.: 6.1.1

14) Find the mean and standard error of the sample proportion of international students in random samples of size 100. Use four decimal places when reporting the standard error.

Diff: 2 Type: SA Var: 1

L.O.: 6.1.1

15) Find the mean and standard error of the sample proportion of international students in random samples of size 200. Use four decimal places when reporting the standard error.

Diff: 2 Type: SA Var: 1

L.O.: 6.1.1

16) For which sample sizes (n = 50, n = 100, and n = 200) would the Central Limit Theorem apply?

A) n = 200

B) n = 50 and n = 100

C) n = 100 and n = 200

D) all three sample sizes

Diff: 2 Type: BI Var: 1

L.O.: 6.1.2

17) What proportion of samples of 200 randomly selected students will have at least 8% international students? Use three decimal places when reporting your answer.

Diff: 2 Type: SA Var: 1

L.O.: 5.1.0;6.1.0

Use the following to answer the questions below:

A study to investigate the dominant paws in cats was described in the scientific journal Animal Behaviour. The researchers used a random sample of 42 domestic cats. In this study, each cat was shown a treat (5 grams of tuna), and while the cat watched, the food was placed inside a jar. The opening of the jar was small enough that the cat could not stick its head inside to remove the treat. The researcher recorded the paw that was first used by the cat to try to retrieve the treat. This was repeated 100 times for each cat (over a span of several days). The paw used most often was deemed the dominant paw (note that one cat used both paws equally and was classified as "ambidextrous"). Of the 42 cats studied, 20 were classified as "left-pawed."

18) Verify that the sample is large enough to use the normal formula to find a confidence interval for the proportion of domestic cats that are "left-pawed."

Diff: 2 Type: SA Var: 1

L.O.: 6.2.0

19) Construct a 95% confidence interval for the proportion of domestic cats that are "left-pawed." Use three decimal places in your margin of error.

Diff: 2 Type: SA Var: 1

L.O.: 6.2.1

20) Construct a 95% confidence interval for the proportion of domestic cats that are "left-pawed." Provide an interpretation of your interval in the context of this data situation.

Diff: 2 Type: ES Var: 1

L.O.: 3.2.4;6.2.0

21) Another researcher wants to conduct a similar study to more precisely estimate the proportion of cats that are "left-pawed." They want to construct a 95% confidence interval that has a margin of error of 6%. How many cats does she need to use in her sample?

A) 267 cats

B) 266 cats

C) 268 cats

D) 269 cats

Diff: 2 Type: BI Var: 1

L.O.: 6.2.2

Use the following to answer questions 18-21:

In a survey of 7,786 randomly selected adults living in Germany, 5,840 said they exercised for at least 30 minutes three or more times per week.

22) Verify that the sample is large enough to use the normal formula to find a confidence interval for the proportion of Germans who exercises for 30 minutes three or more times a week.

= 5,840/7,786 = 0.75

n = 5,840 > 10, n (1 - ) = 1,946 > 10 <— Yes, we can use the normal formula.

Diff: 2 Type: ES Var: 1

L.O.: 6.2.0

23) Construct a 99% confidence interval for the proportion of Germans who exercise for 30 minutes three or more times a week. Use three decimal places in your margin of error.

A) 0.737 to 0.763

B) 0.733 to 0.767

C) 0.722 to 0.778

D) 0.749 to 0.751

Diff: 2 Type: BI Var: 1

L.O.: 6.2.1

24) Construct a 99% confidence interval for the proportion of Germans who exercise for 30 minutes three or more times a week. Provide an interpretation of your interval in the context of this data situation.

Diff: 2 Type: ES Var: 1

L.O.: 3.2.4;6.2.0

25) Suppose an exercise scientist wants to estimate the proportion of American adults who exercise for 30 minutes three or more times per week. He wants to construct a 90% confidence interval with a margin of error of 1%. Note that Americans are typically thought to not be as active as individuals in other countries, and thus the estimate from Germany is likely not a good estimate for Americans. What sample size does he need?

A) 6,766 people

B) 13,532 people

C) 27,060 people

D) 41 people

Diff: 2 Type: BI Var: 1

L.O.: 6.2.2

Use the following to answer the questions below:

In a Gallup survey of 1,012 randomly selected U.S. adults (age 18 and over), 53% said that they were dissatisfied with the quality of education students receive in kindergarten through grade 12.

26) Verify that the sample is large enough to use the normal formula to find a confidence interval for the proportion of Americans who are dissatisfied with the quality of education students receive in kindergarten through grade 12.

= 0.53

n = 536.36 > 10, n(1 - ) = 475.64 > 10 <— Yes, we can use the normal formula.

Diff: 2 Type: ES Var: 1

L.O.: 6.2.0

27) Construct a 90% confidence interval for the proportion of U.S. adults who are dissatisfied with the quality of education students receive in kindergarten through grade 12. Use three decimal places in your margin of error.

A) 0.504 to 0.556

B) 0.509 to 0.551

C) 0.512 to 0.548

D) 0.497 to 0.563

Diff: 2 Type: BI Var: 1

L.O.: 6.2.1

28) Construct a 90% confidence interval for the proportion of U.S. adults who are dissatisfied with the quality of education students receive in kindergarten through grade 12. Provide an interpretation of your interval in the context of this data situation.

Diff: 2 Type: ES Var: 1

L.O.: 2.3.4;6.2.0

29) Suppose you want to estimate the proportion of local adults who are dissatisfied with the education students receive in kindergarten through grade 12 with 95% confidence and a 5% margin of error. If you suspect that local adults won't differ drastically from those Gallup used, how many people should you sample?

A) You should sample 383 local adults.

B) You should sample 369 local adults.

C) You should sample 343 local adults.

D) You should sample 358 local adults.

Diff: 2 Type: BI Var: 1

L.O.: 6.2.2

30) Test, at the 5% level, if this sample provides evidence that the proportion of Americans who are dissatisfied with education in kindergarten through grade 12 differs significantly from 50%. Be sure to verify that it is appropriate to use a normal distribution to compute the p-value and include all of the details of the test.

: p = 0.50

: p ≠ 0.50

n = n (1 - ) = 1,012*0.5 = 506 > 10

Since both are larger than 10, the sample size is large enough to use the normal distribution to compute the p-value.

Test statistic: z = = 1.909

p-value = 0.056 (two-tail probability, using Statkey)

Since the p-value is larger than the 5% significance level, there is no evidence to reject and thus there is no evidence to conclude that proportion of U.S. adults who are dissatisfied with education in kindergarten through grade 12 differs significantly from 50%.

Diff: 2 Type: ES Var: 1

L.O.: 6.3.1

Use the following to answer the questions below:

In a recent study, the Centers for Disease Control and Prevention reported that in a sample of 4,349 African Americans 31% were Vitamin D deficient. Overall, it is believed that Vitamin D deficiency affects 8% of all U.S. adults.

31) Verify that the sample size is large enough to use a normal distribution to conduct a test comparing the population proportion of African Americans with Vitamin D deficiency to the overall rate of 8%.

n = 4,349*0.08 = 347.92 > 10

n(1 - ) = 4,349*0.92 = 4,001.08 > 10

Since both exceed 10, the sample is large enough to perform a test to compare the population proportion to the overall rate of 8%.

Diff: 2 Type: ES Var: 1

L.O.: 6.3.0

32) Test, at the 1% significance level, if this sample provides evidence that the rate of Vitamin D deficiency among African Americans differs significantly from the overall rate of 8%. Include all of the details of the test.

: p = 0.08

: p ≠ 0.08

Test statistic: z = = 55.9

n = 4,349*0.08 = 347.92 > 10

n(1 - ) = 4,349*0.92 = 4,001.08 > 10

Since both exceed 10, the sample is large enough to use the normal distribution to compute the p-value.

p-value ≈ 0 (two-tail probability using Statkey)

Because the p-value is drastically smaller than the 5% significance, we have very strong evidence to reject and thus have very strong evidence to conclude that the rate of Vitamin D deficiency among African Americans differs significantly from the overall rate of 8% for American adults.

Diff: 2 Type: ES Var: 1

L.O.: 6.3.1

33) Verify that the sample size is large enough to use the normal distribution to construct a confidence interval for the proportion of African Americans with Vitamin D deficiency.

= 0.31

n = 4,349*0.31 = 1,348.19 > 10

n(1 - ) = 4,349*0.69 = 3,000.81 > 10

Since both exceed 10, the sample size is large enough to use the normal distribution.

Diff: 2 Type: ES Var: 1

L.O.: 6.2.0

34) Construct a 99% confidence interval for the proportion of African Americans with Vitamin D deficiency. Use three decimal places in your margin of error.

A) 0.292 to 0.328

B) 0.298 to 0.328

C) 0.296 to 0.324

D) 0.282 to 0.338

Diff: 2 Type: BI Var: 1

L.O.: 6.2.1

Use the following to answer the questions below:

The owner of a small pet supply store wants to open a second store in another city, but he only wants to do so if more than one-third of the city's households have pets (otherwise there won't be enough business). He selects a random sample of 150 households and finds that 64 have pets.

35) Verify that the sample size is large enough to perform a test to compare the population proportion of households in the city with pets to the target.

n = 150(1/3) = 50 > 10

n(1 - ) = 150(2/3) = 100 > 10

Since both exceed 10, the sample size is large enough to use a normal distribution to perform this test about the population proportion.

Diff: 2 Type: ES Var: 1

L.O.: 6.3.0

36) Test, at the 5% level, if this sample provides evidence that significantly more than one-third of the city's households have a pet. Include all of the details of the test.

: p = 1/3

: p > 1/3

= = 0.427

Test statistic: z = = 2.433

n = 150(1/3) = 50 > 10

n(1 - ) = 150(2/3) = 100 > 10

Since both exceed 10, the sample size is large enough to use a normal distribution to perform this test about the population proportion.

p-value = 0.0075

Because the p-value is smaller than the 5% significance level, we have strong evidence to reject and thus have strong evidence to conclude that the proportion of the city's households that have pets is significantly higher than one-third.

Diff: 2 Type: ES Var: 1

L.O.: 6.3.1

37) Verify that the sample size is large enough to use the normal distribution to construct a confidence interval for the proportion of the city's households that own pets.

= = 0.427

n = 64 >10

n(1 - ) = 86 > 10

Since there are more than 10 successes and failures, the sample size is large enough to use a normal distribution.

Diff: 2 Type: ES Var: 1

L.O.: 6.2.0

38) Construct a 95% confidence interval for the proportion of the city's households that own pets. Round the sample proportion and margin of error to three decimal places.

A) 0.348 to 0.506

B) 0.361 to 0.493

C) 0.323 to 0.531

D) 0.331 to 0.523

Diff: 2 Type: BI Var: 1

L.O.: 6.2.1

Use the following to answer the questions below:

In May 2012 President Obama made history by revealing his support of gay marriage. Around that time the Gallup Organization polled 1,024 U.S. adults about their opinions on gay/lesbian relations and gay marriage. They found that 54% of those sampled viewed gay/lesbian relations as "morally acceptable."

39) Verify that the sample size is large enough to use the normal distribution to construct a confidence interval for the proportion of U.S. adults who consider gay and lesbian relations to be "morally acceptable."

= 0.54

n = 1,024*0.54 = 552.96 > 10

n(1 - ) = 1,024*0.46 = 471.04 > 10

Since there are more than 10 successes and failures in each group, the sample size is large enough to use the normal distribution.

Diff: 2 Type: ES Var: 1

L.O.: 6.2.0

40) Construct a 90% confidence interval for the proportion of U.S. adults who find gay/lesbian relations to be "morally acceptable." Round the margin of error to three decimal places.

A) 0.514 to 0.566

B) 0.509 to 0.571

C) 0.500 to 0.580

D) 0.495 to 0.585

Diff: 2 Type: BI Var: 1

L.O.: 6.2.1

41) What sample size would we need to reduce the margin of error to ± 1.5%?

A) 2,988 U.S. adults

B) 2,905 U.S. adults

C) 2,887 U.S. adults

D) 2,876 U.S. adults

Diff: 2 Type: BI Var: 1

L.O.: 6.2.2

42) Does this sample provide evidence that the majority of U.S. adults (i.e., more than half) believe that gay/lesbian relations are "morally acceptable"? Use a 5% significance level. Verify that the sample size is large enough to use the normal distribution to compute the p-value for this test and include all of the details of the test.

p = proportion of U.S. adults who believe that gay/lesbian relations are morally acceptable

: p = 0.5

: p > 0.5

n = n(1 - ) = 1,024*0.5 = 512

Because both exceed 10, the sample size is large enough to use the normal distribution to compute the p-value for this test.

Test statistic: z = = 2.56

p-value = 0.0052 (right-tail, using Statkey)

Because the p-value is considerably smaller than the 5% significance level, we have very strong evidence to reject and thus have very strong evidence to conclude that the majority (i.e., more than half) of U.S. adults believe that gay/lesbian relations are "morally acceptable."

Diff: 2 Type: ES Var: 1

L.O.: 6.3.1

Use the following to answer the questions below:

In a survey conducted by the Gallup organization, 1,017 adults were asked "In general, how much trust and confidence do you have in the mass media — such as newspapers, TV, and radio — when it comes to reporting the news fully, accurately, and fairly?" Of the 1,017 respondents, 214 said they had "no confidence at all."

43) Test, at the 5% level, if this sample provides evidence that the proportion of U.S. adults who have no confidence in the media differs significantly from 25%. Verify that the sample size is large enough to use the normal distribution to compute the p-value for this test and include all of the details of the test.

p = proportion of U.S. adults who have no confidence in the media

: p = 0.25

: p ≠ 0.25

n = 1,017*0.25 = 254.25 > 10

n(1 - ) = 1,017*0.75 = 762.75 > 10

Since both are greater than 10, the sample size is large enough to use the normal distribution to compute the p-value for this test.

Test statistic: z = = -2.95

p-value = 0.003 (two-sided probability, using Statkey)

Because the p-value is smaller than the 5% significance level, we have evidence to reject and thus have evidence to conclude that the proportion of U.S. adults who have no confidence in the media differs significantly from 25%.

Diff: 2 Type: ES Var: 1

L.O.: 6.3.1

44) Verify that the sample size is large enough to use the normal distribution to construct a confidence interval for the proportion of U.S. adults who have no confidence in the media.

n = 214

n(1 - ) = 803

Since there are more than 10 successes and failures, the sample size is large enough to use the normal distribution to construct a confidence interval.

Diff: 2 Type: ES Var: 1

L.O.: 6.2.0

45) Construct a 90% confidence interval for the proportion of U.S. adults who have no confidence in the media. Round the margin of error to three decimal places.

A) 0.189 to 0.231

B) 0.185 to 0.235

C) 0.177 to 0.243

D) 0.171 to 0.249

Diff: 2 Type: BI Var: 1

L.O.: 6.2.1

46) What sample size is needed to reduce the margin of error to 1%?

A) 4,490 U.S. adults

B) 4,553 U.S. adults

C) 4,567 U.S. adults

D) 4,598 U.S. adults

Diff: 2 Type: BI Var: 1

L.O.: 6.2.2

Use the following information to answer the confidence interval for a population proportion p in the questions below.

47) 0.645 to 0.700

What is the margin of error?

A) 0.055

B) 0.0275

C) 0.01375

D) 0.6725

Diff: 3 Type: BI Var: 1

L.O.: 3.2.0

48) 0.655 to 0.685

What is the best estimate of p?

A) 0.685

B) 0.03

C) 0.655

D) 0.67

Diff: 3 Type: BI Var: 1

L.O.: 3.2.0

6.2 Inference for a Mean

Use the following to answer the questions below:

A sample of 148 college students at a large university reports getting an average of 6.85 hours of sleep last night with a standard deviation of 2.12 hours.

1) Is it reasonable to use the t-distribution to construct a confidence interval for the average amount of sleep students at this university got last night?

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 6.4.2

2) Construct a 98% confidence interval for the average amount of sleep students at this university got last night. Use two decimal places in your margin of error.

A) 6.44 to 7.26 hours

B) 6.49 to 7.21 hours

C) 6.53 to 7.17 hours

D) 6.38 to 7.32 hours

Diff: 2 Type: BI Var: 1

L.O.: 6.4.3;6.5.1

3) Construct a 98% confidence interval for the average amount of sleep students at this university got last night. Provide an interpretation of your interval in the context of this data situation.

Diff: 2 Type: ES Var: 1

L.O.: 3.2.4;6.5.0

4) Suppose you want to conduct a similar study at your university. Assuming that the standard deviation of this sample is a reasonable estimate of the standard deviation of sleep time at your university, how many students do you need to survey to estimate the mean sleep time of students at your university with 95% confidence and a margin of error of 0.5 hours?

A) 70 students

B) 69 students

C) 88 students

D) 89 students

Diff: 2 Type: BI Var: 1

L.O.: 6.5.2

Use the following to answer the questions below:

An Internet provider contacts a random sample of 300 customers and asks how many hours per week the customers use the Internet. The responses are summarized in the provided dotplot. The average amount of time spent on the Internet per week was 7.2 hours, with a standard deviation of 7.9 hours.

5) Is it reasonable to use the t-distribution to construct a confidence interval for the average amount of time customers of this Internet provider spend on the Internet each week?

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 6.4.2

6) Construct a 95% confidence interval for the average amount of time customers of this Internet provider spend on the Internet each week. Round the margin of error to one decimal place.

A) 6.3 to 8.1 hours

B) 6.0 to 8.4 hours

C) 5.7 to 8.7 hours

D) 5.5 to 8.9 hours

Diff: 2 Type: BI Var: 1

L.O.: 6.4.3;6.5.1

7) Construct a 95% confidence interval for the average amount of time customers of this Internet provider spend on the Internet each week. Provide an interpretation of your interval in the context of this data situation.

Diff: 2 Type: ES Var: 1

L.O.: 3.2.4;6.5.0

8) If we want a margin of error of 0.5 hours, how large of a sample would we need?

A) 960 people

B) 953 people

C) 952 people

D) 944 people

Diff: 2 Type: BI Var: 1

L.O.: 6.5.2

Use the following to answer the questions below:

According to a National Science Foundation study, individuals who graduated with a doctoral degree had an average of $14,115 graduate debt. Assume that the standard deviation of graduate debt is $26,400. If we take lots of samples of individuals who graduated with a doctoral degree, what would you expect the standard error of the distribution of sample mean graduate debt amounts to be in each case? In each case, use two decimal places when reporting your standard error.

9) n = 200 individuals

Diff: 2 Type: SA Var: 1

L.O.: 6.4.1

10) n = 500 individuals

Diff: 2 Type: SA Var: 1

L.O.: 6.4.1

Use the following to answer the questions below:

For each of the following, assume that the sample is a random sample from a distribution that is reasonably normally distributed and that we are doing inference for a population mean.

11) Find endpoints of a t-distribution with 2.5% beyond them in each tail if the sample has size

A) -2.145 and 2.145

B) -1.533 and 1.533

C) -1.918 and 1.918

D) -2.328 and 2.328

Diff: 1 Type: BI Var: 1

L.O.: 6.4.3

12) Find endpoints of a t-distribution with 10% beyond them in each tail if the sample has size

A) -1.533 and 1.533

B) -2.145 and 2.145

C) -1.918 and 1.918

D) -2.328 and 2.328

Diff: 1 Type: BI Var: 1

L.O.: 6.4.3

13) Find endpoints of a t-distribution with 3% beyond them in each tail if the sample has size

A) -1.918 and 1.918

B) -2.145 and 2.145

C) -1.533 and 1.533

D) -2.328 and 2.328

Diff: 2 Type: BI Var: 1

L.O.: 6.4.3

14) Find endpoints of a t-distribution with 1.5% beyond them in each tail if the sample has size 22.

A) -2.328 and 2.328

B) -2.145 and 2.145

C) -1.533 and 1.533

D) -1.918 and 1.918

Diff: 1 Type: BI Var: 1

L.O.: 6.4.3

15) Find the area in a t-distribution to the right of 2.6 if the sample has size

Diff: 1 Type: SA Var: 1

L.O.: 6.4.3

16) Find the area in a t-distribution to the right of 1.75 if the sample has size

Diff: 2 Type: SA Var: 1

L.O.: 6.4.3

17) Find the area in a t-distribution to the left of -2.7 if the sample has size

Diff: 2 Type: SA Var: 1

L.O.: 6.4.3

18) Find the area in a t-distribution to the left of -0.68 if the sample has size

Diff: 2 Type: SA Var: 1

L.O.: 6.4.3

Use the following to answer the questions below:

For each of the following, find the standard error of the distribution of sample means. Use two decimal places when reporting your standard error.

19) Samples of size 15 from a population with mean 25 and standard deviation 4.

Diff: 2 Type: SA Var: 1

L.O.: 6.4.1

20) Samples of size 50 from a population with mean 450 and standard deviation 75.

Diff: 2 Type: SA Var: 1

L.O.: 6.4.1

21) Samples of size 25 from a population with mean 10 and standard deviation 2.

Diff: 2 Type: SA Var: 1

L.O.: 6.4.1

22) Samples of size 250 from a population with mean 80 and standard deviation of 15.

Diff: 2 Type: SA Var: 1

L.O.: 6.4.1

Use the following to answer the questions below:

A dotplot and the summary statistics for a sample are provided. In each case, indicate whether or not it is appropriate to use the t-distribution.

23) n = 12; = 4.75; s = 1.603

A) Appropriate

B) Not Appropriate

Diff: 2 Type: BI Var: 1

L.O.: 6.4.2

24) n = 10; = 7.80; s = 9.28

A) Appropriate

B) Not Appropriate

Diff: 2 Type: BI Var: 1

L.O.: 6.4.2

25) n = 100; = 9.93; s = 9.247

A) Appropriate

B) Not Appropriate

Diff: 2 Type: BI Var: 1

L.O.: 6.4.2

26) n = 15; = 44; s = 7.32

A) Appropriate

B) Not Appropriate

Diff: 2 Type: BI Var: 1

L.O.: 6.4.2

Use the following to answer the questions below:

Many major television networks air coverage of the incoming election results during primetime hours. The provided boxplot displays the amount of time (in minutes) spent watching election coverage for a random sample of 25 U.S. adults. In this sample, the average time spent watching election coverage was 80.44 minutes with standard deviation of 43.99 minutes.

27) Is it reasonable to use the t-distribution to construct a confidence interval for the average amount of time spent watching election coverage by U.S. adults?

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 6.4.2

28) Construct a 90% confidence interval for the average amount of time U.S. adults spent watching election coverage. Use two decimal places in your margin of error.

A) 65.39 to 95.49 hours

B) 66.24 to 94.64 hours

C) 68.14 to 92.74 hours

D) 71.33 to 89.55 hours

Diff: 2 Type: BI Var: 1

L.O.: 6.5.1

29) Construct a 90% confidence interval for the average amount of time U.S. adults spent watching election coverage. Provide an interpretation of your interval in the context of this data situation.

Diff: 2 Type: ES Var: 1

L.O.: 3.2.4;6.5.0

30) What sample size would we need to estimate the average amount of time U.S. adults watching election coverage with 99% confidence and a margin of error of ± 5 hours?

A) 514 U.S. adults

B) 23 U.S. adults

C) 22 U.S. adults

D) 513 U.S. adults

Diff: 2 Type: BI Var: 1

L.O.: 6.5.2

Use the following to answer the questions below:

Turkey is a staple at most traditional Thanksgiving dinners. A random sample of 12 grocery store customers were asked about the size of the turkey they were purchasing for Thanksgiving. The average weight was 13.9 pounds with a standard deviation of 2.2 pounds. The boxplot displays the distribution of the sample turkey weights.

31) Is it reasonable to use the t-distribution to construct a confidence interval for the average weight of turkeys purchased at this store?

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 6.4.2

32) Construct a 99% confidence interval for the average weight of turkeys purchased at this store. Round your margin of error to two decimal places.

A) 11.93 to 15.87 pounds

B) 13.33 to 14.47 pounds

C) 12.57 to 15.23 pounds

D) 10.80 to 17.01 pounds

Diff: 2 Type: BI Var: 1

L.O.: 6.5.1

33) Construct a 99% confidence interval for the average weight of turkeys purchased at this store. Provide an interpretation of your interval in the context of this data situation.

Diff: 2 Type: ES Var: 1

L.O.: 3.2.4;6.5.0

34) What sample size would we need to reduce the margin of error to ±1 pound?

A) 33 turkeys (customers purchasing turkeys).

B) 15 turkeys (customers purchasing turkeys).

C) 13 turkeys (customers purchasing turkeys).

D) 24 turkeys (customers purchasing turkeys).

Diff: 2 Type: BI Var: 1

L.O.: 6.5.2

35) According to the Minnesota Turkey Growers Association's website, the average weight of turkeys purchased for Thanksgiving dinner is 15 pounds. Test, at the 5% level, if this sample provides evidence that the average weight of turkeys purchased at this store differs from 15 pounds. Include all of the details of the test.

: μ = 15

: μ ≠ 15

Test Statistic: t = = -1.732

The sample data look roughly symmetric with no outliers, so we can use the t-distribution with 11 degrees of freedom to compute the p-value.

p-value = 0.111

Because the p-value is larger than the 5% significance level, we have no evidence to reject and thus have no evidence to conclude that the average weight of turkeys purchased at this store differs significantly from the 15 pounds reported by the Minnesota Turkey Growers Association.

Diff: 2 Type: ES Var: 1

L.O.: 6.6.1

Use the following to answer the questions below:

On August 8, 2012, the national average price for a gallon of regular unleaded gasoline was $3.63. The prices for a random sample of gas stations in the state of Illinois were recorded at that time. The mean price for the sampled gas stations was $3.975, with standard deviation $0.2266. A boxplot of the data is provided.

36) Is it reasonable to use the t-distribution to perform a test about the average gas price in Illinois (on August 8, 2012)?

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 6.6.0

37) Test, at the 5% level, if there is evidence that the average gas price in Illinois (on August 8, 2012) was significantly higher than the national average. Include all of the details of the test.

: μ = 3.63

: μ > 3.63

Test statistic: t = = 4.815

n = 10, so df = 9

Even though the sample size is small, we can use the t-distribution because the sample data are symmetric and thus look like they could reasonably be from a normal population.

p-value = 0.00048 (right tail, using Statkey)

Because the p-value is much smaller than the 5% significance level, we have very strong evidence to reject the null hypothesis and thus have very strong evidence to conclude that the average gas price in Illinois (on August 8, 2012) was significantly higher than the national average of $3.63.

Diff: 2 Type: ES Var: 1

L.O.: 6.6.1

38) Construct a 95% confidence interval for the mean gas price in Illinois (on August 8, 2012). Round your margin of error to three decimal places.

A) 3.813 to 4.137

B) 3.822 to 4.128

C) 3.807 to 4.143

D) 3.791 to 4.159

Diff: 2 Type: BI Var: 1

L.O.: 6.5.1

Use the following to answer the questions below:

A certain species of tree has an average life span of 130 years. A researcher has noticed a large number of trees of this species washing up along a beach as driftwood. She takes core samples from 27 of those trees, selected at random, to count the number of rings and measure the widths of the rings. Counting the rings allows the researcher to determine the age of each tree. The mean age of the sampled driftwood is 119 years old, with standard deviation 46.92 years. The sample data are plotted in the provided dotplot. One of her interests is determining if this sample provides evidence that the average age of the driftwood is less than the 130 year life span expected for this type of tree. If the average age is less than 130 years it might suggest that the trees have died from unusual causes, such as invasive beetles or logging.

39) Verify that it is reasonable to use the t-distribution to perform a test about the average age of driftwood along this beach.

Diff: 2 Type: ES Var: 1

L.O.: 6.6.0

40) Test, at the 5% level, if there is evidence that the average age of driftwood along this beach is significantly below 130 years. Include all of the details of the test.

: μ = 130

: μ < 130

Test statistic: t = = -1.218

n = 27, so df = 26

Because the data are roughly symmetric with no outliers, we can use the t-distribution to compute the p-value.

p-value = 0.117

Because the p-value is larger than the 5% significance level, we have no evidence to reject and thus have no evidence to conclude that the average age of driftwood along this beach is significantly less than 130 years.

Diff: 2 Type: ES Var: 1

L.O.: 6.6.1

Use the following to answer the questions below:

A random sample of 48 students at a large university reported getting an average of 7 hours of sleep on weeknights, with standard deviation 1.62 hours. A dotplot of the data is provided.

41) Explain why it is reasonable to use a t-distribution to perform inference about the mean amount of weeknight sleep for students at this university.

Diff: 2 Type: ES Var: 1

L.O.: 6.5.0

42) It is recommended, for most college age students, to get 8 hours of sleep each night. Does this sample provide evidence, at the 5% level, that college students at this university get significantly less sleep, on average, than what is recommended? Include all of the details of the test.

: μ = 8

: μ < 8

Test statistic: t = = -4.28

Because the sample size is so large, we can use the t-distribution with 47 degrees of freedom to compute the p-value.

p-value = 0.000046 (left-tail, using Statkey)

Because the p-value is considerably smaller than the 5% significance level, we have very strong evidence to reject and thus have very strong evidence to conclude that students at this university, on average, get significantly less sleep than the recommended 8 hours.

Diff: 2 Type: ES Var: 1

L.O.: 6.6.1

43) Construct a 95% confidence interval for the average amount of weeknight sleep for students at this university. Round the margin of error to two decimal places.

A) 6.53 to 7.47 hours

B) 6.54 to 7.46 hours

C) 6.40 to 7.60 hours

D) 6.48 to 7.52 hours

Diff: 2 Type: BI Var: 1

L.O.: 6.5.1

6.3 Inference for a Difference in Proportions

Use the following to answer the questions below:

A study published in the American Journal of Health Promotion by researchers at the University of Minnesota (U of M) found that 124 out of 1,923 U of M females had over $6,000 in credit card debt while 61 out of 1,236 males had over $6,000 in credit card debt.

1) Verify that the sample size is large enough in each group to use the normal distribution to construct a confidence interval for a difference in two proportions.

(1 - ) = 1,799 > 10

= 61 > 10

(1 - ) = 1,175 > 10

Since all are greater than 10, the sample size is large enough in each group to use the normal distribution to construct the confidence interval.

Diff: 2 Type: ES Var: 1

L.O.: 6.8.0

2) Construct a 95% confidence interval for the difference between the proportions of female and male University of Minnesota students who have more than $6,000 in credit card debt Round your sample proportions and margin of error to four decimal places.

A) -0.0012 to 0.0314

B) -0.0300 to 0.0365

C) -0.0074 to 0.0376

D) 0.0068 to 0.0234

Diff: 2 Type: BI Var: 1

L.O.: 6.8.1

3) Test, at the 5% level, if there is evidence that the proportion of female students at U of M with more than $6,000 credit card debt is greater than the proportion of males at U of M with more than $6,000 credit card debt. Include all details of the test.

: =

: >

= proportion of female U of M students with more than $6,000 credit card debt

= proportion of male U of M students with more than $6,000 credit card debt

Pooled proportion (for standard error): = = = 0.0586

Test statistic: z = = 1.763

We can use the normal distribution to compute the p-value because both samples have at least 10 successes and failures.

p-value = 0.039 (Right tail probability found using Statkey)

Because the p-value is less than the 5% significance level, we have evidence to reject the null hypothesis and conclude that the proportion of female U of M students with more than $6,000 in credit card debt is significantly higher than the proportion of male U of M students with more than $6,000 in credit card debt.

Diff: 2 Type: ES Var: 1

L.O.: 6.9.1

Use the following to answer the questions below:

Every year since the 1957-58 academic year, the National Science Foundation (NSF) conducts its Survey of Earned Doctorates (SED) of all individuals receiving research doctoral degrees from accredited U.S. institutions. The results from the 2010 survey published on the NSF website indicate that 78.2% of individuals earning their doctorate in the physical sciences have no graduate debt while 48.3% of those earning their doctorate in the social sciences have no graduate debt. Of the 48,069 research doctorates granted in 2010, 93% completed the SED, thus the information collected by the NSF can be good approximations of the population parameters.

4) Suppose we take random samples of 100 individuals who earned a doctorate in the physical sciences (in 2010) and 100 individuals who earned a doctorate in the social sciences (in 2010). Find the mean and standard error (using four decimal places) of the distribution of differences in sample proportions and indicate if the sample sizes are large enough to use the Central Limit Theorem.

mean = 0.782 - 0.483 = 0.299

SE = = 0.0648

= 100*0.782 = 78.2 > 10

(1 - ) = 100*(1 - 0.782) = 21.8 > 10

= 100*0.483 = 48.3 > 10

(1 - ) = 100*(1-0.483) = 51.7 > 10

Since all are greater than 10, the sample sizes are large enough to use the Central Limit Theorem.

Diff: 2 Type: ES Var: 1

L.O.: 6.7.2;6.7.3

5) Suppose we take random samples of 25 individuals who earned a doctorate in the physical sciences (in 2010) and 50 individuals who earned a doctorate in the social sciences (in 2010). Find the mean and standard error (using four decimal places) of the distribution of differences in sample proportions and indicate if the sample sizes are large enough to use the Central Limit Theorem.

SE = = 0.1087

= 25*0.782 = 19.55 > 10

(1 - ) = 25*(1 - 0.782) = 5.45 < 10 X

= 50*0.483 = 24.15 > 10

(1 - ) = 50*(1 - 0.483) = 25.85 > 10

Since (1 - ) is not greater than 10, the sample sizes are NOT large enough to use the Central Limit Theorem.

Diff: 2 Type: ES Var: 1

L.O.: 6.7.2;6.7.3

6) Suppose we take random samples of 50 individuals who earned a doctorate in the physical sciences (in 2010) and 25 individuals who earned a doctorate in the social sciences (in 2010). Find the mean and standard error (using four decimal places) of the distribution of differences in sample proportions and indicate if the sample sizes are large enough to use the Central Limit Theorem.

SE = = 0.1157

= 50*0.782 = 39.1 > 10

(1 - ) =50*(1 - 0.782) = 10.9 > 10

= 25*0.483 = 12.075 > 10

(1 - ) = 25*(1 - 0.483) = 12.925 > 10

Since all are greater than 10, the sample sizes are large enough to use the Central Limit Theorem.

Diff: 2 Type: ES Var: 1

L.O.: 6.7.2;6.7.3

Use the following to answer the questions below:

Situations comparing two proportions are described. In each case, determine whether the situation involves comparing proportions for two groups or comparing two proportions from the same group.

7) Compare the proportion of U.S. adults who have a positive opinion about the media and the proportion of U.S. adults who have a negative opinion about the media.

A) Comparing proportions for two groups

B) Comparing two proportions from the same group

Diff: 2 Type: BI Var: 1

L.O.: 6.7.1

8) Comparing proportion of milk chocolate M&M's that are blue to the proportion of milk chocolate M&M's that are green.

A) Comparing proportions for two groups

B) Comparing two proportions from the same group

Diff: 2 Type: BI Var: 1

L.O.: 6.7.1

9) Comparing the proportion of milk chocolate M&M's that are blue to the proportion of dark chocolate M&M's that are blue.

A) Comparing proportions for two groups

B) Comparing two proportions from the same group

Diff: 2 Type: BI Var: 1

L.O.: 6.7.1

10) Compare the proportion of female students at a university who play a sport to the proportion of male students at a university who play a sport.

A) Comparing proportions for two groups

B) Comparing two proportions from the same group

Diff: 2 Type: BI Var: 1

L.O.: 6.7.1

11) A study to investigate the dominant paws in cats was described in the scientific journal Animal Behaviour. The researchers used a random sample of 42 domestic cats. In this study, each cat was shown a treat (5 grams of tuna), and while the cat watched, the food was placed inside a jar. The opening of the jar was small enough that the cat could not stick its head inside to remove the treat. The researcher recorded the paw that was first used by the cat to try to retrieve the treat. This was repeated 100 times for each cat (over a span of several days). The paw used most often was deemed the dominant paw (note that one cat used both paws equally and was classified as "ambidextrous"). The researchers were also interested in comparing the proportion of "left-pawed" cats for male and female cats. Of the 21 male cats in the sample, 19 were classified as "left-pawed" while only 1 of the 21 female cats were considered to be "left-pawed". Explain why it would not be appropriate to use the normal distribution to construct a confidence interval for the difference in the proportion of male and female cats that are "left-pawed."

= 19 > 10

(1 - ) = 2 < 10 X

= 20 > 10

(1 - ) = 1 < 10 X

Since there are only two "failures" in the sample of male cats and only 1 "failure" in the sample of female cats, the sample sizes are not large enough to use the Central Limit Theorem.

Diff: 2 Type: ES Var: 1

L.O.: 6.8.1

Use the following to answer the questions below:

February 12, 2009 marked the anniversary of Charles Darwin's birth. To celebrate, Gallup, a national polling organization, surveyed 1,018 randomly selected American adults about their education level and their beliefs about the theory of evolution. In their sample, 325 of their respondents had some college education and 228 were college graduates. Among the 325 respondents with some college education, 133 said that they believed in the theory of evolution. Among the 228 respondents who were college graduates, 121 said that they believed in the theory of evolution.

12) Verify that the sample size is large enough in each group to use the normal distribution to construct a confidence interval for a difference in proportions.

= = 0.409 = sample proportion of those with some college education that believe in evolution

= = 0.531 = sample proportion of the college graduates that believe in evolution

= 133 > 10

(1 - ) = 192 > 10

= 121 > 10

(1 - ) = 107 > 10

Since all are greater than 10, the sample sizes are large enough to apply the Central Limit Theorem and use a normal distribution to construct a confidence interval for a difference in proportions.

Diff: 2 Type: ES Var: 1

L.O.: 6.8.0

13) Construct a 90% confidence interval for the difference between the proportions of college graduates and individuals with some college who believe in the theory of evolution. Round your sample proportions and margin of error to three decimal places.

A) 0.052 to 0.192

B) 0.038 to 0.206

C) 0.079 to 0.165

D) 0.012 to 0.232

Diff: 2 Type: BI Var: 1

L.O.: 6.8.1

14) Test, at the 10% level, if there is evidence that the proportion of college graduates that believe in evolution differs significantly from the proportion of individuals with some college education that believe in evolution. Include all of the details of the test.

= proportion of college graduates that believe in evolution

= proportion of adults with some college education that believe in evolution

: =

: ≠

Pooled proportion (for standard error): = = = 0.459

Test statistic: z = = 2.834

Since both samples have more than 10 successes and failures, the sample sizes are large enough to use a normal distribution for computing the p-value.

p-value = 0.005 (p-value for a two-sided alternative found using Statkey)

Since the p-value is (considerably) smaller than the 10% significance level, we have strong evidence to reject and thus have strong evidence to conclude that there is a significant difference in the proportion of college graduates and proportion of adults with some college who believe in the theory of evolution.

Diff: 2 Type: ES Var: 1

L.O.: 6.9.1

Use the following to answer the questions below:

In a survey, Gallup asked a random sample of U.S. adults if they would prefer to have a job outside the home, or if they would prefer to stay home to care for the family and home. Of the 504 males they surveyed, 391 said that they would prefer to have a job outside of the home. Of the 473 females they surveyed, 254 said that they would prefer a job outside of the home.

15) Verify that the sample size is large enough in each group to use the normal distribution to construct a confidence interval for a difference in proportions.

= = 0.776 = sample proportion of males who would prefer to have a job outside of the home

= = 0.537 = sample proportion of females who would prefer to have a job outside of the home

= 391 > 10

(1 - ) = 113 > 10

= 254 > 10

(1 - ) = 219 > 10

Since all are greater than 10, the sample sizes are large enough to use the normal distribution to construct a confidence interval for a difference in proportions.

Diff: 2 Type: ES Var: 1

L.O.: 6.8.0

16) Construct a 99% confidence interval for the difference between the proportion of men and women who would prefer to have a job outside the home. Use three decimal places when computing the sample proportions and margin of error.

A) 0.163 to 0.315

B) 0.190 to 0.288

C) 0.181 to 0.297

D) 0.197 to 0.295

Diff: 2 Type: BI Var: 1

L.O.: 6.8.1

17) Test, at the 1% level, if there is evidence that the proportion of men who would prefer a job outside of the home is significantly higher than the proportion of women who would prefer a job outside of the home.

= proportion of U.S. adult men who would prefer a job outside of the home

= proportion of U.S. adult women who would prefer a job outside of the home

: =

: >

Pooled proportion (for standard error): = = = 0.660

Test statistic: z = = 7.881

Since both samples have at least 10 successes and failures, the sample sizes are large enough to use the normal distribution to compute the p-value.

p-value ≈ 0 (using Statkey)

Because the p-value is less than the 1% significance level, we have very strong evidence to reject and thus we have very strong evidence that the proportion of men who would prefer a job outside of the home is significantly larger than the proportion of women who would prefer a job outside of the home.

Diff: 2 Type: ES Var: 1

L.O.: 6.9.1

Use the following to answer the questions below:

Consider taking random samples of size 50 from Population A with proportion 0.45 and random samples of size 40 from Population B with proportion 0.38.

18) Find the mean of the distribution of differences in sample proportions, - .

A) 0.45

B) 0.38

C) 0.07

D) -0.07

Diff: 2 Type: BI Var: 1

L.O.: 6.7.2

19) Find the standard error of the distribution of differences in sample proportions, - .

A) 0.0108

B) 0.0704

C) 0.0768

D) 0.1041

Diff: 2 Type: BI Var: 1

L.O.: 6.7.2

20) Are the sample sizes for both groups large enough for the Central Limit Theorem to apply so that the differences in sample proportions follow a normal distribution?

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 6.7.0

Use the following to answer the questions below:

Consider taking random samples of size 30 from Population A with proportion 0.84 and random samples of size 60 from Population B with proportion 0.9.

21) Find the mean of the distribution of differences in sample proportions, - .

A) 0.84

B) 0.90

C) -0.06

D) 0.06

Diff: 2 Type: BI Var: 1

L.O.: 6.7.2

22) Find the standard error of the distribution of differences in sample proportions, - .

A) 0.0060

B) 0.0387

C) 0.0669

D) 0.0773

Diff: 2 Type: BI Var: 1

L.O.: 6.7.2

23) Are the sample sizes for both groups large enough for the Central Limit Theorem to apply so that the differences in sample proportions follows a Normal distribution?

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 6.7.0

Use the following to answer the questions below:

The Gallup organization recently conducted a survey of 1,015 randomly selected U.S. adults about "Black Friday" shopping. They asked the following question:

"As you know, the Friday after Thanksgiving is one of the biggest shopping days of the year.

Looking ahead, do you personally plan on shopping on the Friday after Thanksgiving, or not?"

Of the 515 men who responded, 16% said "Yes." Of the 500 women who responded, 20% said "Yes."

24) Construct a 95% confidence interval for the difference between the proportion of men and women who planned to shop on the Friday after Thanksgiving. Use three decimal places when computing the margin of error.

A) -0.087 to 0.007

B) -0.080 to 0.000

C) -0.102 to 0.022

D) -0.103 to 0.023

Diff: 2 Type: BI Var: 1

L.O.: 6.8.1

25) Test, at the 5% level, if this sample provides evidence that the proportion of women planning to shop on Black Friday differs significantly from the proportion of men planning to shop. Include all of the details of the test.

= proportion of women planning to shop on Black Friday

= proportion of men planning to shop on Black Friday

: =

: ≠

Pooled proportion (for standard error):

= = = 0.1797, or about 0.18

Test Statistic: z = = 1.658

Because both samples have more than 10 successes and 10 failures, the sample sizes are large enough to use the normal distribution to compute the p-value.

p-value = 0.097

Because the p-value is larger than the 5% significance level, we have no evidence to reject and thus have no evidence to conclude that the proportions of men and women planning to shop the Friday after Thanksgiving are significantly different.

Diff: 3 Type: ES Var: 1

L.O.: 6.9.1

6.4 Inference for a Difference in Means

Use the following to answer the questions below:

Consider taking random samples of size 50 from Population A with mean 15 and standard deviation 3 and random samples of size 75 from Population B with mean 10 and standard deviation 5.

1) Find the mean of the distribution of differences in sample means,

A) -5

B) 5

C) 10

D) 15

Diff: 2 Type: BI Var: 1

L.O.: 6.10.1

2) Find the standard error of the distribution of differences in sample means,

A) 0.1267

B) 0.3559

C) 0.5133

D) 0.7165

Diff: 2 Type: BI Var: 1

L.O.: 6.10.1

3) How many degrees of freedom should be used when conducting inference for with samples of this size?

A) 49

B) 50

C) 74

D) 75

Diff: 2 Type: BI Var: 1

L.O.: 6.10.0

Use the following to answer the questions below:

Consider taking random samples of size 100 from Population A with mean 85 and standard deviation of 15 and random samples of size 60 from Population B with mean 78 and standard deviation 12.

4) Find the mean of the distribution of differences in sample means,

A) -7

B) 7

C) 78

D) 85

Diff: 2 Type: BI Var: 1

L.O.: 6.10.1

5) Find the standard error of the distribution of differences in sample means,

A) 0.35

B) 0.5916

C) 2.156

D) 4.65

Diff: 2 Type: BI Var: 1

L.O.: 6.10.1

6) How many degrees of freedom should be used when conducting inference for with samples of this size?

A) 99

B) 100

C) 59

D) 60

Diff: 2 Type: BI Var: 1

L.O.: 6.10.0

Use the following to answer the questions below:

Students in a large lecture class want to know who has, on average, more Facebook friends, male or female students. The data for the students are displayed in the provided dotplots and summary statistics are available in the provided table.

Gender	n		s
Male	68	678.9	317.6
Female	91	616.1	330.6

7) Is it reasonable to use a t-distribution for inference about the difference in mean number of Facebook friends for male and female students at this university?

A) Yes

B) No

Diff: 2 Type: MC Var: 1

L.O.: 6.11.0;6.12.0

8) Construct a 95% confidence interval for the difference in mean number of Facebook friends for male and female students at this university. Use two decimal places in your margin of error.

A) -40.62 to 166.22

B) -22.43 to 148.03

C) -38.75 to 164.35

D) -33.95 to 159.55

Diff: 2 Type: BI Var: 1

L.O.: 6.11.1

9) Test, at the 5% level, if this sample provides evidence of a significant difference in the mean number of Facebook friends for male and female students at this university. Include all of the details of the test.

= mean number of Facebook friends for male students at this university

= mean number of Facebook friends for female students at this university

: =

: ≠

Test Statistic: t = = 1.212

Because of the large sample sizes, we can use the t-distribution (with 67 degrees of freedom) to compute the p-value.

p-value = 0.23 (two-tail, using Statkey)

Because the p-value is much larger than the 5% significance level, we have no evidence to reject and thus have no evidence to conclude that there is a significant difference in the mean number of Facebook friends between male and female students at this university.

Diff: 2 Type: ES Var: 1

L.O.: 6.12.1

Use the following to answer the questions below:

As part of a course project, a statistics student surveyed random samples of 50 student athletes and 50 student non-athletes at his university, with the goal of comparing the heights of the two groups. His summary statistics are displayed in the provided table.

	n		s
Athletes	50	68.96	4.25
Non-athletes	50	67.28	3.46

10) Construct a 99% confidence interval for the difference in mean heights between student athletes and non-athletes at this university. Use two decimal places in your margin of error.

A) -0.40 to 3.76

B) -0.32 to 3.68

C) -0.16 to 3.20

D) -0.20 to 3.56

Diff: 2 Type: BI Var: 1

L.O.: 6.11.1

11) Test, at the 5% level, if student athletes at this university are significantly taller, on average, than student non-athletes. Include all of the details.

= mean height of student athletes at the university

= mean height of student non-athletes at the university

: =

: >

Test Statistic: t = = 2.168

This is a reasonably large sample size (and no real reason to suspect that heights are drastically skewed), so we can use a t-distribution to compute the p-value for the test. Because we use

p-value = 0.018 (right-tail, using Statkey)

Because the p-value is smaller than the 5% significance level, we have evidence to reject and thus have evidence to conclude that student athletes are significantly taller, on average, than student non-athletes at this university.

Diff: 2 Type: ES Var: 1

L.O.: 6.12.1

Use the following to answer the questions below:

A professor with a large introductory statistics class noticed that nearly half of his students missed class the day before a long break (like Thanksgiving Break or Spring Break). He randomly called on students and found 10 students in attendance and 10 students who had skipped class. Later in his office, he examined the current course grades for the 20 students he had selected. A plot of his findings and summary statistics are provided. Note that the grades were entered as proportions, and thus a grade of 0.925 is a 92.5% in the course.

	n		s
Attending	10	0.8669	0.0765
Not Attending	10	0.7811	0.0805

12) Verify that it is reasonable to use a t-distribution to construct a confidence interval, or perform a test about, the difference in mean grades for the two groups of students.

Diff: 2 Type: ES Var: 1

L.O.: 6.11.0;6.12.0

13) Construct a 98% confidence interval for the difference between the mean grades for students attending and not attending class the day before break. Use four decimal places in your margin of error.

A) -0.0132 to 0.1848

B) 0.0278 to 0.1438

C) -0.0042 to 0.1758

D) -0.0095 to 0.1811

Diff: 2 Type: BI Var: 1

L.O.: 6.11.1

14) Test, at the 5% level, if there is evidence that students who attended class before break have, on average, a significantly higher course grade than those who skipped. Include all of the details of the test.

= mean course grade for students who attended the class before break

= mean course grade for students who skipped the class before break

: =

: >

Test Statistic: t = = 2.44

While both sample sizes are small, the dotplots appear reasonably symmetric and don't have any major outliers, so we can use a t-distribution to compute the p-value (with 9 degrees of freedom since both samples are of size 10).

p-value = 0.019 (right tail, using Statkey)

Because the p-value is smaller than the 5% significance level, we have evidence that students who attended the class before break have, on average, a significantly higher course grade than the students who skipped.

Diff: 2 Type: ES Var: 1

L.O.: 6.12.1

Use the following to answer the questions below:

Consider a test of : = versus : > using the sample results with and with

15) What is the test statistic for this test?

A) -1.688

B) 1.688

C) 4.74

D) 0.7701

Diff: 2 Type: BI Var: 1

L.O.: 6.12.0

16) What are the degrees of freedom for this test?

A) 4

B) 23

C) 25

D) 27

Diff: 2 Type: BI Var: 1

L.O.: 6.12.0

17) What value is closest to the p-value for this test?

A) 0.104

B) 0.948

C) 0.0014

D) 0.052

Diff: 2 Type: BI Var: 1

L.O.: 6.12.0

Use the following to answer the questions below:

Consider constructing a 90% confidence interval for using the sample results with and with

18) What is the best estimate of - ?

A) 7

B) -7

C) 103

D) 96

Diff: 2 Type: BI Var: 1

L.O.: 6.11.0

19) What are the degrees of freedom in this situation?

A) 10

B) 49

C) 35

D) 39

Diff: 2 Type: BI Var: 1

L.O.: 6.11.0

20) What is the t* for the 90% confidence interval?

A) 1.645

B) 1.677

C) 1.685

D) 2.023

Diff: 2 Type: BI Var: 1

L.O.: 6.11.0

21) What is the margin of error for this confidence interval?

A) 8.44

B) 5.01

C) 1.73

D) 11.91

Diff: 2 Type: BI Var: 1

L.O.: 6.11.0

Use the following to answer the questions below:

A small university is trying to monitor its electricity usage. For a random sample of 30 weekend days (Saturdays and Sundays), the student center used an average of 94.26 kilowatt hours (kWh) with standard deviation 43.29. For a random sample of 60 weekdays, (Monday - Friday), the student center used an average of 112.63 kWh with standard deviation 32.07.

22) Test, at the 5% level, if significantly more electricity is used at the student center, on average, on weekdays than weekend days. Include all details of the test.

= mean amount of electricity (in kWh) used on weekdays

= mean amount of electricity (in kWh) used on weekend days

: =

: >

Test Statistic: t = = 2.059

Both sample sizes are large, so we can use the t-distribution with 29 degrees of freedom.

p-value = 0.024 (right tail, using Statkey)

There is evidence to reject and thus there is evidence to conclude that more electricity is used at the student center, on average, on weekdays than on weekend days.

Diff: 2 Type: ES Var: 1

L.O.: 6.12.1

23) Construct a 95% confidence interval for the difference in mean electricity use at the student center between weekdays and weekend days. Use two decimal places in your margin of error.

A) 0.12 to 36.62 kWh

B) -6.80 to 43.54 kWh

C) -4.61 to 41.35 kWh

D) -3.38 to 40.51 kWh

Diff: 2 Type: BI Var: 1

L.O.: 6.11.1

Use the following to answer the questions below:

In a given year, the average score on the Mathematics portion of the ACT for males was 21.3 with standard deviation 5.3. The average score on the Mathematics portion of the ACT for females was 20.2 with standard deviation 4.8.

24) If random samples are taken with 50 males and 70 females, find the standard error of the distribution of differences in sample means, where and represent the sample means for males and females, respectively. Report the standard error with four decimal places.

Diff: 2 Type: SA Var: 1

L.O.: 6.10.1

25) If random samples are taken with 60 males and 60 females, find the standard error of the distribution of differences in sample means, where and represent the sample means for males and females, respectively. Report the standard error with four decimal places.

Diff: 2 Type: SA Var: 1

L.O.: 6.10.1

26) If random samples are taken with 120 males and 105 females, find the standard error of the distribution of differences in sample means, where and represent the sample means for males and females, respectively. Report your standard error with four decimal places.

Diff: 2 Type: SA Var: 1

L.O.: 6.10.1

27) What effect does increasing the sample sizes have on the center of the distribution?

A) It increases the center of the distribution.

B) It decreases the center of the distribution.

C) It has no effect on the center of the distribution.

Diff: 2 Type: BI Var: 1

L.O.: 6.10.1

28) What effect does increasing the sample sizes have on the spread of the distribution?

A) It increases the spread of the distribution.

B) It decreases the spread of the distribution.

C) It has no effect on the spread of the distribution.

Diff: 2 Type: BI Var: 1

L.O.: 6.10.1

6.5 Paired Difference in Means

Use the following to answer the questions below:

Students in a small statistics class were asked to count the number of scars both on their "dominant" hand (the one they use most often) and on their "off" hand. The summary statistics are provided. It is of interest to compare the average number of scars on the dominant and off hands.

	n		s
Dominant	25	1.92	2.326
Off	25	2.72	3.007
Difference (D - O)	25	-0.8	2.363

1) Why is it appropriate to use paired data in this analysis? Explain briefly.

Diff: 2 Type: ES Var: 1

L.O.: 6.13.1

2) Boxplots of the raw data are provided. Would it be appropriate to use a t-distribution to construct a confidence interval for, or perform a test about, the difference in the mean number of scars on dominant and off hands? Specifically mention which boxplot(s) you are using to justify your answer.

Diff: 2 Type: ES Var: 1

L.O.: 6.13.0;6.13.3;6.13.4

3) Construct a 90% confidence interval for the difference in mean number of scars on dominant and off hands. Round your margin of error to two decimal places.

A) -1.61 to 0.01

B) -1.67 to 0.07

C) -1.71 to 0.11

D) -1.81 to 0.21

Diff: 2 Type: BI Var: 1

L.O.: 6.13.3

4) Test to see if the mean number of scars on dominant hands is significantly different from the mean number of scars on off hands. Use a 10% significance level. Include all of the details of the test.

= difference in mean number of scars on dominant and off hands (dominant - off) (Note - = mean number of scars on dominant hand - mean number of scars on off hand would also be acceptable.)

: = 0

: ≠ 0

Test statistic: t = = -1.693

n = 25, so df = 24

p-value = 0.103 (two-sided p-value found in Statkey using a t distribution with 24 degrees of freedom)

Since the p-value is larger than the 10% significance level, we have no evidence to reject and thus have no evidence conclude that there is a significant difference in the mean number of scars on dominant and off hands.

Diff: 2 Type: ES Var: 1

L.O.: 6.13.4

Use the following to answer the questions below:

The Math and Verbal SAT scores for a random sample of 10 students from a large introductory statistics course are provided.

Student	1	2	3	4	5	6	7	8	9	10		s
Math SAT	640	620	700	680	600	720	770	580	660	580	655	63.1
Verbal SAT	580	690	560	560	600	700	620	620	710	500	614	68.8
Differences (Math - Verbal)	60	-70	140	120	0	20	150	-40	-50	80	41	81.0

5) Which data analysis method is more appropriate in this situation: paired data difference in means or difference in means with two separate groups?

A) Paired data difference in means

B) Difference in means with two separate groups

Diff: 2 Type: BI Var: 1

L.O.: 6.13.1

6) Boxplots of the raw data are provided. Would it be appropriate to use a t-distribution to construct a confidence interval for, or perform a test about, the difference in the mean Math and Verbal SAT scores? Specifically mention which boxplot(s) you are using to justify your answer.

Diff: 2 Type: ES Var: 1

L.O.: 6.13.0

7) Construct a 90% confidence interval for the difference in mean Math and Verbal SAT scores for students in the class. Use two decimal places in your margin of error.

A) -5.95 to 87.95

B) -9.20 to 91.20

C) -10.13 to 92.13

D) -8.439 to 90.44

Diff: 2 Type: BI Var: 1

L.O.: 6.13.3

8) Test, at the 10% level, if Math SAT scores are significantly higher, on average, than Verbal SAT scores for students in the class. Include all of the details of the test.

= mean Math SAT score for students in the class

= mean Verbal SAT score for students in the class

: =

: >

Test statistic: t = = 1.60

Because the differences are relatively symmetric, we can use the t-distribution with 9 degrees of freedom to compute the p-value.

p-value = 0.072 (right tail, using Statkey)

Because the p-value is smaller than the 10% significance level, there is some (somewhat weak) evidence to reject and thus there is some (somewhat weak) evidence to conclude that Math SAT scores are, on average, significantly higher than Verbal SAT scores for students in the class.

Diff: 2 Type: ES Var: 1

L.O.: 6.13.4

Use the following to answer the questions below:

As part of a course project, a statistics student surveyed random samples of 50 student athletes and 50 student non-athletes at his university, with the goal of comparing the heights of the two groups. His summary statistics are displayed in the provided table.

	n		s
Athletes	50	68.96	4.25
Non-athletes	50	67.28	3.46

9) Which data analysis method is more appropriate in this situation: paired data difference in means or difference in means with two separate groups?

A) Difference in means with two separate groups

B) Paired data difference in means

Diff: 2 Type: BI Var: 1

L.O.: 6.13.1

Use the following to answer the questions below:

"Black Friday," which occurs annually the day after Thanksgiving, is one of the biggest shopping days of the year. During the holiday season, many stores created controversy by starting their mega-sales on Thanksgiving itself. In a random sample of 25 individuals who shopped during the Black Friday four-day weekend (Thursday - Sunday), the average amount spent was $399.40 with standard deviation $171.10. The data are displayed in the provided dotplot.

10) Construct a 95% confidence interval for the average amount spent by individuals who shopped over the Black Friday weekend. Use two decimal places in your margin of error.

A) $328.77 to $470.03

B) $311.28 to $487.52

C) $332.33 to $466.47

D) $324.11 to $474.69

Diff: 2 Type: BI Var: 1

L.O.: 6.5.1

11) Construct a 95% confidence interval for the average amount spent by individuals who shopped over the Black Friday weekend. Provide an interpretation of your interval in the context of this data situation.

Diff: 2 Type: ES Var: 1

L.O.: 3.2.4;6.5.0

12) A natural question would be if more money was spent over the Black Friday weekend than over the previous year's Black Friday weekend (which did not start on Thursday). What information would be necessary to address this question using the paired data difference in means method? How could the data be collected so that the difference in means for two separate groups method would be most appropriate?

In order to analyze the data using two separate groups, we would just need responses from a sample of individuals about how much they spent in the previous year—they can be completely different people.

Diff: 3 Type: ES Var: 1

L.O.: 6.13.1

13) Suppose we know that in a random sample of n = 22 individuals who shopped over Black Friday weekend in 2011 the average amount spent was $381.30 with standard deviation $119.80. Construct a 95% confidence interval for the difference in the mean amount spent between the 2012 and 2011 Black Friday weekends. Round the margin of error to two decimal places.

Recall that for the 2012 sample of 25 individuals, the average amount spent was $399.40 with standard deviation $171.10. Dotplots of both samples are provided.

A) -$70.72 to $106.92

B) -$91.86 to $128.06

C) -$65.59 to $101.79

D) -$76.07 to $112.27

Diff: 2 Type: BI Var: 1

L.O.: 6.11.1

14) Would the 95% confidence interval provide evidence that the average amount spent over the 2011 Black Friday weekend differs from the average amount spent over the 2012 Black Friday weekend?

A) No

B) Yes

Diff: 2 Type: MC Var: 1

L.O.: 6.11.0

15) Suppose we know that in a random sample of n = 22 individuals who shopped over Black Friday weekend in 2011 the average amount spent was $381.30 with standard deviation $119.80. Recall that for the 2012 sample of 25 individuals, the average amount spent was $399.40 with standard deviation $171.10. Dotplots of both samples are provided.

Test, at the 5% level, if the samples provide evidence that Black Friday shoppers spent more, on average, in 2012 than they did in 2011. Include all of the details of the test.

= mean amount spent by shoppers over Black Friday weekend in 2012

= mean amount spent by shoppers over Black Friday weekend in 2011

: =

: >

Test Statistic: t = = 0.424

Neither of the two samples appears to be severely skewed or have major outliers, and thus it is reasonable to use a t-distribution to compute the p-value.

p-value = 0.338 (right-tail, using Statkey)

Because the p-value is so much larger than the 5% significance level, we have no evidence to reject and thus have no evidence to conclude that shoppers spent significantly more, on average, over the 2012 Black Friday weekend than they did over the 2011 Black Friday weekend.

Diff: 2 Type: ES Var: 1

L.O.: 6.12.1

Use the following to answer the questions below:

Zumba, often described as a Latin-inspired dance fitness party, is currently one of the most popular group fitness classes, but its health benefits have been little studied. An exercise science professor at a large university conducted a study to investigate some of the health benefits of Zumba. He recorded the weight of 9 female college students before they began a six week long Zumba program. As part of the program, they took a 60 minute long Zumba class three days a week. At the end of the program, the subjects were weighed again. Of interest is their weight loss, defined as weight before the program started minus weight after completing the program. The results are displayed in the following table.

The mean weight loss for the sample was 2.667 pounds with standard deviation 2.5 pounds. A dotplot of the differences is provided.

16) Construct a 99% confidence interval for the mean weight loss. Use three decimal places in your margin of error.

A) -0.129 to 5.463 pounds

B) -0.316 to 5.081 pounds

C) -0.521 to 4.813 pounds

D) -0.223 to 5.111 pounds

Diff: 2 Type: BI Var: 1

L.O.: 6.13.3

17) Test, at the 1% level, if there is evidence that the Zumba program is effective for weight loss. Include all of the details of the test.

= mean weight before the program began

= mean weight after the program has completed

: =

: >

Test statistic: t = = 3.2004

Even though the sample size is small, the distribution appears reasonably symmetric with no major outliers, and thus we can use the t-distribution with 8 degrees of freedom.

p-value = 0.0063 (right tail, using Statkey)

Because the p-value is smaller than the significance level, there is evidence to reject and thus evidence to conclude that Zumba is effective for weight loss.

Diff: 2 Type: ES Var: 1

L.O.: 6.13.4

Use the following to answer the questions below:

A 1997 study described in the European Journal of Clinical Nutrition compares the growth of vegetarian and omnivorous children, ages 7 - 11, in Northwest England. In the study, each of the 50 vegetarian children in the study was matched with an omnivorous child of the same age with similar demographic characteristics. One of the aspects on which the children were compared was their body mass index (BMI). The differences in BMI for each pair of children (one vegetarian and one omnivore) was computed as vegetarian BMI minus omnivore BMI.

	n		s
Vegetarian	50	16.76	1.91
Omnivorous	50	17.12	2.23
Difference (Vegetarian - Omnivorous)	50	-0.36	2.69

18) Which data analysis method is more appropriate in this situation: paired data difference in means or difference in means with two separate groups?

A) Paired data difference in means

B) Difference in means with two separate groups

Diff: 3 Type: BI Var: 1

L.O.: 6.13.1

19) Construct a 95% confidence interval for the difference in mean BMI between vegetarian and omnivorous children. Use three decimal places in your margin of error.

A) -1.125 to 0.405

B) -1.433 to 0.713

C) -1.340 to 0.620

D) -1.312 to 0.592

Diff: 2 Type: BI Var: 1

L.O.: 6.13.3

20) Test, at the 5% level, if there is evidence that the mean BMI for vegetarian children differs significantly from the mean BMI for omnivorous children. Include all of the details of the test.

= mean BMI for vegetarian children

= mean BMI for omnivorous children

(alternatively, they could define to represent the mean difference, where the differences are defined as

: =

: ≠

Test statistic: t = = -0.946

The sample size is fairly large so we can use the t-distribution with 49 degrees of freedom to compute the p-value for the test.

p-value = 0.349 (two-tail, using Statkey)

Because the p-value is considerably larger than the 5% significance level, we have no evidence to reject and thus have no evidence to conclude that the mean BMI for vegetarian children differs significantly from the mean BMI for omnivorous children.

Diff: 2 Type: ES Var: 1

L.O.: 6.13.4

21) Construct a 96% confidence interval for - using the paired data in the following table. Round all values to three decimal places.

Assume that the results come from random samples from populations that are approximately normal and that the differences are computed using

A) 0.362 to 4.304

B) 0.316 to 4.350

C) 0.492 to 4.174

D) 0.377 to 4.289

Diff: 2 Type: BI Var: 1

L.O.: 6.13.3

22) Report the test statistic (with two decimal places), p-value, and conclusion for a test of versus using the paired data provided in the following table. Use a 5% significance level.

Assume that the results come from random samples from populations that are approximately normal and that the differences are computed using

The mean difference is = -4 and the standard deviation of the differences is .

Test statistic: t = = -3.99

n = 7, so df = 6

p-value = 0.0036

Because the p-value is considerably smaller than the 5% significance level, we have very strong evidence to reject and thus have very strong evidence in favor of

Diff: 2 Type: ES Var: 1

L.O.: 6.13.4

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