chapter 11: confidence intervals and hypothesis tests for...

11-1

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Chapter 11: Confidence Intervals and Hypothesis Tests for Means – Quiz A Name ________________________________________ 11.3.3 Calculate and describe the standard error of the mean. 1. In a metal fabrication process, metal rods are produced to a specified target length of 15 feet. Suppose that the lengths are normally distributed. A quality control specialist collects a random sample of 16 rods and finds the sample mean length to be 14.8 feet and a standard deviation of 0.65 feet. a. Describe the sampling distribution for the sample mean. b. What is the standard error? c. For 95% confidence, what is the margin of error? d. Based on the sample results, create the 95% confidence interval and interpret. 11.5.6 Create and interpret confidence intervals for the mean. 2. A manufacturer of cheese filled ravioli supplies a pizza restaurant chain. Based on data collected from its automatic filling process, the amount of cheese inserted into the ravioli is normally distributed. To make sure that the automatic filling process is on target, quality control inspectors take a sample of 25 ravioli and measure the weight of cheese filling. They find a sample mean weight of 15 grams with a standard deviation of 1.5 grams. a. Describe the sampling distribution for the sample mean. b. What is the standard error? c. What is the margin of error for 99% confidence? d. What is the margin of error for 90% confidence? e. Based on the sample results, find the 99% confidence interval and interpret. f. Based on the sample results, find the 90% confidence interval and interpret. g. For a more accurate determination of the mean weight, the quality control inspectors wish to estimate it within 0.25 grams with 99% confidence. How many ravioli should they sample?

11-2 Chapter 11 Confidence Intervals and Hypothesis Tests for Means


11.5.6 Create and interpret confidence intervals for the mean. 3. Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. Last year the average life expectancy of all policyholders was 77 years. ABI Insurance wants to determine if their clients now have a longer life expectancy, on average, so they randomly sample some of their recently paid policies. The ages of the clients in the sample are shown below.

86 75 83 84 81 77 78 79 79 8176 85 70 76 79 81 73 74 72 83

a. Based on the sample results, find the 90% confidence interval and interpret. b. For more accurate cost determination, ABI Insurance wants to estimate the average life expectancy to within one year with 95% confidence. How many randomly selected recently paid policies would they need to sample? c. Suppose ABI samples 100 recently paid policies. This sample yields a mean of 77.7 years and a standard deviation of 3.6 years. Find a 90% confidence interval and interpret. 11.5.6 Create and interpret confidence intervals for the mean. 4. A sample of students from an introductory business course were polled regarding the number of hours they spent studying for the last exam. All students anonymously submitted the number of hours on a 3 by 5 card. There were 24 individuals in the one section of the course polled. The data was used to make inferences regarding the other students taking the course. The data are shown below:

4.5 22 7 14.5 9 9 3.5 8 11 7.5 18 20 7.5 9 10.5 15 19 2.5 5 9 8.5 14 20 8

a. Based on the sample results, find the 95% confidence interval.

b. Interpret the results.

c. Do you expect a 90% confidence interval to be wider or narrower and why?

d. A previous study of a large cross-section of students in this course showed that students studies 12 hours per week. Are the results of this current study statistically different than the assumption of 12 hours per week studying? Define the hypotheses and compare the t-value to the critical t value, evaluating at α = 0.05.

Quiz A 11-3


Chapter 11: Confidence Intervals and Hypothesis Tests for Means – Quiz A - Key 1. In a metal fabrication process, metal rods are produced to a specified target length of 15 feet. Suppose that the lengths are normally distributed. A quality control specialist collects a random sample of 16 rods and finds the sample mean length to be 14.8 feet and a standard deviation of 0.65 feet. a. Describe the sampling distribution for the sample mean. The sampling distribution for the sample mean can be modeled using the t-distribution with 15 degrees of freedom. b. What is the standard error?

ftnsySE 1625.0

16

65.0)( ===

c. For 95% confidence, what is the margin of error?

ftftySEtn 346.01625.0131.2)(*1 =×=×−

d. Based on the sample results, create the 95% confidence interval and interpret.

346.08.14)(*1 ±=×± − ySEty n

The 95% confidence interval for true mean length is from 14.454 to 15.146 ft. We are 95% confident that the average length of metal rods from this process is between 14.454 and 15.146 ft. 2. A manufacturer of cheese filled ravioli supplies a pizza restaurant chain. Based on data collected from its automatic filling process, the amount of cheese inserted into the ravioli is normally distributed. To make sure that the automatic filling process is on target, quality control inspectors take a sample of 25 ravioli and measure the weight of cheese filling. They find a sample mean weight of 15 grams with a standard deviation of 1.5 grams. a. Describe the sampling distribution for the sample mean. The sampling distribution for the sample mean can be modeled using the t-distribution with 24 degrees of freedom. b. What is the standard error?

gramsnsySE 3.0

25

5.1)( ===



c. What is the margin of error for 99% confidence?

gramsgramsySEtn 8391.03.0797.2)(*1 =×=×−

d. What is the margin of error for 90% confidence?

gramsgramsySEtn 5133.03.0711.1)(*1 =×=×−

e. Based on the sample results, find the 99% confidence interval and interpret.

8391.015)(*1 ±=×± − ySEty n

The 99% confidence interval for true mean weight is 14.16 to 15.84 grams. We are 99% confident that the mean weight of cheese in the ravioli made by this process is between 14.16 and 15.84 grams. f. Based on the sample results, find the 90% confidence interval and interpret.

5133.015)(*1 ±=×± − ySEty n

The 90% confidence interval for true mean weight is 14.49 to 15.51 grams. We are 90% confident that the mean weight of cheese in the ravioli made by this process is between 14.49 and 15.51 grams. g. For a more accurate determination of the mean weight, the quality control inspectors wish to estimate it within .25 grams with 99% confidence. How many ravioli should they sample?

2

22

25.0

)5.1()575.2(=n

n = 238.7 or 239 ravioli

Quiz A 11-5


3. Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. Last year the average life expectancy of all policyholders was 77 years. ABI Insurance wants to determine if their clients now have a longer life expectancy, on average, so they randomly sample some of their recently paid policies. The ages of the clients in the sample are shown below.

86 75 83 84 81 77 78 79 79 8176 85 70 76 79 81 73 74 72 83

a. Based on the sample results, find the 90% confidence interval and interpret. The sample has a mean of 78.6 years and a standard deviation of 4.48 years.

732.16.7820

48.4729.16.78)(*

1 ±=±=×± − ySEty n

The 90% confidence interval for true mean age is 76.87 to 80.33 years. We are 90% confident that the average age of clients with recently paid policies is between 76.87 and 80.33 years. b. For more accurate cost determination, ABI Insurance wants to estimate the average life expectancy to within one year with 95% confidence. How many randomly selected recently paid policies would they need to sample?

2

22

1

)48.4()96.1(=n

n = 77.1 or 78 policies c. Suppose ABI samples 100 recently paid policies. This sample yields a mean of 77.7 years and a standard deviation of 3.6 years. Find a 90% confidence interval and interpret.

100

6.3660.17.77)(*

1 ±=×± − ySEty n

The 90% confidence interval for true mean age is 77.1 to 78.3 years. We are 90% confident that the average age of clients with recently paid policies is between 77.1 years to 78.3 years.



4. A sample of students from an introductory business course were polled regarding the number of hours they spent studying for the last exam. All students anonymously submitted the number of hours on a 3 by 5 card. There were 24 individuals in the one section of the course polled. The data was used to make inferences regarding the other students taking the course. The data are shown below:

4.5 22 7 14.5 9 9 3.5 8 11 7.5 18 20 7.5 9 10.5 15 19 2.5 5 9 8.5 14 20 8

a. Based on the sample results, find the 95% confidence interval. The sample has a mean of 10.92 hours and a standard deviation of 5.60 hours.

1

5 6010 92 2 069 10 92 2 365

24*n

.y t SE( y ) . . . .−± × = ± = ±

The 95% confidence interval for true mean age is 8.5 to 13.3 hours.

b. Interpret the results.

We are 95% confident that the average age of clients with recently paid policies is between 76.87 and 80.33 years.

c. Do you expect a 90% confidence interval to be wider or narrower and why?

A 90% confidence interval will be narrower because the margin of error will be smaller for a lower confidence level.

d. A previous study of a large cross-section of students in this course showed that students studies 12 hours per week. Are the results of this current study statistically different than the assumption of 12 hours per week studying? Define the hypotheses and compare the t-value to the critical t value, evaluating at α = 0.05.

Hypotheses: H0: µ = 12 hrs; HA: µ ≠ 12 hrs

t-test: 24/6.5

1292.10 −=t =-0.945; the critical t value for df = 23 at 0.05 = ±2.0686

The calculated value is less than the critical value so the result is not statistically significant at α = 0.05. There is no evidence that the level of studying has changed.

Quiz B 11-7


Chapter 11: Confidence Intervals and Hypothesis Tests for Means – Quiz B Name ________________________________________ 11.3.3 Calculate and describe the standard error of the mean. 1. A small business ships specialty homemade candies to anywhere in the world. Past records indicate that the weight of orders is normally distributed. Suppose a random sample of 16 orders is selected and each is weighed. The sample mean was found to be 110 grams with a standard deviation of 14 grams. a. Describe the sampling distribution for the sample mean. b. What is the standard error? c. For 90% confidence, what is the margin of error? d. Based on the sample results, create the 90% confidence interval and interpret. 11.5.6 Create and interpret confidence intervals for the mean. 2. Grandma Gertrude’s Chocolates, a family owned business, has an opportunity to supply its product for distribution through a large coffee house chain. However, the coffee house chain has certain specifications regarding cacao content as it wishes to advertise the health benefits (antioxidants) of the chocolate products it sells. In order to determine the mean % cacao in its dark chocolate products, quality inspectors sample 36 pieces. They find a sample mean of 55% with a standard deviation of 4%. a. Describe the sampling distribution for the sample mean. b. What is the standard error? c. What is the margin of error for 90% confidence? d. What is the margin of error for 95% confidence? e. Based on the sample results, find the 90% confidence interval and interpret. f. Based on the sample results, find the 95% confidence interval and interpret. g. For a more accurate determination of the mean weight, the quality control inspectors wish to estimate it within 1% with 95% confidence. How many pieces of dark chocolate should they sample?



11.5.6 Create and interpret confidence intervals for the mean. 3. A large software development firm recently relocated its facilities. Top management is interested in fostering good relations with their new local community and has encouraged their professional employees to engage in local service activities. They wish to determine the average number of hours the firm’s professionals volunteer per month. A random sample of 24 professionals reported the following number of hours:

12 13 14 14 15 15 15 16 16 16 16 16 17 17 17 18 18 18 18 19 19 19 20 21

a. Based on the sample results, find the 95% confidence interval and interpret. b. For a more accurate determination, top management wants to estimate the average number of hours volunteered per month by their professional staff to within one hour with 99% confidence. How many randomly selected professional employees would they need to sample? c. Suppose 40 professional employees are randomly selected. This sample yields a mean of 15.2 hours and a standard deviation of 1.8 hours. Find a 95% confidence interval and interpret. 11.7.8 Perform hypothesis tests for means. 4. A pharmaceutical company wants to answer the question whether it takes longer than 45 seconds for a drug in pill form to dissolve in the gastric juices of the stomach. A sample was taken from patients taken the given drug in pill form and times for the pills to be dissolved were measured as the following:

42.7 43.4 44.6 45.1 45.6 45.9 46.8 47.6 a. State the hypotheses to test this question.

b. State the assumptions of the test.

c. Perform the test and determine if it is statistically significant at α = 0.05. d. Interpret the result.

Quiz B 11-9


Chapter 11: Confidence Intervals and Hypothesis Tests for Means – Quiz B – Key 1. A small business ships specialty homemade candies to anywhere in the world. Past records indicate that the weight of orders is normally distributed. Suppose a random sample of 16 orders is selected and each is weighed. The sample mean was found to be 110 grams with a standard deviation of 14 grams. a. Describe the sampling distribution for the sample mean. The sampling distribution for the sample mean can be modeled using the t-distribution with 15 degrees of freedom. b. What is the standard error?

gramsnsySE 5.3

16

14)( ===

c. For 90% confidence, what is the margin of error?

gramsySEtn 135.65.3753.1)(*1 =×=×−

d. Based on the sample results, create the 90% confidence interval and interpret.

135.6110)(*1 ±=×± − ySEty n

The 90% confidence interval for true mean weight of orders is from 103.87 to 116.14 grams. We are 90% confident that the average weight of candy orders is between 103.87 and 116.14 grams. 2. Grandma Gertrude’s Chocolates, a family owned business, has an opportunity to supply its product for distribution through a large coffee house chain. However, the coffee house chain has certain specifications regarding cacao content as it wishes to advertise the health benefits (antioxidants) of the chocolate products it sells. In order to determine the mean % cacao in its dark chocolate products, quality inspectors sample 36 pieces. They find a sample mean of 55% with a standard deviation of 4%. a. Describe the sampling distribution for the sample mean. The sampling distribution for the sample mean can be modeled using the t-distribution with 35 degrees of freedom. b. What is the standard error?

%67.036

4)( ===

nsySE



c. What is the margin of error for 90% confidence?

%13.167.0690.1)(*1 =×=×− ySEtn

d. What is the margin of error for 95% confidence?

%36.167.0030.2)(*1 =×=×− ySEtn

e. Based on the sample results, find the 90% confidence interval and interpret.

13.155)(*1 ±=×± − ySEty n

The 90% confidence interval for true % cacao is 53.87% to 56.13%. We are 90% confident that the mean percentage of cacao in Grandma Gertrude’s dark chocolate is between 53.87% and 56.13%. f. Based on the sample results, find the 95% confidence interval and interpret.

36.155)(*1 ±=×± − ySEty n

The 95% confidence interval for true % cacao is 53.64% to 56.36%. We are 95% confident that the mean percentage of cacao in Grandma Gertrude’s dark chocolate is between 53.64% and 56.36%. g. For a more accurate determination of the mean weight, the quality control inspectors wish to estimate it within 1% with 95% confidence. How many pieces of dark chocolate should they sample?

2

22

1

)4()96.1(=n

n = 61.46 or 62 pieces of chocolate

Quiz B 11-11


3. A large software development firm recently relocated its facilities. Top management is interested in fostering good relations with their new local community and has encouraged their professional employees to engage in local service activities. They wish to determine the average number of hours the firm’s professionals volunteer per month. A random sample of 24 professionals reported the following number of hours:

12 13 14 14 15 15 15 16 16 16 16 16 17 17 17 18 18 18 18 19 19 19 20 21

a. Based on the sample results, find the 95% confidence interval and interpret. The sample has a mean of 16.6 hours and a standard deviation of 2.22 hours.

938.06.1624

22.2069.26.16)(*

1 ±=±=×± − ySEty n

The 95% confidence interval for true mean number of volunteer hours is 15.66 to 17.54. We are 95% confident that the average number of hours volunteered by professionals employed with this firm is between 15.66 and 17.54. b. For a more accurate determination, top management wants to estimate the average number of hours volunteered per month by their professional staff to within one hour with 99% confidence. How many randomly selected professional employees would they need to sample?

2

22

1

)22.2()575.2(=n

n = 32.67 or 33 professionals c. Suppose 40 professional employees are randomly selected. This sample yields a mean of 15.2 hours and a standard deviation of 1.8 hours. Find a 95% confidence interval and interpret.

576.02.1540

8.1023.22.15)(*

1 ±=±=×± − ySEty n

14.62 to 15.78 hours The 95% confidence interval for true mean number of volunteer hours is 14.62 to 15.78. We are 95% confident that the average number of hours volunteered by professionals employed with this firm is between 14.62 and 15.78.



4. A pharmaceutical company wants to answer the question whether it takes longer than 45 seconds for a drug in pill form to dissolve in the gastric juices of the stomach. A sample was taken from patients taken the given drug in pill form and times for the pills to be dissolved were measured as the following:

42.7 43.4 44.6 45.1 45.6 45.9 46.8 47.6 a. State the hypotheses to test this question.

0 AH : = 45sec; H : > 45secμ μ

b. State the assumptions of the test.

The subjects in the sample must be randomly selected from a population. The sample data must come from a normally distributed population of observations for the variable under study. d. Perform the test and determine if it is statistically significant at α = 0.05. Test of mu = 45 vs > 45 95% Lower Variable N Mean StDev SE Mean Bound T P Time (s) 8 45.212 1.640 0.580 44.114 0.37 0.362

The result is a P value of 0.362 which is not statistically significant at the given level of α = 0.05. c. Interpret the result. There is no evidence to believe that it takes longer than 45 seconds for the pills to dissolve.

Quiz C 11-13


Chapter 11: Confidence Intervals and Hypothesis Tests for Means – Quiz C – Multiple Choice Name ________________________________________ 11.2.1 Examine the sampling distribution of the mean. 1. In a metal fabrication process, metal rods are produced to a specified target length of 15 feet. Suppose that the lengths are normally distributed. A quality control specialist collects a random sample of 16 rods and finds the sample mean length to be 14.8 feet and a standard deviation of 0.65 feet. Which of the following statements is true? A. The sampling distribution for the sample mean follows the t-distribution with 16 degrees of freedom. B. The sampling distribution for the sample mean follows the t-distribution with 15 degrees of freedom. C. The mean of the sampling distribution for the sample mean is 14.8 feet. D. The sampling distribution for the sample mean is Normal with a mean of 14.8 feet and standard deviation of .65 feet. E. The standard error is 0.65 feet. 11.3.3 Calculate and describe the standard error of the mean. 2. In a metal fabrication process, metal rods are produced to a specified target length of 15 feet. Suppose that the lengths are normally distributed. A quality control specialist collects a random sample of 16 rods and finds the sample mean length to be 14.8 feet and a standard deviation of 0.65 feet. The standard error of the mean is A. 0.65 ft. B. 0.346 ft. C. 0.1625 ft. D. 0.0098 ft. E. 1.625 ft. 11.5.6 Create and interpret confidence intervals for the mean. 3. In a metal fabrication process, metal rods are produced to a specified target length of 15 feet. Suppose that the lengths are normally distributed. A quality control specialist collects a random sample of 16 rods and finds the sample mean length to be 14.8 feet and a standard deviation of 0.65 feet. The 95% confidence interval for the true mean length of rods produced by this process is A. 14.544 to 15.056 ft. B. 14.345 to 15.255 ft. C. 13.912 to 15.688 ft. D. 14.454 to 15.146 ft. E. 13.834 to 15.766 ft.



11.4.5 Find critical values of t. 4. A manufacturer of cheese filled ravioli supplies a pizza restaurant chain. Based on data collected from its automatic filling process, the amount of cheese inserted into the ravioli is normally distributed. To make sure that the automatic filling process is on target, quality control inspectors take a sample of 25 ravioli. The correct value of t* to construct a 98% confidence interval for the true mean amount of cheese filling is A. 2.492 B. 3.467 C. 2.797 D. 2.064 E. 3.745 11.5.6 Create and interpret confidence intervals for the mean. 5. A manufacturer of cheese filled ravioli supplies a pizza restaurant chain. Based on data collected from its automatic filling process, the amount of cheese inserted into the ravioli is normally distributed. To make sure that the automatic filling process is on target, quality control inspectors take a sample of 25 ravioli and measure the weight of cheese filling. They find a sample mean weight of 15 grams with a standard deviation of 1.5 grams. What is the margin of error at 90% confidence? A. 0.3 grams B. 0.5133 grams C. 0.8391 grams D. 1.5 grams E. 0.06 grams 11.7.8 Perform hypothesis tests for means. 6. A pharmaceutical company wants to answer the question whether it takes longer than 45 seconds for a drug in pill form to dissolve in the gastric juices of the stomach. A sample was taken from patients taken the given drug in pill form and times for the pills to be dissolved were measured. State the hypotheses to test this question. A. 0 AH : = 45sec; H : > 45secμ μ

B. sec45:Hsec;45:H A0 ≤> μμ

C. sec45:Hsec;45:H A0 ≤= μμ

D. 0 45 45AH : sec;H : secμ ≥ μ <

E. sec45:Hsec;45:H A0 =< μμ

Quiz C 11-15


11.6.7 Check assumptions and conditions for inference. 7. A pharmaceutical company wants to answer the question whether it takes longer than 45 seconds for a drug in pill form to dissolve in the gastric juices of the stomach. A sample was taken from patients taken the given drug in pill form and times for the pills to be dissolved were measured. b. State assumptions of the test. A. The subjects must be part of a medical study group investigating the drug. B. The subjects in the sample must be randomly selected from a population. C. B and D are required conditions. D. The sample data must come from a normally distributed population of observations for the variable under study. E. None of the above. 11.5.6 Create and interpret confidence intervals for the mean. 8. A manufacturer of cheese filled ravioli supplies a pizza restaurant chain. Based on data collected from its automatic filling process, the amount of cheese inserted into the ravioli is normally distributed. To make sure that the automatic filling process is on target, quality control inspectors take a sample of 25 ravioli and measure the weight of cheese filling. They find the 99% confidence interval of 14.16 to 15.84 grams. Which of the following is the correct interpretation? A. We are 99% confident that the mean weight of cheese filling in all ravioli made by this process is between 14.16 and 15.84 grams. B. 99% of all ravioli made by this process will have cheese filling weights between 14.16 and 15.84 grams. C. The weight of cheese filling in the ravioli is between 14.16 and 15.84 grams 99% of the time. D. All of the above. E. None of the above. 11.5.2 Determine sample sizes. 9. A manufacturer of cream filled donuts wants to make sure that its automatic filling process is on target. Based on a sample of 25 donuts the mean weight of cream filling is estimated to be 15 grams with a standard deviation of 1.5 grams. However, the quality control inspectors wish to estimate the mean weight of cream filling more accurately so that they can be 99% confident that it is within 0.25 grams of the true mean. How many donuts should they sample? A. 150 B. 543 C. 212 D. 410 E. 239



11.7.8 Perform hypothesis tests for means. 10. A pharmaceutical company wants to answer the question whether it takes longer than 45 seconds for a drug in pill form to dissolve in the gastric juices of the stomach. A sample was taken from 8 patients taking the given drug in pill form and times for the pills to be dissolved were measured. The mean was 45.212 sec for the sample data with a standard error of 0.580. Determine the P-value for this test. A. 0.181 B. 0.724 C. 0.649 D. 0.362 E. 0.366

Quiz C 11-17


Chapter 11: Confidence Intervals and Hypothesis Tests for Means – Quiz C – Key

1. B 2. C 3. D 4. A 5. B 6. A 7. C 8. A 9. E 10. D



Chapter 11: Confidence Intervals and Hypothesis Tests for Means – Quiz D – Multiple Choice Name ________________________________________ 11.1.1 Examine the sampling distribution of the mean. 1. A small business ships specialty homemade candies to anywhere in the world. Past records indicate that the weight of orders is normally distributed. Suppose a random sample of 16 orders is selected and each is weighed. The sample mean was found to be 110 grams with a standard deviation of 14 grams. Which of the following statements is true? A. The sampling distribution for the sample mean follows the t-distribution with 16 degrees of freedom. B. The sampling distribution for the sample mean follows the t-distribution with 15 degrees of freedom. C. The mean of the sampling distribution for the sample mean is 110 grams. D. The sampling distribution for the sample mean is Normal with a mean of 110 grams and standard deviation of 14 grams. E. The standard error is 14 grams. 11.2.3 Calculate and describe the standard error of the mean. 2. A small business ships specialty homemade candies to anywhere in the world. Past records indicate that the weight of orders is normally distributed. Suppose a random sample of 16 orders is selected and each is weighed. The sample mean was found to be 110 grams with a standard deviation of 14 grams. The standard error of the mean is A. 14 grams B. 0.875 grams C. .35 grams D. 3.5 grams E. 27.5 grams 11.5.6 Create and interpret confidence intervals for the mean. 3. A small business ships specialty homemade candies to anywhere in the world. Past records indicate that the weight of orders is normally distributed. Suppose a random sample of 16 orders is selected and each is weighed. The sample mean was found to be 110 grams with a standard deviation of 14 grams. The 90% confidence interval for the true mean weight of orders is A. 103.87 to 116.14 grams. B. 86.046 to 133.954 grams. C. 99.5 to 120.5 grams. D. 102.55 to 117.45 grams. E. 103.14 to 116.86 grams.

Quiz D 11-19


11.4.5 Find critical values of t. 4. Grandma Gertrude’s Chocolates, a family owned business, has an opportunity to supply its product for distribution through a large coffee house chain. However, the coffee house chain has certain specifications regarding cacao content as it wishes to advertise the health benefits (antioxidants) of the chocolate products it sells. In order to determine the mean % cacao in its dark chocolate products, quality inspectors sample 36 pieces. They find a sample mean of 55% with a standard deviation of 4%. The correct value of t* to construct a 90% confidence interval for the true mean % cacao is A. 2.797 B. 1.711 C. 1.690 D. 2.030 E. 1.318 11.5.6 Create and interpret confidence intervals for the mean. 5. Grandma Gertrude’s Chocolates, a family owned business, has an opportunity to supply its product for distribution through a large coffee house chain. However, the coffee house chain has certain specifications regarding cacao content as it wishes to advertise the health benefits (antioxidants) of the chocolate products it sells. In order to determine the mean % cacao in its dark chocolate products, quality inspectors sample 36 pieces. They find a sample mean of 55% with a standard deviation of 4%. What is the margin of error at 95% confidence? A. 1.13% B. 1.36% C. 0.67% D. 4% E. 1.67% 11.5.6 Create and interpret confidence intervals for the mean. 6. Grandma Gertrude’s Chocolates, a family owned business, has an opportunity to supply its product for distribution through a large coffee house chain. However, the coffee house chain has certain specifications regarding cacao content as it wishes to advertise the health benefits (antioxidants) of the chocolate products it sells. In order to determine the mean % cacao in its dark chocolate products, quality inspectors sample 36 pieces. They find a sample mean of 55% with a standard deviation of 4%. The 90% confidence interval for the true mean % cacao is A. 53.87% to 56.13%. B. 51% to 59%. C. 53.33% to 56.67%. D. 53.64% to 56.36% E. 54.33% to 55.67%.



11.5.6 Create and interpret confidence intervals for the mean. 7. A large software development firm recently relocated its facilities. Top management is interested in fostering good relations with their new local community and has encouraged their professional employees to engage in local service activities. They wish to determine the average number of hours the firm’s professionals volunteer per month. They find the 95% confidence interval of 14.62 to 15.78. Which of the following is the correct interpretation? A. We are 95% confident that the average number of hours per month volunteered by professionals employed with this firm is between 14.62 and 15.78. B. 95% of all firm’s professional employees volunteer between 14.62 and 15.78 hours per month. C. The number of hours per month volunteered by the firm’s professional employees is between 14.62 and 15.78 95% of the time. D. All of the above. E. None of the above. 11.5.6 Create and interpret confidence intervals for the mean. 8. Marcy’s Consignment shop is based in Port Angeles, Washington. It offers both male and female never or slightly used clothing and accessories on a consignment basis. Marcy’s recently redesigned their website and wants to show that sales have increased. The sales manager selects 30 sales records at random from the store’s online client list and she finds a mean increase in spending of $3.50 with a standard deviation of $7.20. The resulting P-value for anα of 0.05 is A. 0.012 B. 0.315 C. 0.630 D. 0.006 E. 0.024 11.5.2 Determine sample sizes. 9. Top management at a large software company wishes to estimate the average number of hours its firm’s professional employees volunteer in the local community. Based on past similar studies, the standard deviation was found to be 2.22 hours. If top management wants to estimate the average number of hours volunteered per month by their professional staff to within one hour with 99% confidence, how many randomly selected professional employees would they need to sample? A. 19 B. 25 C. 54 D. 44 E. 33

Quiz D 11-21


11.7.8 Perform hypothesis tests for means. 10. For a one-sided alternative hypothesis, the critical t value for an α of 0.02 and df of 19 is A. 1.734 B. 2.093 C. 2.540 D. 1.729 E. 3.883



Chapter 11: Confidence Intervals and Hypothesis Tests for Means – Quiz D – Key

1. B 2. D 3. A 4. C 5. B 6. A 7. A 8. D 9. E 10. B

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