ECON-E 370 Exam 3 Prep (HWs 6,7,8; quizzes; quick checks)

Exhibit: Job Outsourcing. Among the trends reshaping the U.S. workplace, a large proportion of Americans see outsourcing of jobs as a negative rather than a positive force when it comes to their livelihoods. To check this trend, a researcher selected a sample of 200 workers and conducted a survey. In the survey, she asked how the outsourcing of jobs to other countries affected the jobs or careers of the surveyed individuals. The responses were recorded as "No Effect" if outsourcing of jobs hasn't made much of a difference, "Hurt" if it hurt a job or career, or "Help" if outsourcing was seen as a positive trend helping a job. - 60 Hurt - 14 Help - 126 No Effect Compute the 90% confidence interval for p, the population proportion of those who view outsourcing of jobs as a negative trend hurting their jobs. Provide the lower bound of the interval below (round your answer to 4 decimal places).

0.2467

Suppose that you are working on an upper tail test. As the test statistic becomes larger, the p-value - becomes larger - becomes negative - stays the same, since the sample size has not been changed - gets smaller

gets smaller

Suppose that you set up the hypothesis test: H0:μ≤800; H1:μ>800 Given the population standard deviation, the sample size of 59 and the sample mean of 879, you calculate the value of the test statistic zx¯ and determine the p-value = 0.017 for the test. This p-value means that - if the population mean was equal to 800, then there is a 1.7% probability of obtaining the sample mean of 879 or lower in the sample of 59 observations. - there is a 1.7% probability that the population mean is above 800. - if the population mean was equal to 800, then there is a 1.7% probability of obtaining the sample mean of 879 or greater in the sample of 59 observations. - there is a 1.7% probability that the population mean is below 800.

if the population mean was equal to 800, then there is a 1.7% probability of obtaining the sample mean of 879 or greater in the sample of 59 observations.

Think about the hypothesis test: H0:μ≥42; H1:μ<42. A sample of 49 observations provides a sample mean of 38 and a sample standard deviation of 7. The null hypothesis will be rejected if - zx¯<−zα - tx¯<−tα - tx¯>−tα - tx¯>tα

tx¯<−tα

A university is interested in promoting graduates of its honors program by establishing that the mean GPA of these graduates exceeds 3.50. A sample of 36 honors students found to have a mean GPA equal to 3.60 and the standard deviation equal to 0.40. You are looking to conduct a hypothesis test to establish that the mean GPA exceeds 3.50. If you used a critical value approach to test the hypothesis in this problem, the null hypothesis would be rejected if 1. tx¯>ta 2. tx¯<−tα 3. zx¯>zα 4. zx¯<zα

tx¯>ta with an upper-tail test, we reject the null if the test statistic is ABOVE the critical value

In a one-tailed hypothesis test (upper tail), the z test statistic is determined to be 0.95. The p-value for this test is

0.1711

Exhibit: Average Income. The Bureau of Labor Statistics reported that the average yearly income of dentists in the year 2019 was $130,000. A sample of 36 dentists, which was taken in 2020, showed an average yearly income of $125,000. Assume the standard deviation of the population of dentists in 2020 is $25,000. Round your solutions for this Exhibit to 4 decimal places. Compute the test statistic.

-1.2

Exhibit: Female Employees. Last year, 50% of MNM, Inc. employees were female. It is believed that there has been a reduction in the percentage of females in the company. This year, in a random sample of 500 employees, 230 were female. Based on this information, you want to determine if there has been a decrease in the percentage of females in the company (at the significance level of 5%). Round your solutions for this Exhibit to 4 decimal places. Compute the test statistic.

-1.7889

Consider the following hypothesis test: H0:μ≥42; H1:μ<42 A sample of 49 observations provides a sample mean of 38 and a sample standard deviation of 7. Compute the value of the test statistic. - not possible because population standard deviation is not known - 4 - -4 - -2

-4

In a one-tailed hypothesis test (lower tail), the t test statistic is determined to be -2.75. The sample size is 45. The p-value for this test is

0.0043

Exhibit: Costco Customers. Customers at Costco spend an average of $150 per trip. One of Costco's rivals would like to determine whether Costco's customers spend more per trip. A survey of the receipts of 28 customers found that the sample mean was $156. Assume that the population standard deviation of spending is $13.50 and the spending follows a normal distribution (use the significance level 0.1). Round your solutions for this Exhibit to 4 decimal places. Calculate the p-value for the test.

0.0093

In a two-tailed hypothesis test, the t test statistic is determined to be -2.7. The degrees of freedom for the test is equal 15. The p-value for this test is

0.0165

In a two-tailed hypothesis test, the t test statistic is determined to be 2.45. The sample size is 25. The p-value for this test is

0.022

In a one-tailed hypothesis test (lower tail), the z test statistic is determined to be -1.9. The p-value for this test is

0.0287

In a one-tailed hypothesis test (upper tail), the t test statistic is determined to be 1.95. The sample size is 25. The p-value for this test is

0.0315

If an interval estimate is said to be constructed at the 95% confidence level, the significance level is - 0.975 - 0.95 - 0.025 - 0.05

0.05

Exhibit: Average Debt Load. The Department of Education would like to check if the average debt load of graduating students with a bachelor's degree is different from $23,000. A random sample of 34 students had an average debt load of $24,500. It is believed that the population standard deviation for student debt load is $4,600. The significance level α is set to 0.02 for the hypothesis test. Round your solutions for this Exhibit to 4 decimal places. Calculate the p-value for the test.

0.0572

Exhibit: Tires Life Expectancy. A tire manufacturer has been producing tires with an average life expectancy of 25,000 miles. Now the company is advertising that its new tires' life expectancy has increased. In order to test the legitimacy of the advertising campaign, an independent testing agency tested a sample of 6 of their tires and has provided the following data. Data: n = 6, mu_H0 = 25, xbar = 26.1667, s = 1.722401 Raw data: 28, 27, 25, 28, 24, 25 Compute the p-value for the test.

0.079

In a two-tailed hypothesis test, the z test statistic is determined to be -1.75. The p-value for this test is

0.0801

Exhibit: Community Banks. The American Bankers Association Community Bank Insurance Survey for 2017 had responses from 123 banks. Of these, 64 were community banks, defined to be banks with assets of $1 billion or less. Compute the margin of error for the 95% confidence interval for the population proportion of community banks (round your answer to 4 decimal places).

0.0883

In a two-tailed hypothesis test, the z test statistic is determined to be 1.5. The p-value for this test is

0.1336

Exhibit: Job Outsourcing. Among the trends reshaping the U.S. workplace, a large proportion of Americans see outsourcing of jobs as a negative rather than a positive force when it comes to their livelihoods. To check this trend, a researcher selected a sample of 200 workers and conducted a survey. In the survey, she asked how the outsourcing of jobs to other countries affected the jobs or careers of the surveyed individuals. The responses were recorded as "No Effect" if outsourcing of jobs hasn't made much of a difference, "Hurt" if it hurt a job or career, or "Help" if outsourcing was seen as a positive trend helping a job. - 60 Hurt - 14 Help - 126 No Effect What is the point estimate of p, the population proportion of those who view outsourcing of jobs as a negative trend hurting their jobs? - Not possible to say because the population proportion is not known - 0.3 - 0.63 - 60

0.3

Exhibit: Job Outsourcing. Among the trends reshaping the U.S. workplace, a large proportion of Americans see outsourcing of jobs as a negative rather than a positive force when it comes to their livelihoods. To check this trend, a researcher selected a sample of 200 workers and conducted a survey. In the survey, she asked how the outsourcing of jobs to other countries affected the jobs or careers of the surveyed individuals. The responses were recorded as "No Effect" if outsourcing of jobs hasn't made much of a difference, "Hurt" if it hurt a job or career, or "Help" if outsourcing was seen as a positive trend helping a job. - 60 Hurt - 14 Help - 126 No Effect Compute the 90% confidence interval for p, the population proportion of those who view outsourcing of jobs as a negative trend hurting their jobs. Provide the upper bound of the interval below (round your answer to 4 decimal places).

0.3533

Exhibit: Community Banks. The American Bankers Association Community Bank Insurance Survey for 2017 had responses from 123 banks. Of these, 64 were community banks, defined to be banks with assets of $1 billion or less. Provide the lower bound of the 95% confidence interval for the population proportion of community banks (round your answer to 4 decimal places).

0.432

Exhibit: Community Banks. The American Bankers Association Community Bank Insurance Survey for 2017 had responses from 123 banks. Of these, 64 were community banks, defined to be banks with assets of $1 billion or less. Provide the upper bound of the 95% confidence interval for the population proportion of community banks (round your answer to 4 decimal places).

0.6086

A random sample of 92 students at a university showed an average age of 22 years and a sample standard deviation of 3 years. Find the margin of error for the 96% confidence interval (round your answer to 4 decimal places).

0.6517

For a one-tail test at 12.45% significance level, the critical value of z is

1.1528

Exhibit: Average Income. The Bureau of Labor Statistics reported that the average yearly income of dentists in the year 2019 was $130,000. A sample of 36 dentists, which was taken in 2020, showed an average yearly income of $125,000. Assume the standard deviation of the population of dentists in 2020 is $25,000. Round your solutions for this Exhibit to 4 decimal places. Calculate the critical value for the test.

1.6449

Exhibit: Female Employees. Last year, 50% of MNM, Inc. employees were female. It is believed that there has been a reduction in the percentage of females in the company. This year, in a random sample of 500 employees, 230 were female. Based on this information, you want to determine if there has been a decrease in the percentage of females in the company (at the significance level of 5%). Round your solutions for this Exhibit to 4 decimal places. Calculate the critical value for the test.

1.6449

Exhibit: Tires Life Expectancy. A tire manufacturer has been producing tires with an average life expectancy of 25,000 miles. Now the company is advertising that its new tires' life expectancy has increased. In order to test the legitimacy of the advertising campaign, an independent testing agency tested a sample of 6 of their tires and has provided the following data. Data: n = 6, mu_H0 = 25, xbar = 26.1667, s = 1.722401 Raw data: 28, 27, 25, 28, 24, 25 Compute the test statistic.

1.6592

For a two-tailed test at 9.47% significance level, the critical value of z is

1.6711

For a one-tailed test and a sample size of 25 observations at 5% significance level, the critical value of t is (Hint: you are given the sample size, not the degrees of freedom)

1.7109

For a two-tailed test and a sample of 15 observations at 10% significance level, the critical value of t is (Hint: you are given the sample size, not the degrees of freedom)

1.7613

Exhibit: Average Debt Load. The Department of Education would like to check if the average debt load of graduating students with a bachelor's degree is different from $23,000. A random sample of 34 students had an average debt load of $24,500. It is believed that the population standard deviation for student debt load is $4,600. The significance level α is set to 0.02 for the hypothesis test. Round your solutions for this Exhibit to 4 decimal places. Compute the test statistic.

1.9014

Exhibit: Check Out Times. The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.2 minutes with a standard deviation of 0.5 minutes. We want to test to determine whether or not the mean waiting time of all customers is any different from 3 minutes (use the significance level 0.05 for the test). Round your solutions for this Exhibit to 4 decimal places. Calculate the critical value for the test.

1.9842

A population has a standard deviation of 90. A random sample of 300 items from this population is selected. The sample mean is determined to be 250. Find the margin of error for the 96% confidence level (round your answer to 4 decimal places).

10.6717

In order to determine a confidence interval for the mean of a population with an unknown standard deviation, a sample of 191 items is selected. The mean of the sample is determined to be 28. The number of degrees of freedom for the critical t value is - 191 - 190 - 27 - 28

190

Exhibit: Average Debt Load. The Department of Education would like to check if the average debt load of graduating students with a bachelor's degree is different from $23,000. A random sample of 34 students had an average debt load of $24,500. It is believed that the population standard deviation for student debt load is $4,600. The significance level α is set to 0.02 for the hypothesis test. Round your solutions for this Exhibit to 4 decimal places. Calculate the critical value for the test.

2.3263

Exhibit: Costco Customers. Customers at Costco spend an average of $150 per trip. One of Costco's rivals would like to determine whether Costco's customers spend more per trip. A survey of the receipts of 28 customers found that the sample mean was $156. Assume that the population standard deviation of spending is $13.50 and the spending follows a normal distribution (use the significance level 0.1). Round your solutions for this Exhibit to 4 decimal places. Compute the test statistic.

2.4

Find the critical z-value for a 98.8% confidence interval (round your answer to 4 decimal places).

2.5121

Find the critical t-value for a 98% confidence interval estimation if the sample size equals 15 (round your answer to 4 decimal places).

2.6245

A simple random sample of 5 observations from a population containing 200 elements was taken, and the following values were obtained: 26 11 26 12 25 A point estimate of the population mean is: - 5 - 20 - 26 - 200

20

The supervisor of a production line wants to check if the average time to assemble an electronic component is different from 14 minutes. Assume that the population of assembly time is normally distributed with a standard deviation of 3.4 minutes. The supervisor times the assembly of 14 components, and finds that the average time for completion is 11.6 minutes. How would you calculate the p-value for the hypothesis test? - 2×P(z<−2.64) - P(z<−2.64) - P(−2.64<z<2.64) - P(z>2.64)

2×P(z<−2.64)

Find the critical t-value for a 98.8% confidence interval estimation using 9 degrees of freedom (round your answer to 4 decimal places).

3.1363

Exhibit: Check Out Times. The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.2 minutes with a standard deviation of 0.5 minutes. We want to test to determine whether or not the mean waiting time of all customers is any different from 3 minutes (use the significance level 0.05 for the test). Round your solutions for this Exhibit to 4 decimal places. Compute the test statistic.

4

Exhibit: Soft Drinks. Last year, a soft drink manufacturer had 21% of the market. In order to increase their portion of the market, the manufacturer has introduced a new flavor in their soft drinks. A sample of 400 individuals participated in the taste test and 120 indicated that they like the taste. We are interested in determining if more than 21% of the population will like the new soft drink at the significance level 0.1. Round your solutions for this Exhibit to 4 decimal places. Compute the test statistic.

4.4193

An accountant claims to be able to complete a standard tax return in under an hour. For a random sample of 27 tax returns, the accountant averaged 69.0 minutes. Assume that the population standard deviation is equal to 10.4 minutes and the population is normally distributed. What is the test statistic for the hypothesis test to check accountant's claim? - 4.5 - -4.5 - 0.87 - -0.87

4.5

The screening process for detecting a rare disease is not perfect. Researchers have developed a blood test that is considered fairly reliable. It gives a positive reaction in 97% of the people who have a disease. However, it erroneously gives a positive reaction for 5% of the people who do not have the disease. Consider the null hypothesis "the individual does not have the disease". What is the probability of Type I error if the new blood test is used? - 97% - 95% - 3% - 5%

5%

Exhibit: One-Bedroom Rental Apartment. Realtor.com contains several hundred postings for one-bedroom apartments in Bloomington, IN. You chose 40 postings at random and calculated a mean monthly rent of $879 and a standard deviation of $168 based on this sample data. (For this exhibit, avoid rounding on the intermediate steps and round your final answer to 4 decimal places) Construct the 90% confidence interval for the mean monthly rent of all advertised one-bedroom apartments in Bloomington and provide the lower bound of the confidence interval.

834.2444

Exhibit: Department Store. A random sample of 84 credit sales in a department store showed an average sale of $90.00. From past data, it is known that the standard deviation of the population of sales is $30.00. (For this exhibit, avoid rounding on the intermediate steps and round your final answer to 4 decimal places) Construct the 85% confidence interval for the population average credit sale and provide the lower bound of the confidence interval.

85.288

A simple random sample of 81 observations was taken from a large population. The sample mean and the standard deviation were determined to be 405 and 81 respectively. The estimated standard error of the mean is - 5 - 1 - 45 - 9

9

Exhibit: One-Bedroom Rental Apartment. Realtor.com contains several hundred postings for one-bedroom apartments in Bloomington, IN. You chose 40 postings at random and calculated a mean monthly rent of $879 and a standard deviation of $168 based on this sample data. (For this exhibit, avoid rounding on the intermediate steps and round your final answer to 4 decimal places) Provide the upper bound of the confidence interval that you calculated above (i.e. the upper bound of the 90% confidence interval for the mean monthly rent for all advertised on-bedroom apartments).

923.7556

Exhibit: Department Store. A random sample of 84 credit sales in a department store showed an average sale of $90.00. From past data, it is known that the standard deviation of the population of sales is $30.00. (For this exhibit, avoid rounding on the intermediate steps and round your final answer to 4 decimal places) Provide the upper bound of the confidence interval that you calculated above (i.e. the upper bound of the 85% confidence interval for the mean credit sale in the department store).

94.712

When the following hypotheses are being tested at a level of significance α H0:μ≥500; H1:μ<500 the null hypothesis will be rejected if the p-value is - <α - >α/2 - ≤1−α/2 - >α

<α

A university is interested in promoting graduates of its honors program by establishing that the mean GPA of these graduates exceeds 3.50. A sample of 36 honors students is taken and is found to have a mean GPA equal to 3.60. The population standard deviation is assumed to equal 0.40. To establish whether the mean GPA exceeds 3.50, you conduct a hypothesis test and calculate the value of the test statistic equal to 1.5. How would you calculate the p-value in Excel? 1. =2*(1 - NORM.S.DIST(1.5, TRUE)) 2. =2*NORM.S.DIST(1.5, TRUE) 3. =1 - NORM.S.DIST(1.5, TRUE) 4. =NORM.S.DIST(1.5, TRUE)

=1 - NORM.S.DIST(1.5, TRUE)

A university is interested in promoting graduates of its honors program by establishing that the mean GPA of these graduates exceeds 3.50. A sample of 36 honors students is taken and is found to have a mean GPA equal to 3.60. The population standard deviation is assumed to equal 0.40. To establish whether the mean GPA exceeds 3.50, you conduct a hypothesis test. How would you calculate the critical value in Excel at a 1% significance level? 1. =1 - NORM.S.INV(0.01) 2. =1 - ABS(NORM.S.INV(0.01)) 3. =ABS(NORM.S.INV(0.01)) 4. =NORM.S.INV(0.01)

=ABS(NORM.S.INV(0.01))

A university interested in tracking its honors program believes that the proportion of graduates with a GPA of 3.00 or below is less than 0.20. In a sample of 200 graduates, 30 students have a GPA of 3.00 or below. In testing the university's belief, you calculated the test statistic to be equal to -1.77. How would you calculate the associated p-value in Excel? 1. =NORM.S.DIST(−1.77,TRUE) 2. =T.DIST(−1.77,29,TRUE) 3. =1 - NORM.S.DIST(−1.77,TRUE) 4. =T.DIST(−1.77,199,TRUE)

=NORM.S.DIST(−1.77,TRUE)

A university is interested in promoting graduates of its honors program by establishing that the mean GPA of these graduates exceeds 3.50. A sample of 36 honors students found to have a mean GPA equal to 3.60 and the standard deviation equal to 0.40. You are looking to conduct a hypothesis test to establish that the mean GPA exceeds 3.50. Based on the given information, you calculate the value of the test statistic equal to 1.5. How would you calculate the p-value in Excel? 1. =1 - NORM.S.DIST(1.5, TRUE) 2. =NORM.S.DIST(1.5, TRUE) 3. =T.DIST.RT(1.5, 35) 4. =T.DIST(1.5, 36)

=T.DIST.RT(1.5, 35)

Suppose that you set up the hypothesis test: H0:μ≤20; H1:μ>20 A sample of 26 observations provides a sample mean of 22.4. Assume that the population standard deviation is known and equals 3. The hypothesis test can lead to an erroneous conclusion because - All suggested alternatives are correct - The sample size is not large enough to apply the Central Limit Theorem - The sampling distribution of x¯ may be not a normal distribution - The population may have a distribution different from the normal distribution

All suggested alternatives are correct

Yes, the population data needs to be normally distributed because the sample is too small to apply CLT.

Exhibit: Business Bankruptcies in Canada. Business bankruptcies in Canada are monitored by the Office of the Superintendent of Bankruptcy Canada (OSB). Included in each report are the assets and liabilities the company declared at the time of bankruptcy filing. Suppose a sample of 9 reports from the province of Ontario was selected. The companies in the sample declared the following debt (liabilities minus assets): Are there any additional assumptions needed in this problem to ensure that the computed confidence interval is reliable? - No additional assumptions needed. - Yes, the sample size has to be increased. With the current sample size the confidence interval cannot be reliable. - Yes, the population standard deviation should be known. - Yes, the population data needs to be normally distributed because the sample is too small to apply CLT.

0.2728

Exhibit: Business Bankruptcies in Canada. Business bankruptcies in Canada are monitored by the Office of the Superintendent of Bankruptcy Canada (OSB). Included in each report are the assets and liabilities the company declared at the time of bankruptcy filing. Suppose a sample of 9 reports from the province of Ontario was selected. The companies in the sample declared the following debt (liabilities minus assets): Compute the 99% confidence interval for the average debt of Ontario companies at the time of filing. Provide the lower bound of the interval.

6.7072

Exhibit: Business Bankruptcies in Canada. Business bankruptcies in Canada are monitored by the Office of the Superintendent of Bankruptcy Canada (OSB). Included in each report are the assets and liabilities the company declared at the time of bankruptcy filing. Suppose a sample of 9 reports from the province of Ontario was selected. The companies in the sample declared the following debt (liabilities minus assets): Compute the 99% confidence interval for the average debt of Ontario companies at the time of filing. Provide the upper bound of the interval.

$ million

Exhibit: Business Bankruptcies in Canada. Business bankruptcies in Canada are monitored by the Office of the Superintendent of Bankruptcy Canada (OSB). Included in each report are the assets and liabilities the company declared at the time of bankruptcy filing. Suppose a sample of 9 reports from the province of Ontario was selected. The companies in the sample declared the following debt (liabilities minus assets): What is the unit of measurement for the confidence interval you calculated above? - $ - % - $ million - unitless

Exhibit: Average Income. The Bureau of Labor Statistics reported that the average yearly income of dentists in the year 2019 was $130,000. A sample of 36 dentists, which was taken in 2020, showed an average yearly income of $125,000. Assume the standard deviation of the population of dentists in 2020 is $25,000. Round your solutions for this Exhibit to 4 decimal places. We want to test to determine if there has been a decrease in the average yearly income of dentists (at the significance level 0.05). Construct the null and the alternative hypotheses for this test choosing the fields below:

H0: μ ≥ 130,000 H1: μ < 130,000

Representatives of a large national union announced that the fraction of women in the union was equal to one-half in the previous year. You are interested in testing whether there has been a change in the fraction of women this year. How would you set up the hypothesis test? - H0:p≤0.5; H1:p>0.5 - H0:p≠0.5; H1:p=0.5 - H0:μ≥0.5; H1:μ<0.5 - H0:p=0.5; H1:p≠0.5

H0:p=0.5; H1:p≠0.5

Exhibit: Soft Drinks. Last year, a soft drink manufacturer had 21% of the market. In order to increase their portion of the market, the manufacturer has introduced a new flavor in their soft drinks. A sample of 400 individuals participated in the taste test and 120 indicated that they like the taste. We are interested in determining if more than 21% of the population will like the new soft drink at the significance level 0.1. Round your solutions for this Exhibit to 4 decimal places. Provide the null and the alternative hypotheses. - H0:p≥0.21;H1:p<0.21 - H0:p≤0.21;H1:p>0.21 - H0:p<0.21;H1:p≥0.21 - H0:p>0.21;H1:p≤0.21

H0:p≤0.21;H1:p>0.21

The National Association of Realtors reported that 28% of home buyers in the state of Florida were foreigners in 2019. In 2021, a group of students from Indiana University conducted a study and, based on a sample of 100 people, concluded that 30% of home buyers in the state of Florida are foreigners. So, the members of the group think that the proportion has increased since then. The correct hypothesis statement for the student's group to test is - H0:p≤0.28;H1:p>0.28 - H0:p≤0.30;H1:p>0.30 - H0:p≥0.28;H1:p<0.28 - H0:p≥0.30;H1:p<0.30

H0:p≤0.28;H1:p>0.28

A university interested in tracking its honors program believes that the proportion of graduates with a GPA of 3.00 or below is less than 0.20. In a sample of 200 graduates, 30 students have a GPA of 3.00 or below. In testing the university's belief, the appropriate hypotheses are - H0:p≥0.2; H1:p<0.2 - H0:p≤0.2; H1:p>0.2 - H0:p¯≥0.2; H1:p¯<0.2 - H0:μ≤0.2; H1:μ>0.2

H0:p≥0.2; H1:p<0.2 NOT pbar

Exhibit: Female Employees. Last year, 50% of MNM, Inc. employees were female. It is believed that there has been a reduction in the percentage of females in the company. This year, in a random sample of 500 employees, 230 were female. Based on this information, you want to determine if there has been a decrease in the percentage of females in the company (at the significance level of 5%). Round your solutions for this Exhibit to 4 decimal places. Provide the null and the alternative hypotheses. - H0:μ≥0.5;H1:μ<0.5 - H0:p≥50;H1:p<50 - H0:p≥0.5;H1:p<0.5 - H0:p¯≥0.46;H1:p¯<0.46

H0:p≥0.5;H1:p<0.5

Exhibit: Average Debt Load. The Department of Education would like to check if the average debt load of graduating students with a bachelor's degree is different from $23,000. A random sample of 34 students had an average debt load of $24,500. It is believed that the population standard deviation for student debt load is $4,600. The significance level α is set to 0.02 for the hypothesis test. Round your solutions for this Exhibit to 4 decimal places. Provide the null and the alternative hypotheses. - H0:μ=$24,500;H1:μ≠$24,500 - H0:μ≠$23,000;H1:μ=$23,000 - H0:μ≠$24,500;H1:μ=$24,500 - H0:μ=$23,000;H1:μ≠$23,000

H0:μ=$23,000;H1:μ≠$23,000

Travelocity would like to test if the average roundtrip airfare between New York and London differs from $1,400. The correct hypothesis statement for this test would be - H0:μ=1,400; H1:μ≠1,400 - H0:μ≠1,400; H1:μ=1,400 - H0:μ<1,400; H1:μ≥1,400 - H0:μ≥1,400; H1:μ<1,400

H0:μ=1,400; H1:μ≠1,400

A production line operation is designed to fill cartons with laundry detergent to a mean weight of 32 ounces. A sample of cartons is periodically selected and weighted to determine whether overfilling or underfilling is occurring. If the sample data led to a conclusion of underfilling or overfilling, the production line would be shut down and adjusted to obtain a proper filling. The correct hypothesis statement for this test would be - H0:μ=32; H1:μ≠32 - We cannot develop a hypothesis test to help determine whether overfilling or underfilling is occurring - H0:μ≠32; H1:μ=32 - Either H0:μ=32; H1:μ≠32 or

H0:μ=32; H1:μ≠32

Exhibit: Check Out Times. The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.2 minutes with a standard deviation of 0.5 minutes. We want to test to determine whether or not the mean waiting time of all customers is any different from 3 minutes (use the significance level 0.05 for the test). Round your solutions for this Exhibit to 4 decimal places. Provide the null and the alternative hypotheses. - H0:μ≠3;H1:μ=3 - H0:μ=3;H1:μ≠3 - H0:μ≠3.1;H1:μ=3.1 - H0:μ=3.1;H1:μ≠3.1

H0:μ=3;H1:μ≠3

Which of the following statements are valid null and alternative hypotheses? - H0:μ≤10; H1:μ>10 - H0: xbar ≤10; H1: xbar >10 - H0:μ≤10; H1:μ>12 - H0:μ≤10; H1:μ≥10

H0:μ≤10; H1:μ>10

Exhibit: Costco Customers. Customers at Costco spend an average of $150 per trip. One of Costco's rivals would like to determine whether Costco's customers spend more per trip. A survey of the receipts of 28 customers found that the sample mean was $156. Assume that the population standard deviation of spending is $13.50 and the spending follows a normal distribution (use the significance level 0.1). Round your solutions for this Exhibit to 4 decimal places. Provide the null and the alternative hypotheses. - H0:μ≥150;H1:μ<150 - H0:μ≤156;H1:μ>156 - H0:μ≤150;H1:μ>150 - H0:μ=150;H1:μ≠150

H0:μ≤150;H1:μ>150

Exhibit: Tires Life Expectancy. A tire manufacturer has been producing tires with an average life expectancy of 25,000 miles. Now the company is advertising that its new tires' life expectancy has increased. In order to test the legitimacy of the advertising campaign, an independent testing agency tested a sample of 6 of their tires and has provided the following data. Data: n = 6, mu_H0 = 25, xbar = 26.1667, s = 1.722401 Raw data: 28, 27, 25, 28, 24, 25 Provide the null and the alternative hypotheses. - H0:μ<25;H1:μ≥25 - H0:μ≤25;H1:μ>25 - H0:μ=25;H1:μ≠25 - H0:μ≥25;H1:μ<25

H0:μ≤25;H1:μ>25

A university is interested in promoting graduates of its honors program by establishing that the mean GPA of these graduates exceeds 3.50. A sample of 36 honors students is taken and is found to have a mean GPA equal to 3.60. To establish whether the mean GPA exceeds 3.50, the appropriate hypotheses are - H0:μ≥3.50; H1:μ<3.50 - H0:μ≤3.50; H1:μ>3.50 - H0:μ<3.50; H1:μ≥3.50 - H0:x¯≤3.50; H1:x¯>3.50

H0:μ≤3.50; H1:μ>3.50

A golfer claims that his average golf score at the course he plays regularly is less than 90. The correct hypothesis statement for this golfer to prove his claim would be - H0:μ≠90; H1:μ=90 - H0:μ≥90; H1:μ<90 - H0:μ=90; H1:μ≠90 - H0:μ≤90; H1:μ≥90

H0:μ≥90; H1:μ<90

A university interested in tracking its honors program believes that the proportion of graduates with a GPA of 3.00 or below is less than 0.20. In a sample of 200 graduates, 30 students have a GPA of 3.00 or below. In testing the university's belief, you calculated the p-value to be equal to 0.0385. How would you interpret this p-value? - If the actual proportion of graduates with a GPA of 3.00 or below was 0.20, then there would be a 3.85% chance of observing that a sample of 200 graduates has a proportion of graduates with GPA of 3.00 or below equal to 15% or less. - If the actual proportion of graduates with a GPA of 3.00 or below was 0.20, then there would be a 3.85% chance of observing that a sample of 200 graduates has a proportion of graduates with GPA of 3.00 or below equal to 15% or more. - There is a 3.85% chance that the population proportion of graduates with GPA of 3.00 or below is less than 0.2. - There is a 3.85% chance that the population proportion of graduates with GPA of 3.00 or below is greater than 0.2.

If the actual proportion of graduates with a GPA of 3.00 or below was 0.20, then there would be a 3.85% chance of observing that a sample of 200 graduates has a proportion of graduates with GPA of 3.00 or below equal to 15% or less.

A university is interested in promoting graduates of its honors program by establishing that the mean GPA of these graduates exceeds 3.50. A sample of 36 honors students is taken and is found to have a mean GPA equal to 3.60. The population standard deviation is assumed to equal 0.40. Suppose that, with the given information, you calculated that the p-value for the test is equal to 0.067. How would you interpret this p-value? - There is a 6.7% chance that the mean GPA is smaller 3.50. - If the mean GPA was equal to 3.50, then there would be a 6.7% chance of observing a mean GPA in a sample of 36 students equal to 3.60 or more. - If the mean GPA was equal to 3.50, then there would be a 6.7% chance of observing a mean GPA in a sample of 36 students equal to 3.60 or less. - There is a 6.7% chance that the mean GPA exceeds 3.50.

If the mean GPA was equal to 3.50, then there would be a 6.7% chance of observing a mean GPA in a sample of 36 students equal to 3.60 or more. "Recall that the p-value is the probability of observing the sample results at least as extreme as a selected sample (in the direction of the alternative) if the null hypothesis is true with equality."

For a given confidence level and sample size, which of the following is true in the interval estimation of the population mean when σ is known? - If the population standard deviation is larger, the interval is wider. - If the population is larger, the interval is narrower. - If the sample standard deviation is smaller, the interval is wider. - If the population standard deviation is smaller, the interval is wider.

If the population standard deviation is larger, the interval is wider.

Exhibit: Female Employees. Last year, 50% of MNM, Inc. employees were female. It is believed that there has been a reduction in the percentage of females in the company. This year, in a random sample of 500 employees, 230 were female. Based on this information, you want to determine if there has been a decrease in the percentage of females in the company (at the significance level of 5%). Round your solutions for this Exhibit to 4 decimal places. Suppose that, instead of the critical value approach, you decided to use the p-value approach and found that the p-value for the test in this problem is equal to 0.0368. This p-value is best interpreted as the following: - If the true proportion of female employees in the company was 50%, then there would be a 3.68% chance of observing a proportion of females equal to or above 46% in the sample of 500 individuals. - There is a 3.68% chance that the proportion of females in the population is below 50%. - There is a 3.68% chance of observing a proportion of females equal to or below 50% in the sample of 500 individuals . - If the true proportion of female employees in the company was 50%, then there would be a 3.68% chance of observing a proportion of females equal to or below 46% in the sample of 500 individuals.

If the true proportion of female employees in the company was 50%, then there would be a 3.68% chance of observing a proportion of females equal to or below 46% in the sample of 500 individuals.

[$60.1875; $69.8125]

In an examination of purchasing patterns of shoppers, a sample of 16 shoppers revealed that they spent, on average, $65 per hour of shopping with a standard deviation of $11. Assuming that the amount spent per hour of shopping is normally distributed, find a 90% confidence interval for the mean amount. Consult the following table to find the critical value: - [$60.4763; $69.5238] - [$61.315; $68.685] - [$61.4745; $68.5255] - [$60.1875; $69.8125]

After computing a confidence interval, a researcher believes the results are meaningless because the width of the interval is too large. Which one of the following is the best recommendation for this researcher? - Increase the sample size. - Decrease the sample size. - Reduce the population variance. - Increase the level of confidence for the interval.

Increase the sample size.

A professor of statistics wants to test that the average amount of money a typical college student spends per day during spring break is less than $70. Based upon previous research, the population standard deviation is estimated to be $17.32. The professor surveys 35 students and finds that the mean spending is $72.43. How would you calculate the p-value for this test? - P(z<0.83) - P(z>0.83) - P(z<−0.83) - P(−0.83<z<0.83)

P(z<0.83)

What is the most typical form of a calculated confidence interval - Point estimate ± Margin of error - Population parameter ± Standard error - Population parameter ± Margin of error - Point estimate ± Standard error

Point estimate ± Margin of error

Exhibit: Average Debt Load. The Department of Education would like to check if the average debt load of graduating students with a bachelor's degree is different from $23,000. A random sample of 34 students had an average debt load of $24,500. It is believed that the population standard deviation for student debt load is $4,600. The significance level α is set to 0.02 for the hypothesis test. Round your solutions for this Exhibit to 4 decimal places. Which of the following statements is the most accurate given your p-value? Note: the choices refer to significance levels different from 0.02. So, the conclusion might be different from what you stated earlier. - Reject the null hypothesis at α = 0.10, but not at α = 0.05 - Reject the null hypothesis at α = 0.05 - Fail to reject the null hypothesis at α = 0.10 - Fail to reject the null hypothesis at α ≤ 0.10

Reject the null hypothesis at α = 0.10, but not at α = 0.05

The manager of an automobile dealership is trying to understand if a new bonus plan has increased sales. Previously, the mean sales rate per salesperson was five automobiles per month. Suppose that, the manager conducts a hypothesis test and, based on the hypothesis test, she determines that the p-value equals 0.0062. Using the significance level α=0.05, how would you state the conclusion for the test? - Since the p-value = 0.0062 < 0.05, do not reject H0. Therefore, there is an evidence that the new bonus plan is more efficient than the old one - Since the p-value = 0.0062 < 0.05, reject H0. Therefore, there is an evidence that the new bonus plan is more efficient than the old one. - Since the p-value = 0.0062 < 0.05, do not reject H0. Therefore, there is no evidence that the new bonus plan is more efficient than the old one. - Since the p-value = 0.0062 < 0.05, reject H0. Therefore, there is an evidence that the new bonus plan is no more efficient than the old one.

Since the p-value = 0.0062 < 0.05, reject H0. Therefore, there is an evidence that the new bonus plan is more efficient than the old one.

Medicare would like to test if the average monthly rate for one-bedroom assisted-living facility is different from $3,300. Suppose that you collect sample information and, based on the hypothesis test, determine that the z test statistic is equal to 1.98 while the critical value of z is 2.05. How would you state the conclusion? - Since −zα/2=−2.05<zx¯=1.98<zα/2=2.05, do not reject H0. Therefore, there is not enough evidence to conclude that the monthly rate for one-bedroom assisted-living facility is different from $3,300. - Since −zα/2=−2.05<zx¯=1.98<zα/2=2.05, reject H0. Therefore, there is not enough evidence to conclude that the monthly rate for one-bedroom assisted-living facility is different from $3,300. - Since −zα/2=−2.05<zx¯=1.98<zα/2=2.05, do not reject H0. Therefore, there is an evidence that the monthly rate for one-bedroom assisted-living facility is different from $3,300. - Since −zα/2=−2.05<zx¯=1.98<zα/2=2.05, reject H0. Therefore, there is an evidence that the monthly rate for one-bedroom assisted-living facility is different from $3,300.

Since −zα/2=−2.05<zx¯=1.98<zα/2=2.05, do not reject H0. Therefore, there is not enough evidence to conclude that the monthly rate for one-bedroom assisted-living facility is different from $3,300.

Exhibit: One-Bedroom Rental Apartment. Realtor.com contains several hundred postings for one-bedroom apartments in Bloomington, IN. You chose 40 postings at random and calculated a mean monthly rent of $879 and a standard deviation of $168 based on this sample data. (For this exhibit, avoid rounding on the intermediate steps and round your final answer to 4 decimal places) Without calculations, can you tell what would happen to the confidence interval if you were asked to calculate the 95% confidence interval instead? - The confidence interval would be narrower because the margin of error would be smaller. - The confidence interval would be wider because the margin of error would be smaller. - The confidence interval would be wider because the margin of error would be bigger. - The confidence interval would be narrower because the margin of error would be bigger.

The confidence interval would be wider because the margin of error would be bigger.

Exhibit: Soft Drinks. Last year, a soft drink manufacturer had 21% of the market. In order to increase their portion of the market, the manufacturer has introduced a new flavor in their soft drinks. A sample of 400 individuals participated in the taste test and 120 indicated that they like the taste. We are interested in determining if more than 21% of the population will like the new soft drink at the significance level 0.1. Round your solutions for this Exhibit to 4 decimal places. The p-value is best interpreted as the following: - The p-value equals the probability that the proportion of people who like the new flavor is above 21%. - The p-value indicates the probability of observing the proportion of people who like new flavor equal to 30% or lower in the sample of 400 individuals, if actual proportion of people who like the new flavor is 21%. - The p-value indicates the probability of observing the proportion of people who like new flavor equal to 21% or greater in the sample of 400 individuals, if actual proportion of people who like the new flavor is 21%. - The p-value indicates the probability of observing the proportion of people who like new flavor equal to 30% or greater in the sample of 400 individuals, if actual proportion of people who like the new flavor is 21%.

The p-value indicates the probability of observing the proportion of people who like new flavor equal to 30% or greater in the sample of 400 individuals, if actual proportion of people who like the new flavor is 21%.

Whenever estimating the confidence interval for the population proportion, which of the following is required? - The population proportion should be known. - The population data should be normally distributed. - The sample size and the population proportion should be such that np ≥ 5 and n(1 - p) ≥ 5. - The sample size should be n ≥ 30.

The sample size and the population proportion should be such that np ≥ 5 and n(1 - p) ≥ 5.

It is known that the population data is not normally distributed. Also, the population standard deviation is not known. A sample of 6 items is selected from this population to develop an interval estimate for the mean of the population (µ). Choose the correct alternative below. - The t distribution with 5 degrees of freedom must be used for constructing the confidence interval. - The sample size must be increased for constructing a reliable confidence interval. - The t distribution with 6 degrees of freedom must be used for constructing the confidence interval. - The standard normal distribution may be used for constructing the confidence interval.

The sample size must be increased for constructing a reliable confidence interval.

Exhibit: Department Store. A random sample of 84 credit sales in a department store showed an average sale of $90.00. From past data, it is known that the standard deviation of the population of sales is $30.00. (For this exhibit, avoid rounding on the intermediate steps and round your final answer to 4 decimal places) Assume that the information about the average sale was lost. You decided to repeat the study and collect information for another sample. This time you obtain a random sample of 400 sales which accidentally resulted in the same average sale of $90.00. Without calculations, can you tell what would be the effect of the new data on the standard of error of the mean sale and the 85% confidence interval? - The standard error of the mean sale will decrease and the confidence interval will become wider. - The standard error of the mean sale will increase and the confidence interval will become wider. - The standard error of the mean sale will increase and the confidence interval will become narrower. - The standard error of the mean sale will decrease and the confidence interval will become narrower.

The standard error of the mean sale will decrease and the confidence interval will become narrower.

Exhibit: Average Income. The Bureau of Labor Statistics reported that the average yearly income of dentists in the year 2019 was $130,000. A sample of 36 dentists, which was taken in 2020, showed an average yearly income of $125,000. Assume the standard deviation of the population of dentists in 2020 is $25,000. Round your solutions for this Exhibit to 4 decimal places. Suppose that, instead of the critical value approach, you decided to use the p-value approach and found that the p-value for the test in this problem is equal to 0.1151. This p-value is best interpreted as the following: - There is a 11.51% probability of observing the average yearly income in the sample of 36 dentists of $125,000 or below if the actual average yearly income of dentists is $130,000. - There is a 11.51% probability of observing the average yearly income in the sample of 36 dentists of $125,000 or above if the actual average yearly income of dentists is $130,000. - There is a 11.51% probability of observing the average yearly income in the sample of 36 dentists of $125,000 or below. - There is a 11.51% probability of observing the average yearly income in the sample of 36 dentists of $130,000 or below.

There is a 11.51% probability of observing the average yearly income in the sample of 36 dentists of $125,000 or below if the actual average yearly income of dentists is $130,000.

A university is interested in promoting graduates of its honors program by establishing that the mean GPA of these graduates exceeds 3.50. A sample of 36 honors students is taken and is found to have a mean GPA equal to 3.60. The population standard deviation is assumed to equal 0.40. To establish whether the mean GPA exceeds 3.50, you conduct a hypothesis test. Suppose you concluded that you do not reject the null at 1% significance level. How would you state the conclusion "in words" in relation to this problem? - There is not enough evidence to conclude that the mean GPA is smaller than 3.50. - There is enough evidence to conclude that the mean GPA is smaller than 3.50. - There is not enough evidence to conclude that the mean GPA is greater than 3.50. - There is enough evidence to conclude that the mean GPA is greater than 3.50.

There is not enough evidence to conclude that the mean GPA is greater than 3.50.

Exhibit: Average Income. The Bureau of Labor Statistics reported that the average yearly income of dentists in the year 2019 was $130,000. A sample of 36 dentists, which was taken in 2020, showed an average yearly income of $125,000. Assume the standard deviation of the population of dentists in 2020 is $25,000. Round your solutions for this Exhibit to 4 decimal places. State your conclusion for the test in words relying on the economic context of the problem. - There is not enough evidence to conclude that the yearly income of dentists has decreased since 2019. - There is enough evidence to conclude that the yearly income of dentists has decreased since 2019. - There is not enough evidence to conclude that the yearly income of dentists has changed since 2019. - There is enough evidence to conclude that the yearly income of dentists has increased since 2019.

There is not enough evidence to conclude that the yearly income of dentists has decreased since 2019.

In the point estimation, - data from the population is used to estimate the population parameter - data from the sample is used to estimate the sample statistic - the mean of the population equals the mean of the sample - data from the sample is used to estimate the population parameter

data from the sample is used to estimate the population parameter

As the sample size n increases, the margin of error - increases - decreases - stays the same - increases or decreases depending on the value of the mean

decreases

The error of rejecting a true null hypothesis is - Type II error - Committed when not enough information is available - the same as β - Type I error

Type I error

Confidence intervals for the population mean may be created for the cases when the population standard deviation is known or unknown. How are these two cases treated differently? - Use the z distribution when σ is known; use the t distribution when σ is unknown. - Use the z distribution when s is unknown; use the t distribution when s is known. - Use the z distribution when s is known; use the t distribution when s is unknown. - Use the z distribution when σ is unknown; use the t distribution when σ is known.

Use the z distribution when σ is known; use the t distribution when σ is unknown.

Which of the following statements is correct? (this question is a little tricky) - We can establish a claim if the sample evidence is consistent with the null hypothesis - We can accept the null hypothesis if the sample evidence is not inconsistent with the null hypothesis - Since the sample evidence cannot be supported by the null hypothesis, we can reject the null - We cannot establish a claim if the null hypothesis is not rejected

We cannot establish a claim if the null hypothesis is not rejected

Given that a 95% confidence interval for the population mean is (6.5, 12.5), we can state the following: - There is a 95% probability that the true population mean falls between 6.5 and 12.5. - We expect that 5% of all possible sample means drawn from the population will produce confidence intervals that include the population's mean (which makes us 95% confident that interval (6.5, 12.5) contains the population mean). - We expect that 95% of all possible sample means drawn from the population will produce confidence intervals that include the population's mean (which makes us 95% confident that interval (6.5, 12.5) contains the population mean). - There is a 95% probability that the sample mean is between 6.5 and 12.5.

We expect that 95% of all possible sample means drawn from the population will produce confidence intervals that include the population's mean (which makes us 95% confident that interval (6.5, 12.5) contains the population mean).

Exhibit: Community Banks. The American Bankers Association Community Bank Insurance Survey for 2017 had responses from 123 banks. Of these, 64 were community banks, defined to be banks with assets of $1 billion or less. How would you interpret the confidence interval constructed in this problem? - If we were to take one more sample of 123 banks, there would be a 95% probability that the proportion of community banks would be within the confidence interval. - We expect that 95% of all possible sample mean assets for the banks will produce confidence intervals that include the population's mean assets (which makes us 95% confident that the interval we constructed contains the population mean bank asset). - There is a 95% probability that the population proportion of community banks falls within the interval we constructed. - We expect that 95% of all possible sample proportions of community banks will produce confidence intervals that include the population proportion of community banks (which makes us 95% confident that the interval we constructed in this problem contains the population proportion).

We expect that 95% of all possible sample proportions of community banks will produce confidence intervals that include the population proportion of community banks (which makes us 95% confident that the interval we constructed in this problem contains the population proportion).

Exhibit: Tires Life Expectancy. A tire manufacturer has been producing tires with an average life expectancy of 25,000 miles. Now the company is advertising that its new tires' life expectancy has increased. In order to test the legitimacy of the advertising campaign, an independent testing agency tested a sample of 6 of their tires and has provided the following data. Data: n = 6, mu_H0 = 25, xbar = 26.1667, s = 1.722401 Raw data: 28, 27, 25, 28, 24, 25 Do we need any additional assumptions about the life expectancy of the tires in the population to make sure that the conclusion stated in the previous question is reliable? - No. The sample standard deviation can be calculated based on the given information, which guarantees that the conclusions derived in the test are reliable. - Yes. The population standard deviation of the life expectancy of the tires should be known to obtain reliable conclusions. - Yes. The life expectancy of the tires should follow the normal distribution in the population which would guarantee that the conclusions derived in the test are reliable. - No. Using the information in the problem, we can calculate the test statistic and the p-value. So, everything is provided to conduct the test and obtain the conclusions without any additional information or any additional assumptions.

Yes. The life expectancy of the tires should follow the normal distribution in the population which would guarantee that the conclusions derived in the test are reliable.

At an academically challenging high school, the average GPA of a high school senior is known to be normally distributed with a variance of 0.45. A sample of 15 seniors is taken and their average GPA is found to be 2.83. Assume that z0.05=1.645 and z0.1=1.282. The 90% confidence interval for the population mean GPA is: - [2.6875, 2.9725] - [2.608, 3.052] - [2.5451, 3.1149] - [2.719, 2.941]

[2.5451, 3.1149]

As the number of degrees of freedom for the t distribution increases, the difference between the t distribution and the standard normal distribution - becomes smaller - becomes larger - none of the suggested alternatives is correct - stays the same

becomes smaller

The value added and subtracted from a point estimate in order to develop an interval estimate of the population parameter is known as the - confidence level - margin of error - interval estimate - parameter estimate

margin of error

If a hypothesis is rejected at 5% level of significance, it (Hint: draw a picture for this question) - will never be rejected at the 1% level - will always be accepted at the 1% level - will always be rejected at the 1% level - may be rejected or not rejected at the 1% level

may be rejected or not rejected at the 1% level

Exhibit: Soft Drinks. Last year, a soft drink manufacturer had 21% of the market. In order to increase their portion of the market, the manufacturer has introduced a new flavor in their soft drinks. A sample of 400 individuals participated in the taste test and 120 indicated that they like the taste. We are interested in determining if more than 21% of the population will like the new soft drink at the significance level 0.1. Round your solutions for this Exhibit to 4 decimal places. Calculate the p-value for the test.

n/a not: 0.5398, 0.4602

As the confidence level decreases, the confidence interval becomes - narrow - stays the same - wider - wider or narrow depending on whether the population standard deviation is known or not

narrow

As the sample size n increases, the confidence interval becomes - wider or narrow depending on whether the population standard deviation is known or not - stays the same - narrower - wider

narrower

A university is interested in promoting graduates of its honors program by establishing that the mean GPA of these graduates exceeds 3.50. A sample of 36 honors students is taken and is found to have a mean GPA equal to 3.60. The population standard deviation is assumed to equal 0.40. To establish whether the mean GPA exceeds 3.50, you conduct a hypothesis test. You calculate that the value of the test statistic is equal to zx¯=1.5zx¯=1.5 and the critical value at a 1% significance level is equal to zα=2.32zα=2.32. At the significance level of 1%, the decision is to - reject Ho because zx¯<zα - not reject Ho because zx¯>−zα - not reject Ho because zx¯<zα - No conclusion can be made

not reject Ho because zx¯<zα

What conditions must be safisfied in the hypothesis tests about the population proportion? - the sample size should be at least 30 observations - the population data must be normally distributed - the population standard deviation must be known - np ≥ 5 and nq ≥ 5

np ≥ 5 and nq ≥ 5

Exhibit: Job Outsourcing. Among the trends reshaping the U.S. workplace, a large proportion of Americans see outsourcing of jobs as a negative rather than a positive force when it comes to their livelihoods. To check this trend, a researcher selected a sample of 200 workers and conducted a survey. In the survey, she asked how the outsourcing of jobs to other countries affected the jobs or careers of the surveyed individuals. The responses were recorded as "No Effect" if outsourcing of jobs hasn't made much of a difference, "Hurt" if it hurt a job or career, or "Help" if outsourcing was seen as a positive trend helping a job. - 60 Hurt - 14 Help - 126 No Effect What assumptions did you make when constructing the confidence interval? - np≥5 and n(1−p)≥5. - No additional assumptions is needed in this problem. - The population data is normally distributed. - The sample size is greater than 30 observations.

np≥5 and n(1−p)≥5.

Exhibit: Soft Drinks. Last year, a soft drink manufacturer had 21% of the market. In order to increase their portion of the market, the manufacturer has introduced a new flavor in their soft drinks. A sample of 400 individuals participated in the taste test and 120 indicated that they like the taste. We are interested in determining if more than 21% of the population will like the new soft drink at the significance level 0.1. Round your solutions for this Exhibit to 4 decimal places. State your conclusion for the test using the p-value. - p-value < 0.05, so we cannot reject Ho. Therefore, there is not enough evidence to conclude that the proportion of people who like new favor is above 21%. - p-value < 0.05, so we reject Ho. Therefore, there is enough evidence to conclude that the proportion of people who like new favor is below 21%. - p-value < 0.05, so we reject Ho. Therefore, there is enough evidence to conclude that the proportion of people who like new favor is above 21%. - p-value < 0.05, so we cannot reject Ho. Therefore, there is enough evidence to conclude that the proportion of people who like new favor is below 21%.

p-value < 0.05, so we reject Ho. Therefore, there is enough evidence to conclude that the proportion of people who like new favor is above 21%.

Exhibit: Costco Customers. Customers at Costco spend an average of $150 per trip. One of Costco's rivals would like to determine whether Costco's customers spend more per trip. A survey of the receipts of 28 customers found that the sample mean was $156. Assume that the population standard deviation of spending is $13.50 and the spending follows a normal distribution (use the significance level 0.1). Round your solutions for this Exhibit to 4 decimal places. State your conclusion for the test using the p-value. - p-value < 0.1, so we reject Ho. Therefore, there is enough evidence to conclude that Costco's customers spend more than $150 per trip. - p-value < 0.1, so we cannot reject Ho. Therefore, there is enough evidence to conclude that Costco's customers spend more than $150 per trip. - p-value < 0.1, so we cannot reject Ho. Therefore, there is not enough evidence to conclude that Costco's customers spend more than $150 per trip. - p-value < 0.1, so we reject Ho. Therefore, there is not enough evidence to conclude that Costco's customers spend more than $150 per trip.

p-value < 0.1, so we reject Ho. Therefore, there is enough evidence to conclude that Costco's customers spend more than $150 per trip.

Exhibit: Tires Life Expectancy. A tire manufacturer has been producing tires with an average life expectancy of 25,000 miles. Now the company is advertising that its new tires' life expectancy has increased. In order to test the legitimacy of the advertising campaign, an independent testing agency tested a sample of 6 of their tires and has provided the following data. Data: n = 6, mu_H0 = 25, xbar = 26.1667, s = 1.722401 Raw data: 28, 27, 25, 28, 24, 25 State your conclusion for the test using the p-value and the significance level of 3%. - p-value > 0.03, so we do not reject Ho. Therefore, there is enough evidence to conclude that the mean life expectancy of the new tires has decreased. - p-value < 0.03, so we reject Ho. Therefore, there is enough evidence to conclude that the mean life expectancy of the new tires has increased. - p-value > 0.03, so we do not reject Ho. Therefore, there is not enough evidence to conclude that the mean life expectancy of the new tires has increased. - p-value < 0.03, so we reject Ho. Therefore, there is enough evidence to conclude that the mean life expectancy of the new tires has decreased.

p-value > 0.03, so we do not reject Ho. Therefore, there is not enough evidence to conclude that the mean life expectancy of the new tires has increased.

The sample standard deviation, s, is the point estimator of the - sample mean xbar - population standard deviation σ - population mean μ - sample proportion

population standard deviation σ

The level of significance is the - probability of Type II error - same as the confidence level - probability of Type I error - same as the p-value

probability of Type I error

When the p-value is used for hypothesis testing, the null hypothesis is rejected if - p−value=1−α - p−value>α - p−value≤α - It depends on the type of the test

p−value≤α

A university is interested in promoting graduates of its honors program by establishing that the mean GPA of these graduates exceeds 3.50. A sample of 36 honors students found to have a mean GPA equal to 3.60 and the standard deviation equal to 0.40. You are looking to conduct a hypothesis test to establish that the mean GPA exceeds 3.50. Suppose that you calculated the p-value for the test with the given information to be equal to 0.071. At a 10% significance level, the decision is to - reject Ho. So, there is enough evidence to conclude that the mean GPA is greater than 3.50. - reject Ho. So, there is not enough evidence to conclude that the mean GPA is greater than 3.50. - not reject Ho. So, there is not enough evidence to conclude that the mean GPA is greater than 3.50. - not reject Ho. So, there is enough evidence to conclude that the mean GPA is greater than 3.50.

reject Ho. So, there is enough evidence to conclude that the mean GPA is greater than 3.50.

A university is interested in promoting graduates of its honors program by establishing that the mean GPA of these graduates exceeds 3.50. A sample of 36 honors students is taken and is found to have a mean GPA equal to 3.60. The population standard deviation is assumed to equal 0.40. Suppose that you calculated that the p-value for the test with the given information equal to 0.067. At a 10% significance level, the decision is to - reject Ho. So, there is enough evidence to conclude that the mean GPA is greater than 3.50. - reject Ho. So, there is not enough evidence to conclude that the mean GPA is greater than 3.50. - not reject Ho. So, there is not enough evidence to conclude that the mean GPA is greater than 3.50. - not reject Ho. So, there is enough evidence to conclude that the mean GPA is greater than 3.50.

reject Ho. So, there is enough evidence to conclude that the mean GPA is greater than 3.50.

Which of the following represents an estimate? - s=3 - xbar - p - μ=3

s=3

Which is the following represents an estimator? - μ - pbar = 0.4 - s^2 - σ^2

s^2 - s^2 point estimator of the σ^2 - xbar is the point estimator of the μ - pbar has a value that is an estimate

When s is used to estimate σ, the margin of error is computed by using - mean of the population - mean of the sample - t-distribution - the standard normal distribution

t-distribution

When conducting a hypothesis test for the population mean when σ is unknown and the sample size is 30 or more, the distribution we use for computing the critical value and the p-value is - the binomial distribution - not known unless the population distribution is known - the standard normal distribution - the Student's t-distribution

the Student's t-distribution

The Department of Economic and Community Development (DECD) reported that in 2009 the average number of new jobs created per county was 450 (the only information that you know from DECD). Doing a project in your international economics class, you want to determine whether there has been a decrease in the average number of jobs created, and you collect information about 11 countries. To conduct the hypothesis test, what distribution would you use to calculate the critical value and the p-value? - the Student's t-distribution with 11 degrees of freedom - the binomial distribution - the standard normal distribution - the Student's t-distribution with 10 degrees of freedom

the Student's t-distribution with 10 degrees of freedom

In hypothesis testing, if the null hypothesis is rejected, - no conclusions can be drawn from the test - the alternative hypothesis is true - the sample size has been too small - the data must have been accumulated incorrectly

the alternative hypothesis is true

In hypothesis testing, the tentative assumption about the population parameter is - None of the suggested alternatives is correct - the alternative hypothesis - either the null or the alternative - the null hypothesis

the null hypothesis

In order to construct the interval estimate of the population mean µ when σ is known and the sample is very small, - the population must have an area under the probability density function be equal to 1 - the population data must be normally distributed - the population may have any probability distribution - the population must be very large

the population data must be normally distributed

We use the t-distribution in the hypothesis tests about the population mean when - the sample standard deviation is unknown - the sample is small - the population standard deviation is not known - the population does not follow the normal distribution

the population standard deviation is not known

For the confidence interval of the mean when the population standard deviation is known and the sample is large, the proper distribution to use is - the t distribution with n - 1 degrees of freedom - the standard normal distribution - the binomial distribution - the t distribution with n degrees of freedom