An efficiency expert is hired to improve productivity at a company. She administers an efficiency to test a simple random sample of the employees, implements a productivity training course, and administers the efficiency test again one month later to the same employees tested originally. The efficiency test results are below. Employee # #1 #2 #3 #4 #5 #6 #7 #8 #9 #10 #11 #12 #13 #14 #15 Before 21 25 45 28 23 27 22 28 39 35 34 40 27 20 33 After 32 25 36 47 37 30 40 39 28 48 28 41 27 33 28 Test score differences are known to come from a normally distributed population. Use the p-value method and a 8% significance level to test the claim that employee efficiency improved after the productivity training course. What population parameter is being tested?

An efficiency expert is hired to improve productivity at a company. She administers an efficiency to test a simple random sample of the employees, implements a productivity training course, and administers the efficiency test again one month later to the same employees tested originally. The efficiency test results are below. Employee # #1 #2 #3 #4 #5 #6 #7 #8 #9 #10 #11 #12 #13 #14 #15 Before 21 25 45 28 23 27 22 28 39 35 34 40 27 20 33 After 32 25 36 47 37 30 40 39 28 48 28 41 27 33 28 Test score differences are known to come from a normally distributed population. Use the p-value method and a 8% significance level to test the claim that employee efficiency improved after the productivity training course. How many populations are being tested?

An efficiency expert is hired to improve productivity at a company. She administers an efficiency to test a simple random sample of the employees, implements a productivity training course, and administers the efficiency test again one month later to the same employees tested originally. The efficiency test results are below. Employee # #1 #2 #3 #4 #5 #6 #7 #8 #9 #10 #11 #12 #13 #14 #15 Before 21 25 45 28 23 27 22 28 39 35 34 40 27 20 33 After 32 25 36 47 37 30 40 39 28 48 28 41 27 33 28 Test score differences are known to come from a normally distributed population. Use the p-value method and a 8% significance level to test the claim that employee efficiency improved after the productivity training course. Calculate the sample mean difference.

An efficiency expert is hired to improve productivity at a company. She administers an efficiency to test a simple random sample of the employees, implements a productivity training course, and administers the efficiency test again one month later to the same employees tested originally. The efficiency test results are below. Employee # #1 #2 #3 #4 #5 #6 #7 #8 #9 #10 #11 #12 #13 #14 #15 Before 21 25 45 28 23 27 22 28 39 35 34 40 27 20 33 After 32 25 36 47 37 30 40 39 28 48 28 41 27 33 28 Test score differences are known to come from a normally distributed population. Use the p-value method and a 8% significance level to test the claim that employee efficiency improved after the productivity training course. What is the claim? (At this point, you should have already selected the formula that will be used to calculate the test statistic and written it in the test statistic box.)

An efficiency expert is hired to improve productivity at a company. She administers an efficiency to test a simple random sample of the employees, implements a productivity training course, and administers the efficiency test again one month later to the same employees tested originally. The efficiency test results are below. Employee # #1 #2 #3 #4 #5 #6 #7 #8 #9 #10 #11 #12 #13 #14 #15 Before 21 25 45 28 23 27 22 28 39 35 34 40 27 20 33 After 32 25 36 47 37 30 40 39 28 48 28 41 27 33 28 Test score differences are known to come from a normally distributed population. Use the p-value method and a 8% significance level to test the claim that employee efficiency improved after the productivity training course. The claim is the _________ hypothesis.

An efficiency expert is hired to improve productivity at a company. She administers an efficiency to test a simple random sample of the employees, implements a productivity training course, and administers the efficiency test again one month later to the same employees tested originally. The efficiency test results are below. Employee # #1 #2 #3 #4 #5 #6 #7 #8 #9 #10 #11 #12 #13 #14 #15 Before 21 25 45 28 23 27 22 28 39 35 34 40 27 20 33 After 32 25 36 47 37 30 40 39 28 48 28 41 27 33 28 Test score differences are known to come from a normally distributed population. Use the p-value method and a 8% significance level to test the claim that employee efficiency improved after the productivity training course. What is the alternative hypothesis?

An efficiency expert is hired to improve productivity at a company. She administers an efficiency to test a simple random sample of the employees, implements a productivity training course, and administers the efficiency test again one month later to the same employees tested originally. The efficiency test results are below. Employee # #1 #2 #3 #4 #5 #6 #7 #8 #9 #10 #11 #12 #13 #14 #15 Before 21 25 45 28 23 27 22 28 39 35 34 40 27 20 33 After 32 25 36 47 37 30 40 39 28 48 28 41 27 33 28 Test score differences are known to come from a normally distributed population. Use the p-value method and a 8% significance level to test the claim that employee efficiency improved after the productivity training course. What is the test statistic (rounded to the nearest thousandth)?

An efficiency expert is hired to improve productivity at a company. She administers an efficiency to test a simple random sample of the employees, implements a productivity training course, and administers the efficiency test again one month later to the same employees tested originally. The efficiency test results are below. Employee # #1 #2 #3 #4 #5 #6 #7 #8 #9 #10 #11 #12 #13 #14 #15 Before 21 25 45 28 23 27 22 28 39 35 34 40 27 20 33 After 32 25 36 47 37 30 40 39 28 48 28 41 27 33 28 Test score differences are known to come from a normally distributed population. Use the p-value method and a 8% significance level to test the claim that employee efficiency improved after the productivity training course. The critical region is best described as ____________.

An efficiency expert is hired to improve productivity at a company. She administers an efficiency to test a simple random sample of the employees, implements a productivity training course, and administers the efficiency test again one month later to the same employees tested originally. The efficiency test results are below. Employee # #1 #2 #3 #4 #5 #6 #7 #8 #9 #10 #11 #12 #13 #14 #15 Before 21 25 45 28 23 27 22 28 39 35 34 40 27 20 33 After 32 25 36 47 37 30 40 39 28 48 28 41 27 33 28 Test score differences are known to come from a normally distributed population. Use the p-value method and a 8% significance level to test the claim that employee efficiency improved after the productivity training course. What is the p-value (rounded to the nearest ten-thousandth)?

An efficiency expert is hired to improve productivity at a company. She administers an efficiency to test a simple random sample of the employees, implements a productivity training course, and administers the efficiency test again one month later to the same employees tested originally. The efficiency test results are below. Employee # #1 #2 #3 #4 #5 #6 #7 #8 #9 #10 #11 #12 #13 #14 #15 Before 21 25 45 28 23 27 22 28 39 35 34 40 27 20 33 After 32 25 36 47 37 30 40 39 28 48 28 41 27 33 28 Test score differences are known to come from a normally distributed population. Use the p-value method and a 8% significance level to test the claim that employee efficiency improved after the productivity training course. What is the significance level (expressed as a decimal)?

An efficiency expert is hired to improve productivity at a company. She administers an efficiency to test a simple random sample of the employees, implements a productivity training course, and administers the efficiency test again one month later to the same employees tested originally. The efficiency test results are below. Employee # #1 #2 #3 #4 #5 #6 #7 #8 #9 #10 #11 #12 #13 #14 #15 Before 21 25 45 28 23 27 22 28 39 35 34 40 27 20 33 After 32 25 36 47 37 30 40 39 28 48 28 41 27 33 28 Test score differences are known to come from a normally distributed population. Use the p-value method and a 8% significance level to test the claim that employee efficiency improved after the productivity training course. What is the statistical conclusion?

An efficiency expert is hired to improve productivity at a company. She administers an efficiency to test a simple random sample of the employees, implements a productivity training course, and administers the efficiency test again one month later to the same employees tested originally. The efficiency test results are below. Employee # #1 #2 #3 #4 #5 #6 #7 #8 #9 #10 #11 #12 #13 #14 #15 Before 21 25 45 28 23 27 22 28 39 35 34 40 27 20 33 After 32 25 36 47 37 30 40 39 28 48 28 41 27 33 28 Test score differences are known to come from a normally distributed population. Use the p-value method and a 8% significance level to test the claim that employee efficiency improved after the productivity training course. What is the wordy conclusion?

There have been 16 US Supreme Court Chief Justices who have had the following tenures (in years): 18 17 15 7 4 11 8 10 21 14 8 28 34 4 0 5 There have been 93 US Supreme Court Associate Justices who have had a mean tenure of 16.1720 years with a standard deviation of 9.871593 years. Use the critical value method and a 5% significance level to test the claim that the tenures of US Supreme Court Chief Justices have the same variation as the tenures of US Supreme Court Associate Justices. What population parameter is being tested?

There have been 16 US Supreme Court Chief Justices who have had the following tenures (in years): 18 17 15 7 4 11 8 10 21 14 8 28 34 4 0 5 There have been 93 US Supreme Court Associate Justices who have had a mean tenure of 16.1720 years with a standard deviation of 9.871593 years. Use the critical value method and a 5% significance level to test the claim that the tenures of US Supreme Court Chief Justices have the same variation as the tenures of US Supreme Court Associate Justices. How many populations are being tested?

There have been 16 US Supreme Court Chief Justices who have had the following tenures (in years): 18 17 15 7 4 11 8 10 21 14 8 28 34 4 0 5 There have been 93 US Supreme Court Associate Justices who have had a mean tenure of 16.1720 years with a standard deviation of 9.871593 years. Use the critical value method and a 5% significance level to test the claim that the tenures of US Supreme Court Chief Justices have the same variation as the tenures of US Supreme Court Associate Justices. Calculate the sample variance for the tenures of US Supreme Court Chief Justices.

There have been 16 US Supreme Court Chief Justices who have had the following tenures (in years): 18 17 15 7 4 11 8 10 21 14 8 28 34 4 0 5 There have been 93 US Supreme Court Associate Justices who have had a mean tenure of 16.1720 years with a standard deviation of 9.871593 years. Use the critical value method and a 5% significance level to test the claim that the tenures of US Supreme Court Chief Justices have the same variation as the tenures of US Supreme Court Associate Justices. What is the claim? (At this point, you should have already selected the formula that will be used to calculate the test statistic and written it in the test statistic box.)

There have been 16 US Supreme Court Chief Justices who have had the following tenures (in years): 18 17 15 7 4 11 8 10 21 14 8 28 34 4 0 5 There have been 93 US Supreme Court Associate Justices who have had a mean tenure of 16.1720 years with a standard deviation of 9.871593 years. Use the critical value method and a 5% significance level to test the claim that the tenures of US Supreme Court Chief Justices have the same variation as the tenures of US Supreme Court Associate Justices. The claim is the _________ hypothesis.

There have been 16 US Supreme Court Chief Justices who have had the following tenures (in years): 18 17 15 7 4 11 8 10 21 14 8 28 34 4 0 5 There have been 93 US Supreme Court Associate Justices who have had a mean tenure of 16.1720 years with a standard deviation of 9.871593 years. Use the critical value method and a 5% significance level to test the claim that the tenures of US Supreme Court Chief Justices have the same variation as the tenures of US Supreme Court Associate Justices. What is the alternative hypothesis?

There have been 16 US Supreme Court Chief Justices who have had the following tenures (in years): 18 17 15 7 4 11 8 10 21 14 8 28 34 4 0 5 There have been 93 US Supreme Court Associate Justices who have had a mean tenure of 16.1720 years with a standard deviation of 9.871593 years. Use the critical value method and a 5% significance level to test the claim that the tenures of US Supreme Court Chief Justices have the same variation as the tenures of US Supreme Court Associate Justices. The critical region is best described as ____________.

There have been 16 US Supreme Court Chief Justices who have had the following tenures (in years): 18 17 15 7 4 11 8 10 21 14 8 28 34 4 0 5 There have been 93 US Supreme Court Associate Justices who have had a mean tenure of 16.1720 years with a standard deviation of 9.871593 years. Use the critical value method and a 5% significance level to test the claim that the tenures of US Supreme Court Chief Justices have the same variation as the tenures of US Supreme Court Associate Justices. What is the significance level (expressed as a decimal)?

There have been 16 US Supreme Court Chief Justices who have had the following tenures (in years): 18 17 15 7 4 11 8 10 21 14 8 28 34 4 0 5 There have been 93 US Supreme Court Associate Justices who have had a mean tenure of 16.1720 years with a standard deviation of 9.871593 years. Use the critical value method and a 5% significance level to test the claim that the tenures of US Supreme Court Chief Justices have the same variation as the tenures of US Supreme Court Associate Justices. What is the critical value (rounded to the nearest ten-thousandth)?

There have been 16 US Supreme Court Chief Justices who have had the following tenures (in years): 18 17 15 7 4 11 8 10 21 14 8 28 34 4 0 5 There have been 93 US Supreme Court Associate Justices who have had a mean tenure of 16.1720 years with a standard deviation of 9.871593 years. Use the critical value method and a 5% significance level to test the claim that the tenures of US Supreme Court Chief Justices have the same variation as the tenures of US Supreme Court Associate Justices. What is the test statistic (rounded to the nearest ten-thousandth)?

There have been 16 US Supreme Court Chief Justices who have had the following tenures (in years): 18 17 15 7 4 11 8 10 21 14 8 28 34 4 0 5 There have been 93 US Supreme Court Associate Justices who have had a mean tenure of 16.1720 years with a standard deviation of 9.871593 years. Use the critical value method and a 5% significance level to test the claim that the tenures of US Supreme Court Chief Justices have the same variation as the tenures of US Supreme Court Associate Justices. What is the statistical conclusion?

There have been 16 US Supreme Court Chief Justices who have had the following tenures (in years): 18 17 15 7 4 11 8 10 21 14 8 28 34 4 0 5 There have been 93 US Supreme Court Associate Justices who have had a mean tenure of 16.1720 years with a standard deviation of 9.871593 years. Use the critical value method and a 5% significance level to test the claim that the tenures of US Supreme Court Chief Justices have the same variation as the tenures of US Supreme Court Associate Justices. What is the wordy conclusion?

Based on a simple random sample of 52 juice-fast dieters, juice-fast dieters mean net weight change after three months is -21 pounds with a standard deviation of 12.4 pounds. Based on a simple random sample of 61 paleo dieters, paleo dieters mean net weight change after three months is -25 pounds with a standard deviation of 16.7 pounds. Use the p-value method and a 3% significance level to test the claim that the weight of paleo dieters varies more than the weight of juice-fast dieters after three months. What population parameter is being tested?

Based on a simple random sample of 52 juice-fast dieters, juice-fast dieters mean net weight change after three months is -21 pounds with a standard deviation of 12.4 pounds. Based on a simple random sample of 61 paleo dieters, paleo dieters mean net weight change after three months is -25 pounds with a standard deviation of 16.7 pounds. Use the p-value method and a 3% significance level to test the claim that the weight of paleo dieters varies more than the weight of juice-fast dieters after three months. How many populations are being tested?

Based on a simple random sample of 52 juice-fast dieters, juice-fast dieters mean net weight change after three months is -21 pounds with a standard deviation of 12.4 pounds. Based on a simple random sample of 61 paleo dieters, paleo dieters mean net weight change after three months is -25 pounds with a standard deviation of 16.7 pounds. Use the p-value method and a 3% significance level to test the claim that the weight of paleo dieters varies more than the weight of juice-fast dieters after three months. Calculate the sample variance of paleo dieters net weight change.

Based on a simple random sample of 52 juice-fast dieters, juice-fast dieters mean net weight change after three months is -21 pounds with a standard deviation of 12.4 pounds. Based on a simple random sample of 61 paleo dieters, paleo dieters mean net weight change after three months is -25 pounds with a standard deviation of 16.7 pounds. Use the p-value method and a 3% significance level to test the claim that the weight of paleo dieters varies more than the weight of juice-fast dieters after three months. What is the claim? (At this point, you should have already selected the formula that will be used to calculate the test statistic and written it in the test statistic box.)

Based on a simple random sample of 52 juice-fast dieters, juice-fast dieters mean net weight change after three months is -21 pounds with a standard deviation of 12.4 pounds. Based on a simple random sample of 61 paleo dieters, paleo dieters mean net weight change after three months is -25 pounds with a standard deviation of 16.7 pounds. Use the p-value method and a 3% significance level to test the claim that the weight of paleo dieters varies more than the weight of juice-fast dieters after three months. The claim is the _________ hypothesis.

Based on a simple random sample of 52 juice-fast dieters, juice-fast dieters mean net weight change after three months is -21 pounds with a standard deviation of 12.4 pounds. Based on a simple random sample of 61 paleo dieters, paleo dieters mean net weight change after three months is -25 pounds with a standard deviation of 16.7 pounds. Use the p-value method and a 3% significance level to test the claim that the weight of paleo dieters varies more than the weight of juice-fast dieters after three months. What is the alternative hypothesis?

Based on a simple random sample of 52 juice-fast dieters, juice-fast dieters mean net weight change after three months is -21 pounds with a standard deviation of 12.4 pounds. Based on a simple random sample of 61 paleo dieters, paleo dieters mean net weight change after three months is -25 pounds with a standard deviation of 16.7 pounds. Use the p-value method and a 3% significance level to test the claim that the weight of paleo dieters varies more than the weight of juice-fast dieters after three months. What is the test statistic (rounded to the nearest ten-thousandth)?

Based on a simple random sample of 52 juice-fast dieters, juice-fast dieters mean net weight change after three months is -21 pounds with a standard deviation of 12.4 pounds. Based on a simple random sample of 61 paleo dieters, paleo dieters mean net weight change after three months is -25 pounds with a standard deviation of 16.7 pounds. Use the p-value method and a 3% significance level to test the claim that the weight of paleo dieters varies more than the weight of juice-fast dieters after three months. The critical region is best described as ____________.

Based on a simple random sample of 52 juice-fast dieters, juice-fast dieters mean net weight change after three months is -21 pounds with a standard deviation of 12.4 pounds. Based on a simple random sample of 61 paleo dieters, paleo dieters mean net weight change after three months is -25 pounds with a standard deviation of 16.7 pounds. Use the p-value method and a 3% significance level to test the claim that the weight of paleo dieters varies more than the weight of juice-fast dieters after three months. What is the p-value (rounded to the nearest ten-thousandth)?

Based on a simple random sample of 52 juice-fast dieters, juice-fast dieters mean net weight change after three months is -21 pounds with a standard deviation of 12.4 pounds. Based on a simple random sample of 61 paleo dieters, paleo dieters mean net weight change after three months is -25 pounds with a standard deviation of 16.7 pounds. Use the p-value method and a 3% significance level to test the claim that the weight of paleo dieters varies more than the weight of juice-fast dieters after three months. What is the significance level (expressed as a decimal)?

Based on a simple random sample of 52 juice-fast dieters, juice-fast dieters mean net weight change after three months is -21 pounds with a standard deviation of 12.4 pounds. Based on a simple random sample of 61 paleo dieters, paleo dieters mean net weight change after three months is -25 pounds with a standard deviation of 16.7 pounds. Use the p-value method and a 3% significance level to test the claim that the weight of paleo dieters varies more than the weight of juice-fast dieters after three months. What is the statistical conclusion?

Based on a simple random sample of 52 juice-fast dieters, juice-fast dieters mean net weight change after three months is -21 pounds with a standard deviation of 12.4 pounds. Based on a simple random sample of 61 paleo dieters, paleo dieters mean net weight change after three months is -25 pounds with a standard deviation of 16.7 pounds. Use the p-value method and a 3% significance level to test the claim that the weight of paleo dieters varies more than the weight of juice-fast dieters after three months. What is the wordy conclusion?

STATDISK has the ability to find test statistics, critical values, and p values when testing hypotheses about two matched pair population means. The next questions are designed to introduce you to these functions within STATDISK. The two population matched pair hypothesis test function is located under the "analysis," "hypothesis testing," and "mean matched pairs" sub menu. The claim being tested is μd=μ1-μ2>0 with a 4.71% significance level. Matched pair data are below: First | Second (col 1)|(col 2) in STATDISK 9 | 11 12 | 11 15 | 13 8 | 6 7 | 5 13 | 11 13 | 10 7 | 8 5 | 3 What is the p-value? (Round to the nearest ten-thousandth.)

STATDISK has the ability to calculate the cumulative area to the right under the F distribution when supplied an F value. This function is equivalent to using Table 5 and Table 6 to find a range of possible areas corresponding to an F value between two F values or more extreme than an "edge" F value. Remember, Table 5 and Table 6 do not provide a robust method for finding the area under the F distribution when supplied any F value. The next questions are designed to introduce you to these functions within STATDISK. The F distribution function is located under the "analysis," "probability distributions," and "F distribution" sub menu. What is the cumulative area to the right under the F distribution with 49 numerator degrees of freedom and 69 denominator degrees of freedom to the F value of 1.5896? (Round to the nearest ten-thousandth.)

What is the cumulative area from the left under the F distribution with 138 numerator degrees of freedom and 18 denominator degrees of freedom to the F value of 1.0025? (Round to the nearest ten-thousandth.)

STATDISK also has the ability to calculate the F value when supplied the cumulative area to the right under the F distribution. This function is equivalent to using Table 5 and Table 6; however, the STATDISK function has the ability to supply additional F values not present on the table. Remember, Table 5 and Table 6 does not provide a robust method for finding F values for any area under the F distribution. The next questions are designed to introduce you to these functions within STATDISK. The F distribution function is located under the "analysis," "probability distributions," and "F distribution" sub menu. What is the F value corresponding to the cumulative area from the left under the F distribution of 0.9573 with 256 numerator degrees of freedom and 397 denominator degrees of freedom? (Round to the nearest ten-thousandth.)

What is the F value corresponding to the cumulative area to the right under the F distribution of 0.0107 with 10 numerator degrees of freedom and 5 denominator degrees of freedom? (Round to the nearest ten-thousandth.)

STATDISK has the ability to find test statistics, critical values, and p values when testing hypotheses about two population standard deviations and variances. The next questions are designed to introduce you to these functions within STATDISK. The two population standard deviation hypothesis test function is located under the "analysis," "hypothesis testing," and "standard deviation two samples" sub menu. The claim being tested is σ1=σ2 with a 9.17% significance level. A first sample of size 96 with a sample mean of 270 and a sample standard deviation of 40 and a second sample of size 259 with a sample mean of 256 and a sample standard deviation of 32 are used to test the claim. What is the p-value? (Round to the nearest ten-thousandth.)

The claim being tested is σ1>σ2 with a 1.38% significance level. A first sample (enter in column 3 of STATDISK) and the second sample (enter in column 4 of STATDISK) are below: First: 2, 9, 5, 6, 7, 8, 5, 2 Second: 3, 0, 5, 8, 7, 1, 8, 2, 5, 8, 7 What is the p-value? (Round to the nearest ten-thousandth.)