a For given hypothesis test, if we do not reject H0, and H0 is true. Your Answer: Choice Type I error has been committed. No error has been committed. Type III error has been committed. Type II error has been committed. Selected Correct
Question 2: Score 0/1
The manager of the quality department for a tire manufacturing company wants to know the average tensile strength of rubber used in making a certain brand of radial tire. The population is normally distributed and the population standard deviation is known. She uses a Z test to test the null hypothesis that the mean tensile strength is less than or equal to 800 pounds per square inch. The calculated Z test statistic is a positive value that leads to a p-value of .067 for the test.
Reference: 08_01
If the significance level is .05, the null hypothesis would be rejected. Your Answer: Choice Selected Correct True False
Question 3: Score 0/1
In forming a large sample confidence interval for , two assumptions are required: independent samples and sample sizes of at least 30. Your Answer: Choice Selected Correct True False
Question 4: Score 0/1
In order to test the effectiveness of a drug called XZR designed to reduce cholesterol levels, 9 heart patients' cholesterol levels are measured before they are given the drug. The same 9 patients use XZR for two continuous months. After two months of continuous use the 9 patients' cholesterol levels are measured again. The comparison of cholesterol levels before vs. after administering the drug is an example of testing the difference between: Your Answer: Choice Two population proportions. Matched pairs from two dependent populations. Two means from independent populations. Two population variances from independent populations. Selected Correct
Exercise 8.29 We consider using a random sample of 49 measurements to test H 0: = 20 versus Ha: < 20. If = 15.5 and = 7, calculate the value of the test statistic z. How much evidence is there that H0: = 20 is false and Ha: 20 is true? Round answers to 4 decimal places. Since p-value = 0 with a tolerance of 0.004 is less than .05, there is strong evidence. Recall that the television network will permit National Motors to claim that the ZX-900 achieves a shorter mean stopping distance than a competitor if H 0: > 60 can be rejected in favor of H1: < 60 by setting equal to .05. If the stopping distances of a random sample of n = 81 ZX-900s have a mean of = 56.6 ft, will National Motors be allowed to run the commercial? Assume here that = 6.22 . Calculate a 95 percent confidence interval for . Do the point estimate of and confidence interval for indicate that might be far enough below 60 feet to suggest that we have a practically important result? Round answers to 2 decimal places. Round confidence interval to 3 decimal places. A negative sign should be used instead of parentheses. H0: > 60 versus Ha : < 60. = 56.6 , = 6.22 , z = -4.92 with a tolerance of 0.01 so reject H0, run the commercial. 95% CI: [ 55.245 with a tolerance of 0.1 , 57.955 with a tolerance of 0.1] There might be some practical importance, because might be enough ( less than 60) to prevent some accidents.
Fortune magazine has periodically reported on the rise of fees and expenses charged by stock funds. Suppose that 10 years ago the average annual expense for stock funds was 1.19 percent. Let be the current mean annual expense for all stock funds, and assume that stock fund annual expenses are approximately normally distributed. If a random sample of 12 stock funds gives a sample mean annual expense of = 1.43 % with a standard deviation of s = .31%, use rejection points and this sample information to test H0: < 1.19% versus Ha: > 1.19% by setting a equal to .10, .05, .01, and .001. How much evidence is there that the current mean annual expense for stock funds exceeds the average of 10 years ago? Round t score to 3 decimal places. H0: = 1.19%, H0: > 1.19% t = 2.682 with a tolerance of 0.01 Reject H0 at = 0.1 and 0.05. The further the hypothesized mean is from the actual mean the greater the power of the test. Your Answer: Choice Selected Correct True False
Question 2: Score 0/1
The manager of the quality department for a tire manufacturing company wants to know the average tensile strength of rubber used in making a certain brand of radial tire. The population is normally distributed and the population standard deviation is known. She uses a Z test to test the null hypothesis that the mean tensile strength is 800 pounds per square inch. The calculated Z test statistic is a positive value that leads to a p-value of .045 for the test.
Reference: 08_02
If the significance level ( ) is .01, the null hypothesis would be rejected. Your Answer: Choice Selected Correct True False
Question 3: Score 0/1
When testing the difference between two proportions selected from populations with large independent samples, the Z test statistic is used. Your Answer: Choice Selected Correct True False
Question 4: Score 0/1
In order to test the effectiveness of a drug called XZR designed to reduce cholesterol levels, 9 heart patients' cholesterol levels are measured before they are given the drug. The same 9 patients use XZR for two continuous months. After two months of continuous use the 9 patients' cholesterol levels are measured again. The comparison of cholesterol levels before vs. after administering the drug is an example of testing the difference between: Your Answer: Choice Two population proportions. Matched pairs from two dependent populations. Two means from independent populations. Two population variances from independent populations. National Motors has equipped the ZX-900 with a new disk brake system. We define the stopping distance for a ZX-900 as the distance (in feet) required to bring the automobile to a complete stop from a speed of 38 mph under normal driving conditions using this new brake system. In addition, we define to be the mean stopping distance of all ZX-900s. One of the ZX-900's major competitors is advertised to achieve a mean stopping distance of 65 ft. National Motors would like to claim in a new television commercial that the ZX-900 achieves a shorter mean stopping distance. The standards and practices division of a major television network will permit National Motors to run the commercial if H o: > 65 can be rejected in favor of Ha: < 65 by setting a = 0.01 . Interpret what it means to set at 0.01 . There is a 0.01 probability that the network will advertise the ZX-900 achieves a shorter mean stopping distance than its competitor when it really does not. Selected Correct
Consider the automobile parts supplier in Exercise 8.10. Suppose that a problem-solving team will be assigned to rectify the process producing cylindrical engine parts if the null hypothesis H0: = 3 can be rejected in favor of H : 3 by setting equal to .05. a. A sample of 40 parts yields a sample mean diameter of = 3.006 inches. Assuming equals 0.016 , use rejection points and a p-value to test H0 versus H by setting equal to .05. Should the problem-solving team be assigned? b. Suppose that product specifications state that each and every part must have a diameter between 2.95 and 3.05 inches--that is, the specifications are 3" .05". Use the sample information given in part a to estimate an interval that contains almost all (99.73 percent) of the diameters. Compare this estimated interval with the specification limits. Are the specification limits being met, or are some diameters outside the specification limits? Explain. a. H0: = 3 versus H : 3 z = 2.37 with a tolerance of 0.01 za/2 = z.025 Reject H0 with = .05. p-value = 0.0178 with a tolerance of 0.004, Reject H0 with = .05. The problem-solving team should be assigned. b. 3 = [ 2.958 with a tolerance of 0.01, 3.054 with a tolerance of 0.01] Some diameters will exceed the upper specification limit of 3.05 inches. The auditor for a large corporation routinely monitors cash disbursements. As part of this process, the auditor examines check request forms to determine whether they have been properly approved. Improper approval can occur in several ways. For instance, the check may have no approval, the check request might be missing, the approval might be written by an unauthorized person, or the dollar limit of the authorizing person might be exceeded. a. Last year the corporation experienced a 5 percent improper check request approval rate. Since this was considered unacceptable, efforts were made to reduce the rate of improper approvals. Letting p be the proportion of all checks that are now improperly approved, set up the null and alternative hypotheses needed to attempt to demonstrate that the current rate of improper approvals is lower than last year's rate of 5 percent. b. Suppose that the auditor selects a random sample of 625 checks that have been approved in the last month. The auditor finds that 24 of these 625 checks have been improperly approved. Use rejection points and this sample information to test the hypotheses you set up in part a at the .10, .05, .01, and .001 levels of significance. How much evidence is there that the rate of improper approvals has been reduced below last year's 5 percent rate? c. Find the p-value for the test of part b. Use the p-value to carry out the test by setting equal to .10, .05, .01, and .001. Interpret your results. Round answers to 4 decimal places. a. H0: p > 0.05 versus Ha: p < 0.05. b. Reject H0 at = 0.1; some.
c. p-value = 0.0918 with a tolerance of 0.004 Reject H0 at = 0.1, but not at = 0.1, 0.05 and 0.01. Suppose two independent random samples of sizes n1 = 5 and n2 = 16 that have been taken from two normally distributed populations having variances 21 and 22 give sample standard deviations of s1 = 7 and s2 = 11. a. Test H0:21 = 22 versus Ha: 21 22 with = .05. What do you conclude? b. Test H0:21 > 22 versus Ha: 21 < 22 with = .05. What do you conclude?
Round answers to 2 decimal places. a. F = 2.47
Do not reject H0:
2 1
= 22 > 22
b. F = 2.47
Do not reject H0:
2 1
For a given hypothesis test, if we do not reject H0, and H0 is true. Your Answer: Choice Type II error has been committed. No error has been committed. Type III error has been committed. Type I error has been committed. Selected
Question 2: Score 1/1
When testing a null hypothesis about a single population mean and the population standard deviation is unknown, if the sample size is less than 30, one compares the computed test statistic for significance with a value from the ___________ distribution. Your Answer: Choice Chi-square t Z Selected
Binomial
Question 3: Score 1/1
In testing for the equality of means from two independent populations, if the hypothesis of equal population means is rejected at = .01, it will __________ be rejected at = .05. Your Answer: Choice Always Sometimes Never Selected
Question 4: Score 0/1
In testing the equality of population variances, two assumptions are required: independent samples and normally distributed populations. Your Answer: Choice Selected Correct True False Consider the following table from Exercise 8.56. | t-statistic Hypothesis Test: Mean vs. Hypothesized Value 750.000 Hypothesized value 13.97139 std. dev. 5 n 8.61 t | 8.6104 803.8 Mean Hourly Yield 8.786 std. error 4 df 0 p-value (two-tailed) | p-value | 0.001 Use the p-value to test H0: = 750 versus H: 750 by setting a equal to .10, .05, .01, and .001. How much evidence is there that the new catalyst changes the mean hourly yield? Round answers to 4 decimal places. p-value = 0.001 so reject H0 at = 0.1, 0.05 and 0.01 and do not reject at = 0.001. Very strong evidence.
For each of the following sample sizes and hypothesized values of the population proportion p, determine whether the sample size is large enough to use the large sample test about p given in this section: a. n = 800 and p0 = 0.9. e. n = 288 and p0 = 0.1. b. n = 200 and p0 = 0.03. f. n = 900 and p0 = 0.94. c. n = 20,000 and p0 = 0.03. g. n = 4,000 and p0 = 0.94. d. n = 100 and p0 = 0.25. h. n = 65 and p0 = 0.5. a. Yes, large enough. b. Yes, large enough. c. Yes, large enough. d. Yes, large enough. e. Yes, large enough. f. Yes, large enough. g. Yes, large enough. h.
Yes, large enough.
How safe are child car seats? Consumer Reports (May 2005) tested the safety of child car seats in 30 mph crashes. They found slim safety margins for some child car seats. Suppose that Consumer Reports simulates the safety of the market-leading child car seat. Their test consists of placing the maximum claimed weight in the car seat and simulating crashes at higher and higher miles per hour until a problem occurs. The following data identify the speed at which a problem with the car seat first appeared; such as the strap breaking, seat shell cracked, strap adjuster broke, detached from the base, etc.: 31.0, 29.4, 30.4, 28.9, 29.7, 30.1, 40.5, 31.7, 35.4, 29.1, 31.2, 30.2. Let denote the true mean speed at which a problem with the car seat first appears. The following MINITAB output gives the results of using the sample data to test H 0: = 30 versus Ha: > 30.
Test of mu = 30 vs > 30 Variable N mph Mean StDev SE Mean T P
12 31.4667 3.3249 How much evidence is there that exceeds 30 mph? b. Round answer to 4 decimal places.
0.5156
1.528
0.0774
Since p = 0.0774 there is some evidence. What impact did the September 11 terrorist attack have on U.S. airline demand? An analysis was conducted by Ito and Lee, Assessing the impact of the September 11 terrorist attacks on U.S. airline demand, in the Journal of Economics and Business (January--February 2005). They found a negative short-term effect of over 30% and an ongoing negative impact of over 7%. Suppose that we wish to test the impact by taking a random sample of 12 airline routes before and after 9/11. Passenger miles (millions of passenger miles) for the same routes were tracked for the 12 months prior to and the 12 months immediately following 9/11. Assume that the population of all possible paired differences is normally distributed. a. Set up the null and alternative hypotheses needed to determine whether there was a reduction in mean airline passenger demand. b. Below we present the MINITAB output for the paired differences test. Use the output and rejection points to test the hypotheses at the .10, .05, and .01 levels of significance. Has the true mean airline demand been reduced?
Paired T-Test and CI: Before 911, After 911 Paired T for Before 911 - After 911 N 12 12 12 Mean 117.333 95.583 21.75 StDev 26.976 25.518 10.3056 SE Mean 7.787 7.366 2.9750
Before 911 After 911 Difference
T-Test of mean difference = 0 (vs > 0): T-Value = 7.31 P-Value = 2E-4 c. Use the p-value to test the hypotheses at the .10, .05, and .01 levels of significance. How much evidence is there against the null hypothesis? Round your answer to t in part b to 2 decimal places, to p value in part c to 4 decimal places. Leave no cells blank. You must enter a "0" for the answer to grade correctly. a. H0: < 0versus Ha : > 0 b. t = 7.31 Reject H0 at = 0.1, 0.05, 0.01 and 0.001, but not at = no test values; extremely strong evidence. p-value = 2E-4 Reject H0 at = 0.1, 0.05, 0.01 and 0.001, but not at = no test values; extremely strong evidence.
c.
If a null hypothesis is rejected at a significance level of .01, it will ______ be rejected at a significance level of .05 Your Answer: Choice Always Sometimes Never Selected
Question 2: Score 1/1
For a given hypothesis test, if we do not reject H0, and H0 is true. Your Answer:
Choice Type III error has been committed. Type II error has been committed. No error has been committed. Type I error has been committed.
Selected
Question 3: Score 1/1
In testing the equality of population variances, two assumptions are required: independent samples and normally distributed populations. Your Answer: Choice Selected True False
Question 4: Score 1/1
In order to test the effectiveness of a drug called XZR designed to reduce cholesterol levels, 9 heart patients' cholesterol levels are measured before they are given the drug. The same 9 patients use XZR for two continuous months. After two months of continuous use the 9 patients' cholesterol levels are measured again. The comparison of cholesterol levels before vs. after administering the drug is an example of testing the difference between: Your Answer: Choice Two means from independent populations. Matched pairs from two dependent populations. Two population proportions. Two population variances from independent populations. We consider using a random sample of 49 measurements to test H 0: = 20 versus Ha: < 20. If = 14 and = 3.5 , calculate the value of the test statistic z. Use a rejection point to test H0 versus Ha by setting equal to .001. Selected
Round answers to 2 decimal places. = .001 z = -12 with a tolerance of 0.01 Reject H0 with = .001. How safe are child car seats? Consumer Reports (May 2005) tested the safety of child car seats in 30 mph crashes. They found slim safety margins for some child car seats. Suppose that Consumer Reports simulates the safety of the market-leading child car seat. Their test consists of placing the maximum claimed weight in the car seat and simulating crashes at higher and higher miles per hour until a problem occurs. The following data identify the speed at which a problem with the car seat first appeared; such as the strap breaking, seat shell cracked, strap adjuster broke, detached from the base, etc.: 31.0, 29.4, 30.4, 28.9, 29.7, 30.1, 37.5, 31.7, 35.4, 29.1, 31.2, 30.2. Let denote the true mean speed at which a problem with the car seat first appears. The following MINITAB output gives the results of using the sample data to test H0: = 30 versus Ha: > 30.
Test of mu = 30 vs > 30 Variable mph N Mean StDev SE Mean T P
12 31.2167 2.6226 0.5156 How much evidence is there that exceeds 30 mph? Round answer to 4 decimal places.
1.607
0.0682
Since p = 0.0682 there is some evidence. Suppose we have taken independent, random samples of sizes n1 = 7 and n2 = 7 from two normally distributed populations having means 1 and 2, suppose we obtain 1 = 230 , 2 = 225 , s1 = 5, s 2 = 6. Use rejection points to test the null hypothesis H0: 1 - 2 < 20 versus the alternative hypothesis Ha: 1 - 2 > 20 by setting a equal to .10, .05, .01, and .001. How much evidence is there that the difference between 1 and 2 exceeds 20? A negative sign should be used instead of parentheses. Round your answer to 3 decimal places. Leave no cells blank. You must enter a "0" for the answer to grade correctly. t = -5.081 with a tolerance of 0.01 Reject H0 at = no test values, weak evidence. If a null hypothesis is rejected at a significance level of .01, it will ______ be rejected at a significance level of .05 Your Answer: Choice Always Sometimes Selected
Never
Question 2: Score 0/1
The manager of the quality department for a tire manufacturing company wants to know the average tensile strength of rubber used in making a certain brand of radial tire. The population is normally distributed and the population standard deviation is known. She uses a Z test to test the null hypothesis that the mean tensile strength is less than or equal to 800 pounds per square inch. The calculated Z test statistic is a positive value that leads to a p-value of .067 for the test.
Reference: 08_01
If the significance level is .10, the null hypothesis would be rejected. Your Answer: Choice Selected Correct True False
Question 3: Score 1/1
In an experiment involving matched pairs, a sample of 12 pairs of observations is collected. The degree of freedom for the t statistic is 10. Your Answer: Choice Selected True False
Question 4: Score 1/1
In testing for the equality of means from two independent populations, if the hypothesis of equal population means is rejected at = .01, it will __________ be rejected at = .05. Your Answer:
Choice Always Never Sometimes
Selected
Suppose that a random sample of 16 measurements from a normally distributed population gives a sample mean of = 10.5 and a sample standard deviation of = 8 . Use rejection points to test H0: <10 versus H: >10 using levels of significance = .10, = .05, = .01, and = .001. What do you conclude at each value of ? Round answers to 3 decimal places. t = .25 t.10 = 1.341 with a tolerance of 0.01; Do not reject H0 at = .10 t.05 = 1.753 with a tolerance of 0.01; Do not reject H0 at = .05 t.01 = 2.602 with a tolerance of 0.01; Do not reject H0 at = .01 t.001 = 3.733 with a tolerance of 0.01; Do not reject H0 at = .001 In the Consolidated Power hypothesis test of H0: < 60 versus Ha: > 60 (as discussed in Exercise 8.79) find the sample size needed to make the probability of a Type I error equal to .025 and the probability of a Type II error corresponding to the alternative value a = 60.5 equal to .025. Here, assume equals 5 . Round answers up to the nearest integer. H0 < 60 versus Ha: > 60, 0 = 60 = .025 and = .025 for a = 60.5 z* = z = z.025 = 1.96 z = z.025 = 1.96 n = 1,537 with a tolerance of 1 The auditor for a large corporation routinely monitors cash disbursements. As part of this process, the auditor examines check request forms to determine whether they have been properly approved. Improper approval can occur in several ways. For instance, the check may have no approval, the check request might be missing, the approval might be written by an unauthorized person, or the dollar limit of the authorizing person might be exceeded. a. Last year the corporation experienced a 5 percent improper check request approval rate. Since this was considered unacceptable, efforts were made to reduce the rate of improper approvals. Letting p be the proportion of all checks that are now improperly approved, set up the null and alternative hypotheses needed to attempt to demonstrate that the current rate of improper approvals is lower than last year's rate of 5 percent. b. Suppose that the auditor selects a random sample of 625 checks that have been approved in the last month. The auditor finds that 20 of these 625 checks have
been improperly approved. Use rejection points and this sample information to test the hypotheses you set up in part a at the .10, .05, .01, and .001 levels of significance. How much evidence is there that the rate of improper approvals has been reduced below last year's 5 percent rate? c. Find the p-value for the test of part b. Use the p-value to carry out the test by setting equal to .10, .05, .01, and .001. Interpret your results. Round answers to 4 decimal places. a H : p > 0.05 versus Ha: p < 0.05. . 0 b Reject H0 at = 0.1 and 0.05; strong. . c p-value = 0.0197 with a tolerance of 0.004 . Reject H0 at = 0.1 and 0.05, but not at = 0.01 and 0.001. Do students reduce study time in classes where they achieve a higher midterm score? In a Journal of Economic Education article (Winter 2005), Gregory Krohn and Catherine O'Connor studied student effort and performance in a class over a semester. In an intermediate macroeconomics course, they found that students respond to higher midterm scores by reducing the number of hours they subsequently allocate to studying for the course. Suppose that a random sample of n = 8 students who performed well on the midterm exam was taken and weekly study time before and after the exam were compared. The resulting data are given in Table 9.6. Assume that the population of all possible paired differences is normally distributed. a. Set up the null and alternative hypotheses to test whether there is a difference in the true mean study time before and after the midterm exam. b. Below we present the MINITAB output for the paired differences test. Use the output and rejection points to test the hypotheses at the .10, .05 and .01 levels of significance. Has the true mean study time changed? Table 9.6 Weekly Study Time Data for Students Who Perform Well on the MidTerm Student 1 2 3 4 5 6 7 Before 13 14 17 17 15 14 11 After 7 9 11 10 19 10 12 Paired T-Test and CI: Study Before, Study After Paired T for Study Before - Study After N Mean StudyBefore 8 14.625 8 16 10
StDev 2.065879
SE Mean 0.7055
StudyAfter Difference
8 8
11 3.625
3.545621
1.2101
95% CI for mean difference: (0.312266 , 6.937734 ) T-Test of mean difference = 0 (vs not = 0): T-Value = 2.59 Round answers to 2 decimal places. a. H0: d = 0 versus Ha: d 0 b. t = 2.59 with a tolerance of 0.01 We have strong evidence. If a null hypothesis is rejected at a significance level of .01, it will ______ be rejected at a significance level of .05 Your Answer: Choice Always Never Sometimes Selected
Question 2: Score 1/1
When testing a null hypothesis about a single population mean and the population standard deviation is unknown, if the sample size is less than 30, one compares the computed test statistic for significance with a value from the ___________ distribution. Your Answer: Choice t Binomial Chi-square Z Selected
Question 3: Score 1/1
In testing for the equality of means from two independent populations, if the hypothesis of equal population means is rejected at = .01, it will __________ be rejected at = .05. Your Answer: Choice Always Sometimes Never Selected
Question 4: Score 1/1
When testing the difference between two proportions selected from populations with large independent samples, the Z test statistic is used. Your Answer: Choice Selected True False We consider using a random sample of 49 measurements to test H 0: = 20 versus Ha: < 20. If = 14 and = 3.5 , calculate the value of the test statistic z. Use a rejection point to test H0 versus Ha by setting equal to .001. Round answers to 2 decimal places. = .001 z = -12 with a tolerance of 0.01 Reject H0 with = .001. We consider using a random sample of 100 measurements to test H 0 : = 80 versus Ha: > 80. Use = 85 and = 20, to calculate the value of the test statistic z. Use a rejection point to test versus H0 Ha by setting equal to .01. Round the z value to one decimal place. = .01 z = 2.5 with a tolerance of 0.01 Reject H0 with =.01. The auditor for a large corporation routinely monitors cash disbursements. As part of this process, the auditor examines check request forms to determine whether they have been properly approved. Improper approval can occur in several ways. For instance, the check may have no approval, the check request might be missing, the approval might be written by an
unauthorized person, or the dollar limit of the authorizing person might be exceeded. a. Last year the corporation experienced a 5 percent improper check request approval rate. Since this was considered unacceptable, efforts were made to reduce the rate of improper approvals. Letting p be the proportion of all checks that are now improperly approved, set up the null and alternative hypotheses needed to attempt to demonstrate that the current rate of improper approvals is lower than last year's rate of 5 percent. b. Suppose that the auditor selects a random sample of 625 checks that have been approved in the last month. The auditor finds that 24 of these 625 checks have been improperly approved. Use rejection points and this sample information to test the hypotheses you set up in part a at the .10, .05, .01, and .001 levels of significance. How much evidence is there that the rate of improper approvals has been reduced below last year's 5 percent rate? c. Find the p-value for the test of part b. Use the p-value to carry out the test by setting equal to .10, .05, .01, and .001. Interpret your results. Round answers to 4 decimal places. a. H0: p > 0.05 versus Ha: p < 0.05. b. Reject H0 at = 0.1; some. c. p-value = 0.0918 with a tolerance of 0.004 Reject H0 at = 0.1, but not at = 0.1, 0.05 and 0.01. Assume that we have selected two independent random samples from populations having proportions p1 and p2 and that 1 = 800 /1000 = 0.8 and 2 = 750 /1000 = 0.75 . Test H0: p1 - p2 = 0 versus Ha: p1 - p2 0 by using rejection points and by setting a equal to .10, .05, .01, and .001. How much evidence is there that p1 and p2 differ? Explain. Hint: z.0005 = 3.29. Round answers to 2 decimal places. A negative sign should be used instead of parentheses. H0: p1 - p2 = 0 versus Ha: p1 - p2 0 z = 2.68 with a tolerance of 0.01 Reject H0 at = no values, but not at = 0.1, 0.05, 0.01 and 0.001; weak evidence. In the Consolidated Power hypothesis test of H0: < 60 versus Ha: > 60 (as discussed in Exercise 8.79) find the sample size needed to make the probability of a Type I error equal to .025 and the probability of a Type II error corresponding to the alternative value a = 60.5 equal to .025. Here, assume equals 4 . Round answers up to the nearest integer. H0 < 60 versus Ha: > 60, 0 = 60 = .025 and = .025 for a = 60.5
z* = z = z.025 = 1.96 z = z.025 = 1.96 n = 984 with a tolerance of 1 Consider the automobile parts supplier in Exercise 8.10. Suppose that a problem-solving team will be assigned to rectify the process producing cylindrical engine parts if the null hypothesis H0: = 3 can be rejected in favor of H : 3 by setting equal to .05. a. A sample of 40 parts yields a sample mean diameter of = 3.005 inches. Assuming equals 0.016 , use rejection points and a p-value to test H0 versus H by setting equal to .05. Should the problem-solving team be assigned? b. Suppose that product specifications state that each and every part must have a diameter between 2.95 and 3.05 inches--that is, the specifications are 3" .05". Use the sample information given in part a to estimate an interval that contains almost all (99.73 percent) of the diameters. Compare this estimated interval with the specification limits. Are the specification limits being met, or are some diameters outside the specification limits? Explain. a. H0: = 3 versus H : 3 z = 1.98 with a tolerance of 0.01 za/2 = z.025 Reject H0 with = .05. p-value = 0.0478 with a tolerance of 0.004, Reject H0 with = .05. The problem-solving team should be assigned. b. 3 = [ 2.957, 3.053] Some diameters will exceed the upper specification limit of 3.05 inches. Fortune magazine has periodically reported on the rise of fees and expenses charged by stock funds. Suppose that 10 years ago the average annual expense for stock funds was 1.19 percent. Let be the current mean annual expense for all stock funds, and assume that stock fund annual expenses are approximately normally distributed. If a random sample of 12 stock funds gives a sample mean annual expense of = 1.23 % with a standard deviation of s = .31%, use rejection points and this sample information to test H0: < 1.19% versus Ha: > 1.19% by setting a equal to .10, .05, .01, and .001. How much evidence is there that the current mean annual expense for stock funds exceeds the average of 10 years ago? Round t score to 3 decimal places. H0: = 1.19%, H0: > 1.19% t = 0.447 with a tolerance of 0.01 Reject H0 at = no values. Suppose two independent random samples of sizes n1 = 5 and n2 = 16 that have been taken from two normally distributed populations having variances 21 and 22 give sample standard deviations of s1 = 6 and s2 = 9. a. Test H0:21 = 22 versus Ha: 21 22 with = .05. What do you conclude? b. Test H0:21 > 22 versus Ha: 21 < 22 with = .05. What do you conclude?
Round answers to 2 decimal places. a. F = 2.25
Do not reject H0:
2 1
= 22
b. F = 2.25 > 22 When testing the null hypothesis about a single population variance, one compares the computed test statistic for significance with a value from the ___________ distribution.
Do not reject H0:
1 2
Your Answer: Choice Chi-square Z t Binomial Selected Correct
Question 2: Score 0/1
Which statement is incorrect? Your Answer: Choice When a false null hypothesis is not rejected, a Type II error has occurred. If the null hypothesis is rejected, it is concluded that the alternative hypothesis is true. The null hypothesis contains the equality sign. If we fail to reject the null hypothesis, then it is proven that null hypothesis is true. Selected Correct
Question 3: Score 1/1
In testing for the equality of means from two independent populations, if the null hypothesis is rejected, the test could result in: Your Answer: Choice Both a Type I error and a Type II error Neither a Type I error or a Type II error. A Type II error Either a Type I error or a Type II error. A Type I error. Selected
Question 4: Score 1/1
In order to test the effectiveness of a drug called XZR designed to reduce cholesterol levels, 9 heart patients' cholesterol levels are measured before they are given the drug. The same 9 patients use XZR for two continuous months. After two months of continuous use the 9 patients' cholesterol levels are measured again. The comparison of cholesterol levels before vs. after administering the drug is an example of testing the difference between: Your Answer: Choice Two population variances from independent populations. Two means from independent populations. Matched pairs from two dependent populations. Two population proportions. We consider using a random sample of 100 measurements to test H 0: = 80 versus Ha: > 80. Use = 88 and = 20, to calculate the value of the test statistic z. Calculate the p-value and use it to test H0 versus Ha at each of = .10, .05, .01, and .001. Round answers to 4 decimal places. z=4 p-value = .0001 Reject H0 at = all values. Selected
Question 6: Score 4/4
We consider using a random sample of n = 81 measurements to test H0: = 40 versus H0: 40. If = 33 and = 18 . Use rejection points to test H0 versus Ha by setting equal to .10. Round answers to 2 decimal places. = .10 z = -3.5
Reject
The manager of the quality department for a tire manufacturing company wants to know the average tensile strength of rubber used in making a certain brand of radial tire. She knows the population standard deviation and uses a Z test to test the null hypothesis that the mean tensile strength is 800 pounds per square inch. The calculated Z test statistic is a positive value that leads to a p-value of .067 for the test. If the significance level is .10, the null hypothesis would be rejected. Assume that the population of pressure values is normally distributed. Your Answer: Choice Selected Correct True False We consider using a random sample of n = 81 measurements to test H0: = 40 versus H0: 40. If = 39 and = 18 . Calculate the p-value and use it to test H0 versus H at each of = .10, .05, .01, and .001. Round the z value to two decimal places. Round the p-value to four decimal places. z = -0.5 p-value = .6171 reject H0 at = no values. The auditor for a large corporation routinely monitors cash disbursements. As part of this process, the auditor examines check request forms to determine whether they have been properly approved. Improper approval can occur in several ways. For instance, the check may have no approval, the check request might be missing, the approval might be written by an unauthorized person, or the dollar limit of the authorizing person might be exceeded. a. Last year the corporation experienced a 5 percent improper check request approval rate. Since this was considered unacceptable, efforts were made to reduce the rate of improper approvals. Letting p be the proportion of all checks that are now improperly
approved, set up the null and alternative hypotheses needed to attempt to demonstrate that the current rate of improper approvals is lower than last year's rate of 5 percent. b. Suppose that the auditor selects a random sample of 625 checks that have been approved in the last month. The auditor finds that 28 of these 625 checks have been improperly approved. Use rejection points and this sample information to test the hypotheses you set up in part a at the .10, .05, .01, and .001 levels of significance. How much evidence is there that the rate of improper approvals has been reduced below last year's 5 percent rate? c. Find the p-value for the test of part b. Use the p-value to carry out the test by setting equal to .10, .05, .01, and .001. Interpret your results. Round answers to 4 decimal places. a. H0: p > .05 versus Ha: p < .05. b. Reject H0 at = no values; no. c. p-value = 0.2743 with a tolerance of 0.004 Reject H0 at = no values, but not at = 0.1, 0.05, 0.01 and 0.001. Do students reduce study time in classes where they achieve a higher midterm score? In a Journal of Economic Education article (Winter 2005), Gregory Krohn and Catherine O'Connor studied student effort and performance in a class over a semester. In an intermediate macroeconomics course, they found that students respond to higher midterm scores by reducing the number of hours they subsequently allocate to studying for the course. Suppose that a random sample of n = 8 students who performed well on the midterm exam was taken and weekly study time before and after the exam were compared. The resulting data are given in Table 9.6. Assume that the population of all possible paired differences is normally distributed. a. Set up the null and alternative hypotheses to test whether there is a difference in the true mean study time before and after the midterm exam. b. Below we present the MINITAB output for the paired differences test. Use the output and rejection points to test the hypotheses at the .10, .05 and .01 levels of significance. Has the true mean study time changed? Table 9.6 Weekly Study Time Data for Students Who Perform Well on the MidTerm Student 1 2 3 4 5 6 7 Before 16 14 15 17 18 14 15 After 8 9 12 10 18 10 12 Paired T-Test and CI: Study Before, Study After Paired T for Study Before - Study After N Mean StudyBefore 8 15.625 8 16 10
StDev 1.407886
SE Mean 0.7055
StudyAfter Difference
8 8
11.125 4.5
3.090885
1.2101
95% CI for mean difference: (2.356537 , 6.643463 ) T-Test of mean difference = 0 (vs not = 0): T-Value = 4.97 Round answers to 2 decimal places. a. H0: d = 0 versus Ha: d 0 b. t = 4.97 with a tolerance of 0.01 We have extremely strong evidence. A new company is in the process of evaluating its customer service. The company offers two types of sales: 1. Internet sales; 2. Store sales. The marketing research manager believes that the Internet sales are more than 10% higher than store sales. The null hypothesis would be: Your Answer: Choice Pinternet Pstore .10 Pinternet Pstore = .10 Pinternet Pstore .10 Pinternet Pstore > .10 Pinternet Pstore < .10 In the Consolidated Power hypothesis test of H0: < 60 versus Ha: > 60 (as discussed in Exercise 8.79) find the sample size needed to make the probability of a Type I error equal to .025 and the probability of a Type II error corresponding to the alternative value a = 60.5 equal to .025. Here, assume equals 5 . Round answers up to the nearest integer. H0 < 60 versus Ha: > 60, 0 = 60 = .025 and = .025 for a = 60.5 z* = z = z.025 = 1.96 z = z.025 = 1.96 n = 1,537 with a tolerance of 1 In the book International Marketing, Philip R. Cateora reports the results of an MTV commissioned study of the lifestyles and spending habits of the 1434 age group in six countries. The survey results are given in Table 9.7. a. As shown in Table 9.7, 93 percent of the 14- to 34-year-olds surveyed in the United Selected Correct
States had purchased soft drinks in the last three months, while 81 percent of the 14- to 34-year-olds surveyed in Australia had done the same. Assuming that these results were obtained from independent random samples of 500 respondents in each country, carry out a hypothesis test that tests the equality of the population proportions of 14- to 34year-olds in the United States and in Australia who have purchased soft drinks in the last three months. Also, calculate a 95 percent confidence interval for the difference between these two population proportions, and use this interval to estimate the largest and smallest values that the difference between these proportions might be. Which of the Following Have You Purchased in the Past Three Months? Percentage Percentage in United Percentage Percentage Percentage Percentage in United States in Australia in Brazil in Germany in Japan Kingdom 93% 94 59 96 56 46 24 81% 94 40 43 39 50 33 93% 91 54 50 62 60 30 83% 70 33 67 45 46 38 91% 86 30 55 42 57 39 94% 85 49 63 44 57 40
Product Soft drinks Fast food Athletic footwear Blue jeans Beer* Cigarettes*
b. Again as shown in Table 9.7, 43 percent of the 14- to 34-year-olds surveyed in Australia had purchased athletic footwear in the last three months, while 50 percent of the 14- to 34-year-olds surveyed in Brazil had done the same. Assuming that these results were obtained from independent random samples of 500 respondents in each country, carry out a hypothesis test that tests the equality of the population proportions of 14- to 34-year-olds in Australia and in Brazil who have purchased athletic footwear in the last three months. Also, calculate a 95 percent confidence interval for the difference between these two population proportions, and use this interval to estimate the largest and smallest values that the difference between these proportions might be.
Round your answers to 4 decimal places. A negative sign should be used instead of parentheses. a. H0: p1 - p2 = 0 versus H: p1 - p2 0 z = 5.6418 with a tolerance of 1E-4 Reject H0 at equal to 0.001 [ 0.079 with a tolerance of 1E-4, 0.161 with a tolerance of 1E-4] b. H0: p1 - p2 = 0 versus H: p1 - p2 0 z = -2.219 with a tolerance of 1E-4
Reject H0 at equal to no test values
-0.1317 with a tolerance of 1E-4, -0.0083 with a tolerance of 1E-4]
In testing for the equality of variances from two independent populations, if the null hypothesis is false, the test could result in: Your Answer: Choice Both a Type I error and a Type II error A Type II error Neither a Type I error or a Type II error. Either a Type I error or a Type II error. A Type I error. Selected Correct
National Motors has equipped the ZX-900 with a new disk brake system. We define the stopping distance for a ZX-900 as the distance (in feet) required to bring the automobile to a complete stop from a speed of 38 mph under normal driving conditions using this new brake system. In addition, we define to be the mean stopping distance of all ZX-900s. One of the ZX-900's major competitors is advertised to achieve a mean stopping distance of 62 ft. National Motors would like to claim in a new television commercial that the ZX-900 achieves a shorter mean stopping distance. The standards and practices division of a major television network will permit National Motors to run the commercial if H o: > 62 can be rejected in favor of Ha: < 62 by setting a = 0.001 . Interpret what it means to set at 0.001 . There is a 0.001 probability that the network will advertise the ZX-900 achieves a shorter mean stopping distance than its competitor when it really does not.
Fortune magazine has periodically reported on the rise of fees and expenses charged by stock funds. Suppose that 10 years ago the average annual expense for stock funds was 1.19 percent. Let be the current mean annual expense for all stock funds, and assume that stock fund annual expenses are approximately normally distributed. If a random sample of 12 stock funds gives a sample mean annual expense of = 1.83 % with a standard deviation of s = .31%, use rejection points and this sample information to test H0: < 1.19% versus Ha: > 1.19% by setting a equal to .10, .05, .01, and .001. How much evidence is there that the current mean annual expense for stock funds exceeds the average of 10 years ago? Round t score to 3 decimal places.
H0: = 1.19%, H0: > 1.19% t = 7.152 Reject H0 at = 0.1, 0.05, 0.01 and 0.001.
In an article in the Journal of Retailing, Kumar, Kerwin, and Pereira study factors affecting merger and acquisition activity in retaining by comparing "target firms" and "bidder firms" with respect to several financial and marketing-related variables. If we consider two of the financial variables included in the study, suppose a random sample of 36 "target firms" gives a mean earnings per share of 2.62 with a standard deviation of 0.92 , and that this sample gives a mean debt-to-equity ratio of 1.66 with a standard deviation of 0.82. Furthermore, an independent random sample of 36 "bidder firms" gives a mean earnings per share of 1 with a standard deviation of 0.74 , and this sample gives a mean debt-to-equity ratio of 1.58 with a standard deviation of 0.81. a. Set up the null and alternative hypotheses needed to test whether the mean earnings per share for all target firms differs from the mean earnings per share for all bidder firms. Test these hypotheses at the .10, .05, .01, and .001 levels of significance. How much evidence is there that these means differ? Explain. b. Calculate a 95 percent confidence interval for the difference between the mean earnings per share for target firms and bidder firms.
Round answers to 2 decimal places. A negative sign should be used instead of parentheses. a. H0: T - B = 0 versus Ha: T - B 0 t = 8.23 with a tolerance of 0.01 Reject H0 at equal to 0.1, 0.05, 0.01 and 0.001; extremely strong evidence. b. [ 1.23 with a tolerance of 0.01 , 2.01 with a tolerance of 0.01 ] The manager of the quality department for a tire manufacturing company wants to know the average tensile strength of rubber used in making a certain brand of radial tire. The population is normally distributed and the population standard deviation is known. She uses a Z test to test the null hypothesis that the mean tensile strength is less than or equal to 800 pounds per square inch. The calculated Z test statistic is a positive value that leads to a p-value of .067 for the test.
Reference: 08_01
If the significance level is .10, the null hypothesis would be rejected. Your Answer: Choice Selected Correct True False
We consider using a random sample of n = 81 measurements to test H0: = 40 versus H0: 40. If = 38 and = 18 . Use rejection points to test H0 versus Ha by setting equal to .10. Round answers to 2 decimal places. = .10 z = -1
Do not reject
Use the situation in Exercises 8.11 and 8.46 and a confidence interval to test H0: = 16 versus H: 16 by setting equal to .05. Round answers to 4 decimal places. n = 36, = 16.03 , = 0.25 95% C.I. = [ 15.9483 with a tolerance of 0.01, 16.1117 with a tolerance of 0.01] Since = 16 doesfall within the 95% C.I., do not rejectH 0 with = .05.
Suppose we have taken independent, random samples of sizes n1 = 7 and n2 = 7 from two normally distributed populations having means 1 and 2, suppose we obtain 1 = 240 , 2 = 205 , s1 = 5, s 2 = 6. Use rejection points to test the null hypothesis H0: 1 - 2 < 20 versus the alternative hypothesis Ha: 1 - 2 > 20 by setting a equal to .10, .05, .01, and .001. How much evidence is there that the difference between 1 and 2 exceeds 20? A negative sign should be used instead of parentheses. Round your answer to 3 decimal places. Leave no cells blank. You must enter a "0" for the answer to grade correctly. t = 5.081 Reject H0 at = 0.1, 0.05, 0.01 and 0.001, extremely strong evidence. In a conflict of interest scenario presented to a sample of 205 marketing researchers and that 111 of these researchers disapproved of the actions taken. a. Let p be the proportion of all marketing researchers who disapprove of the actions taken in the conflict of interest scenario. Set up the null and alternative hypotheses needed to attempt to provide evidence supporting the claim that a majority (more than 50 percent) of all marketing researchers disapprove of the actions taken. b. Assuming that the sample of 205 marketing researchers has been randomly selected, use rejection points and the previously given sample information to test the hypotheses you set up in part a at the .10, .05, .01, and .001 levels of significance. How much evidence is there that a majority of all marketing researchers disapprove of the actions taken? c. Suppose a random sample of 1,000 marketing researchers reveals that 539 of the researchers disapprove of the actions taken in the conflict of interest scenario. Use rejection points to determine how much evidence there is that a majority of all
marketing researchers disapprove of the actions taken. d. Note that in parts b and c the sample proportion is (essentially) the same. Explain why the results of the hypothesis tests in parts b and c differ. Round answers to 2 decimal places. Round to 4 decimal places for calculations. a. H0: p < 0.5 versus H : p > 0.5. b. = 0.5415 with a tolerance of 0.004 z = 1.19 with a tolerance of 0.01 z.10 = 1.28 Reject H0 at = no values. Weak evidence. c. = 0.539 with a tolerance of 0.004
z = 2.47 with a tolerance of 0.01 z.001 = 3.09 Reject H0 at = .01 and do not reject at = .001. Very strong evidence. d. = .54 Based on a much larger sample provides stronger p is greater than .50. The bad debt ratio for a financial institution is defined to be the dollar value of loans defaulted divided by the total dollar value of all loans made. Suppose that a random sample of seven Ohio banks is selected and that the bad debt ratios (written as percentages) for these banks are 3 %, 6 %, 5 %, 4 %, 3 %, 5 %, and 6 %. a. Banking officials claim that the mean bad debt ratio for all Midwestern banks is 3.5 percent and that the mean bad debt ratio for Ohio banks is higher. Set up the null and alternative hypotheses needed to attempt to provide evidence supporting the claim that the mean bad debt ratio for Ohio banks exceeds 3.5 percent. b. Assuming that bad debt ratios for Ohio banks are approximately normally distributed, use rejection points and the given sample information to test the hypotheses you set up in part a by setting equal to .10, .05, .01, and .001. How much evidence is there that the mean bad debt ratio for Ohio banks exceeds 3.5 percent? What does this say about the banking official's claim? Round answers to 3 decimal places. a. H0: < 3.5 %versus H : > 3.5%. b. 4.571 with a tolerance of 0.001, s = 1.272 with a tolerance of 0.001, n = 7, degrees of freedom = 6, 0 = 3.5 with a tolerance of 0.01 t = 2.228 with a tolerance of 0.004 There is strong evidence that H0 is false. The auditor for a large corporation routinely monitors cash disbursements. As part of this process, the auditor examines check request forms to determine whether they have been properly approved. Improper approval can occur in several ways. For instance, the check may have no approval, the check request might be missing, the approval might be written by an =
unauthorized person, or the dollar limit of the authorizing person might be exceeded. a. Last year the corporation experienced a 5 percent improper check request approval rate. Since this was considered unacceptable, efforts were made to reduce the rate of improper approvals. Letting p be the proportion of all checks that are now improperly approved, set up the null and alternative hypotheses needed to attempt to demonstrate that the current rate of improper approvals is lower than last year's rate of 5 percent. b. Suppose that the auditor selects a random sample of 625 checks that have been approved in the last month. The auditor finds that 22 of these 625 checks have been improperly approved. Use rejection points and this sample information to test the hypotheses you set up in part a at the .10, .05, .01, and .001 levels of significance. How much evidence is there that the rate of improper approvals has been reduced below last year's 5 percent rate? c. Find the p-value for the test of part b. Use the p-value to carry out the test by setting equal to .10, .05, .01, and .001. Interpret your results. Round answers to 4 decimal places. a. H0: p > .05 versus Ha: p < .05. b. Reject H0 at = 0.1 and 0.05; strong. c. p-value = 0.0446 with a tolerance of 0.004 Reject H0 at = 0.1 and 0.05, but not at = 0.01 and 0.001.