Q&A Hero

Q: A manufacturer of industrial plywood has a contract to supply a boat maker with a large amount of plywood. One of the specifications calls for the standard deviation in thickness to not exceed .03 inch. A sample of n = 30 sheets was sampled randomly from a recent production run. The mean thickness was right at the 3/4 inch target thickness and the sample standard deviation was .05 inch. Testing at the 0.05 level of significance, which of the following is true?A) The test statistic is approximately 80.56.B) The critical value is approximately x2 = 43.773.C) The test statistic is approximately 48.333.D) Based on the sample data, there is no evidence to suggest that the plywood is not meeting the specifications.

Q: If a hypothesis test for a single population variance is to be conducted, which of the following statements is true? A) The null hypothesis must be stated in terms of the population variance. B) The chi-square distribution is used. C) If the sample size is increased, the critical value is also increased for a given level of statistical significance. D) All of the above are true.

Q: A fast food restaurant that sells burritos is concerned about the variability in the amount of filling that different employees place in the burritos. To achieve product consistency it needs this variability to be no more than 1.7 ounces. A sample of n = 18 burritos showed a sample variance of 2.89 ounces. Using a 0.10 level of significance, what can you conclude? A) The standards are being met since (test statistic) < (critical value). B) The standards are not being met since (test statistic) > (critical value). C) The standards are being met since (test statistic) > (critical value). D) The standards are not being met since (test statistic) < (critical value).

Q: A potato chip manufacturer has found that in the past the standard deviation of bag weight has been 0.2 ounce. They want to test whether the standard deviation has changed. The null hypothesis is:

Q: When conducting a one-tailed hypothesis test of a population variance using a sample size of n = 24 and a 0.10 level of significance, the critical value is: A) 32.0069 B) 35.1725 C) 33.1962 D) 36.4150

Q: If a hypothesis test for a single population variance is to be conducted using a significance level of 0.10, a sample size of n = 16, and the test is a one-tailed upper-tail test, the critical value is: A) z = 1.28 B) t = 1.345 C) x2 = 22.3071 D) x2 = 24.9958

Q: When a hypothesis test is to be conducted regarding a population variance, the test statistic will be: A) a t-value from the t-distribution. B) an x2 value from the chi-square distribution. C) a z-value from the standard normal distribution. D) a binomial distribution p value.

Q: An analyst plans to test whether the standard deviation for the time it takes bank tellers to provide service to customers exceeds the standard of 1.5 minutes. The correct null and alternative hypotheses for this test are:

Q: The F test statistic for testing whether the variances of two populations are the same is always positive.

Q: There is interest at the American Savings and Loan as to whether there is a difference between average daily balances in checking accounts that are joint accounts (two or more members per account) versus single accounts (one member per account). To test this, a random sample of checking accounts was selected with the following results: Single Accounts Joint Accounts n = 20 n = 30 s = $256 s = $300 = $1,123 = $1,245 Based upon these data, if tested using a significance level equal to 0.10, the assumption of equal population variances should be rejected.

Q: There is interest at the American Savings and Loan as to whether there is a difference between average daily balances in checking accounts that are joint accounts (two or more members per account) versus single accounts (one member per account). To test this, a random sample of checking accounts was selected with the following results: Single Accounts Joint Accounts n = 20 n = 30 s = $256 s = $300 = $1,123 = $1,245 Based upon these data, the test statistic for testing whether the two populations have equal variances is F = 1.3733.

Q: A first step in testing whether two populations have the same mean value using the t-distribution is to use the chi-square distribution to test whether the populations have equal variances.

Q: Because of the way the F-distribution is formed, all F-tests are one-tailed tests.

Q: The F-distribution can only have positive values.

Q: A frozen food company that makes burritos currently has employees making burritos by hand. It is considering purchasing equipment to automate the process and wants to determine if the automated process would result in lower variability of burrito weights. It takes a random sample from each process as shown below. Process 1 (by hand) Process 2 (automated) n = 10 n = 8 s = 0.22 ounces s = 0.16 ounces In conducting the hypothesis test, the test statistic is F = 1.375.

Q: A frozen food company that makes burritos currently has employees making burritos by hand. It is considering purchasing equipment to automate the process and wants to determine if the automated process would result in lower variability of burrito weights. It takes a random sample from each process as shown below. Process 1 (by hand) Process 2 (automated) n = 10 n = 8 s = 0.2 ounces s = 0.17 ounces To conduct a hypothesis test using a 0.05 level of significance, the critical value is 3.347.

Q: A frozen food company that makes burritos currently has employees making burritos by hand. It is considering purchasing equipment to automate the process and wants to determine if the automated process would result in lower variability of burrito weights. To conduct a hypothesis test using a 0.05 level of significance, the proper format for the null and alternative hypotheses is (where the current by-hand process is process 1 and the automated process is process 2).H0 : Ha : >

Q: One of the key quality characteristics in many service environments is that the variation in service time be reasonably small. Recently, a major amusement park company initiated a new line system at one of its parks. It then wished to compare this new system with the old system in place at a comparable park in another state. At issue is whether the standard deviation in waiting time is less under the new line system than under the old line system. The following information was collected: System 1 (old) System 2 (new) n1 = 6 customers n2 = 8 customers 1 = 15.6 minutes 2 = 17.8 minutes s1 = 4.0 minutes s2 = 3.2 minutes Assuming that it wishes to conduct the test using a 0.05 level of significance, the null hypothesis should be rejected since the test statistic exceeds the F-critical value from the F-distribution table.

Q: One of the key quality characteristics in many service environments is that the variation in service time be reasonably small. Recently, a major amusement park company initiated a new line system at one of its parks. It then wished to compare this new system with the old system in place at a comparable park in another state. At issue is whether the standard deviation in waiting time is less under the new line system than under the old line system. The following information was collected: System 1 (old) System 2 (new) n1 = 6 customers n2 = 8 customers 1 = 15.6 minutes 2 = 17.8 minutes s1 = 4.0 minutes s2 = 3.2 minutes Assuming that it wishes to conduct the test using a 0.05 level of significance, the test statistic will be .

Q: One of the key quality characteristics in many service environments is that the variation in service time be reasonably small. Recently, a major amusement park company initiated a new line system at one of its parks. It then wished to compare this new system with the old system in place at a comparable park in another state. At issue is whether the standard deviation in waiting time is less under the new line system than under the old line system. The following information was collected: System 1 (old)System 2 (new)n1 = 6 customersn2 = 8 customers1 = 15.6 minutes2 = 17.8 minutess1 = 4.0 minutess2 = 3.2 minutesAssuming that it wishes to conduct the test using a 0.05 level of significance, the correct null and alternative hypotheses would be:H0 : Ha : =

Q: The logic behind the F-test for testing whether two populations have equal variances is to determine whether sample variances computed from random samples selected from the two populations differ due to sampling error, or whether the difference is more than can be attributed to sampling error alone, in which case, we conclude that the populations have different variances.

Q: In a two-tailed hypothesis test involving two population variances, if the null hypothesis is true then the F-test statistic should be approximately equal to 1.0.

Q: One of the major automobile makers has developed two new engines. At question is whether the two engines have the same variability with respect to miles per gallon. To test this using a 0.10 level of significance, the following information is available: Engine 1 Engine 2 n1 = 7 n2 = 9 1 = 28.7 2 = 33.4 s1 = 3.4 s2 = 4.12 Based on this situation and the information provided, the null hypothesis cannot be rejected and it is possible that the two engines produce the same variation in mpg.

Q: One of the major automobile makers has developed two new engines. At question is whether the two engines have the same variability with respect to miles per gallon. To test this, the following information is available: Engine 1 Engine 2 n1 = 7 n2 = 9 1 = 28.7 2 = 33.4 s1 = 3.4 s2 = 4.12 Based on this situation and the information provided, the test statistic is 1.2118.

Q: One of the major automobile makers has developed two new engines. At question is whether the two engines have the same variability with respect to miles per gallon. To test this using a significance level equal to 0.10, the following information is available: Engine 1 Engine 2 n1 = 7 n2 = 9 1 = 28.7 2 = 33.4 s1 = 3.4 s2 = 4.12 Based on this situation and the information provided, the critical value is F = 4.147 .

Q: One of the major automobile makers has developed two new engines. At question is whether the two engines have the same variability with respect to miles per gallon. To test this, the following information is available: Engine 1Engine 2n1 = 7n2 = 91 = 28.72 = 33.4s1 = 3.4s2 = 4.12Based on this situation and the information provided, the appropriate null and alternative hypotheses are:H0 : Ha : =

Q: A two-tailed test for two population variances could have a null hypothesis like the following: H0 : =

Q: If a one-tailed F-test is employed when testing a null hypothesis about two population variances, the test statistic is an F-value formed by taking the ratio of the two sample variances so that the sample variance predicted to be larger is placed in the numerator.

Q: A potato chip manufacturer has two packaging lines and wants to determine if the variances differ between the two lines. They take samples of n1 = 10 bags from line 1 and n2 = 8 bags from line 2. To perform the hypothesis test at the 0.05 level of significance, the critical value is F = 3.68.

Q: The null hypothesis that two population variances are equal will tend to be rejected if the ratio of the sample variances from each population is substantially larger than 1.0.

Q: In a test for determining whether two population variances are the same or different, the larger the sample sizes from the two populations, the lower will be the chance of making a Type I statistical error.

Q: In a hypothesis test for the equality of two variances, the lower-tail critical value does not need to be found as long as the larger sample variance is placed in the denominator of the test statistic.

Q: The managers for a vegetable canning facility claim the standard deviation for the ounces per can on the new automated line is less than for the older manual line. Given this, the correct null and alternative hypotheses for performing the statistical test are:H0 : 1 = 2Ha : 1 2

Q: In a two-tailed test for the equality of two variances, the critical value is determined by going to the F-distribution table with an upper-tail area equal to alpha divided by two.

Q: A potato chip manufacturer has two packaging lines and wants to determine if the variances differ between the two lines. They take a sample of n= 15 bags from each line and find the following: The value of the test statistic is F = 1.5

Q: In a two-tailed hypothesis test for the difference between two population variances, if s1 = 3 and s2 = 5, then the test statistic is F = 2.7778.

Q: In a two-tailed hypothesis test for the difference between two population variances, if s1 = 3 and s2 = 5, then the test statistic is F = 1.6667.

Q: In a two-tailed hypothesis test for the difference between two population variances, the test statistic is an F-ratio formed by putting the larger sample variance in numerator.

Q: In a hypothesis test involving two population variances, if the null hypothesis states that the two variances are strictly equal, then the test statistic is a chi-square statistic.

Q: The F-distribution is used to test whether two sample variances are equal.

Q: A sample of n observations is taken from a normally distributed population to estimate the population variance. The degrees of freedom for the chi-square distribution are n-2.

Q: When using a chi-square test for the variance of one population, we are assuming that the population is normally distributed.

Q: For a given significance level, increasing the sample size will tend to increase the chi-square critical value used in testing the null hypothesis about a population variance.

Q: One of the most important aspects of quality improvement is the idea of reducing the variability in a product or service. For instance, a major bank has worked to reduce the variability in the service time at the drive-through. The managers believe that the standard deviation in service time should not exceed 30 seconds. To test whether this goal is being achieved, a random sample of n = 25 cars is selected each week and the service time for each car is measured. Last week, the mean time was 345 seconds with a standard deviation equal to 38 seconds. Given this information, if the significance level is 0.10, the critical value from the chi-square table is about 34.3.

Q: Assume a sample of size n = 12 has been collected. To perform a hypothesis test of a population variance using a 0.05 level of significance, where the null hypothesis is:H0: 2 = 25The upper tail critical value is 21.92.

Q: The variance in the diameter of a bolt should not exceed 0.500 mm. A random sample of n = 12 bolts showed a sample variance of 0.505 mm. The test statistic is X2 = 11.11.

Q: If we are interested in performing a one-tailed, upper-tail hypothesis test about a population variance where the level of significance is .05 and the sample size is n = 20, the critical chi-square value to be used is 30.1435.

Q: A machine that is used to fill soda pop cans with pop has an adjustable mean fill setting, but the standard deviation is not supposed to exceed 0.18 ounces. To make sure that this is the case, the managers at the beverage company each day select a random sample of n = 6 cans and measure the fill volume carefully. In one such case, the following data (ounces per can) were observed. 12.2911.8812.0312.2211.7611.98Based on these sample data, the test statistic is approximately x2 = 5.01.

Q: A contract calls for the strength of a steel rod to stand up to pressure of 200 lbs per square inch on average. The contract also requires that the variability in strength for individual steel rods be no more than 5 pounds per square inch. If a random sample of n = 15 rods is selected and the sample standard deviation is 6.7 pounds, the test statistic is approximately x2 = 25.138.

Q: A one-tailed hypothesis test for a population variance always has the rejection region in the upper tail.

Q: The test statistic that is used when testing a null hypothesis for a population variance is the standard normal z-value.

Q: When estimating the difference between two population means, when should the normal distribution be used and when should the t-distribution be used?

Q: A market research firm has come up with two coupon designs that could be inserted into the envelopes that go out with the monthly statement to credit card customers. The coupons offer the customer an opportunity to become a member of a travel club. The research firm is interested in estimating the difference in proportion of customers who will join the club after receiving one or the other of the two coupons. To obtain this estimate, the research firm has sent out coupon design 1 to a random sample of 100 customers. A second random sample of 100 customers received coupon 2. For the first coupon, 11 customers joined the travel club, while 15 customers who received coupon 2 joined. Based upon this sample information, develop and interpret the desired 95 percent confidence interval estimate.

Q: A maker of toothpaste is interested in testing whether the proportion of adults (over age 18) who use its toothpaste and have no cavities within a six-month period is any different from the proportion of children (18 and under) who use the toothpaste and have no cavities within a six-month period. To test this, it has selected a sample of adults and a sample of children randomly from the population of those customers who use their toothpaste. The following results were observed. Adults ChildrenSample size 100 200Number with 0 cavities 83 165Based on these sample data and using a significance level of 0.05, what conclusion should be reached? Use the p-value approach to conduct the test.

Q: The National Football League (NFL) is interested in testing to see whether there is a difference in the proportion of male fans that prefer instant replay to review officials' calls and the proportion of female fans that prefer instant replay. It is believed that males tend to favor the practice to a higher degree than do females. To test this, random samples of male fans and female fans were selected and the following results were obtained: Male Fans Female FansSample size 300 100Number Preferring 234 57Using a significance level equal to 0.05, what conclusion should be reached based on the sample data?

Q: A real estate agent believes that home with swimming pools take longer to sell than home without swimming pools. A random sample of each type of recently sold homes was taken where the number of days on the market is recorded. Results are: Homes with pool Homes w/o poolSample mean 77 days 65 daysSample Standard deviation 8.4 days 7.2 daysSample size 19 23Assuming that the populations are normally distributed and the variances are equal, conduct the appropriate hypothesis test to determine if the real estate agent is correct. Use the 0.05 level of significance.

Q: A PC company uses two suppliers for rechargeable batteries for its notebook computers. Two factors are important quality features of the batteries: mean use time and variation. It is desirable that the mean use time be high and the variability be low. Recently, the PC maker conducted a test on batteries from the two suppliers. In the test, 9 randomly selected batteries from Supplier 1 were tested and 12 randomly selected batteries from Supplier 2 were tested. The following results were observed: Supplier 1 Supplier 2 n1 = 9 n2 = 12 1 = 67.25 min 2 = 72.4 min. S1= 11.2 min S2 = 9.9 minBased on these sample results, can the PC maker conclude that a difference exists between the two batteries with respect to the population mean use time? Test using a 0.10 level of significance.

Q: A major U.S. oil company has developed two blends of gasoline. Managers are interested in determining whether a difference in mean gasoline mileage will be obtained from using the two blends. As part of their study, they have decided to run a test using the Chevrolet Impala automobile with automatic transmissions. They selected a random sample of 100 Impalas using Blend 1 and another 100 Impalas using Blend 2. Each car was first emptied of all the gasoline in its tank and then filled with the designated blend of the new gasoline. The car was then driven 200 miles on a specified route involving both city and highway roads. The cars were then filled and the actual miles per gallon were recorded. The following summary data were recorded: Blend 1 Blend 2Sample Size 100 100Sample Mean 23.4 mpg 25.7 mpgSample St. Dev. 4.0 mpg 4.2 mpgBased on the sample data, using a 0.05 level of significance, what conclusion should the company reach about whether the population mean mpg is the same or different for the two blends? Use the p-value approach to test the null hypothesis.

Q: A major U.S. oil company has developed two blends of gasoline. Managers are interested in determining whether a difference in mean gasoline mileage will be obtained from using the two blends. As part of their study, they have decided to run a test using the Chevrolet Impala automobile with automatic transmissions. They selected a random sample of 100 Impalas using Blend 1 and another 100 Impalas using Blend 2. Each car was first emptied of all the gasoline in its tank and then filled with the designated blend of the new gasoline. The car was then driven 200 miles on a specified route involving both city and highway roads. The cars were then filled and the actual miles per gallon were recorded. The following summary data were recorded: Blend 1 Blend 2Sample Size 100 100Sample Mean 23.4 mpg 25.7 mpgSample St. Dev. 4.0 mpg 4.2 mpgBased on the sample data, using a 0.05 level of significance, what conclusion should the company reach about whether the population mean mpg is the same or different for the two blends? Use the test statistic approach to test the null hypothesis.

Q: A major U.S. oil company has developed two blends of gasoline. Managers are interested in estimating the difference in mean gasoline mileage that will be obtained from using the two blends. As part of their study, they have decided to run a test using the Chevrolet Impala automobile with automatic transmissions. They selected a random sample of 100 Impalas using Blend 1 and another 100 Impalas using Blend 2. Each car was first emptied of all the gasoline in its tank and then filled with the designated blend of the new gasoline. The car was then driven 200 miles on a specified route involving both city and highway roads. The cars were then filled and the actual miles per gallon were recorded. The following summary data were recorded: Blend 1 Blend 2Sample Size 100 100Sample Mean 23.4 mpg 25.7 mpgSample St. Dev. 4.0 mpg 4.2 mpgBased on these sample data, compute and interpret the 95 percent confidence interval estimate for the difference in mean mpg for the two blends.

Q: The American College Health Association produced the National College Health Assessment (Andy Gardiner, "Surfacing from Depression," February 6, 2006). The assessment indicates that the percentage of U.S. college students who report having been diagnosed with depression has risen from 2000. The assessment surveyed 47,202 students at 74 campuses. It discovered that 10.3% and 14.9% of students indicated that they had been diagnosed with depression in 2000 and 2004, respectively. Assume that half of the students surveyed were surveyed in 2004. Indicate the margin of error for estimating p1 - p2 with 1 - 2. A) 0.04156 B) 0.00121 C) 0.03418 D) 0.00597

Q: The American College Health Association produced the National College Health Assessment (Andy Gardiner, "Surfacing from Depression," February 6, 2006). The assessment indicates that the percentage of U.S. college students who report having been diagnosed with depression has risen from 2000. The assessment surveyed 47,202 students at 74 campuses. It discovered that 10.3% and 14.9% of students indicated that they had been diagnosed with depression in 2000 and 2004, respectively. Assume that half of the students surveyed were surveyed in 2004. Conduct a hypothesis test to determine if there has been more than a 0.04 increase in the proportion of students who indicated they have been diagnosed with depression. Use a significance level of 0.05 and a p-value approach to this test. A) Since p-value = 0.065 > 0.05, do not reject H0. There is not sufficient evidence to conclude that there has been more than a 0.04 increase in the proportion of students that indicate they have been diagnosed with depression. B) Since p-value = 0.025 < 0.05, reject H0. There is sufficient evidence to conclude that there has been more than a 0.04 increase in the proportion of students that indicate they have been diagnosed with depression. C) Since p-value = 0.072 < 0.05, reject H0. There is sufficient evidence to conclude that there has been more than a 0.04 increase in the proportion of students that indicate they have been diagnosed with depression. D) Since p-value = 0.071 > 0.05, do not reject H0. There is not sufficient evidence to conclude that there has been more than a 0.04 increase in the proportion of students that indicate they have been diagnosed with depression.

Q: Suppose as part of a national study of economic competitiveness a marketing research firm randomly sampled 200 adults between the ages of 27 and 35 living in metropolitan Seattle and 180 adults between the ages of 27 and 35 living in metropolitan Minneapolis. Each adult selected in the sample was asked, among other things, whether they had a college degree. From the Seattle sample 66 adults answered yes and from the Minneapolis sample 63 adults answered yes when asked if they had a college degree. Based on the sample data, can we conclude that there is a difference between the population proportions of adults between the ages of 27 and 35 in the two cities with college degrees? Use a level of significance of 0.10 to conduct the appropriate hypothesis test. A) Since the test statistic, 1.8214, is greater than the critical value of 1.645, reject the null hypothesis and conclude that there is a higher proportion of Seattle adults that have a college degree B) Since the test statistic, 2.0112, is greater than the critical value of 1.645, reject the null hypothesis and conclude that there is a higher proportion of Seattle adults that have a college degree. C) Since the test statistic, 0.7001, is not greater than the critical value of 1.645, do not reject the null hypothesis and conclude that there is not a higher proportion of Seattle adults that have a college degree. D) Since the test statistic, 0.8921, is not greater than the critical value of 1.645, do not reject the null hypothesis and conclude that there is not a higher proportion of Seattle adults that have a college degree.

Q: In an article entitled "Childhood Pastimes Are Increasingly Moving Indoors," Dennis Cauchon asserts that there have been huge declines in spontaneous outdoor activities such as bike riding, swimming, and touch football. In the article, he cites separate studies by the national Sporting Goods Association and American Sports Data that indicate bike riding alone is down 31% from 1995 to 2004. According to the surveys, 68% of 7- to 11-year-olds rode a bike at least six times in 1995 and only 47% did in 2004. Assume the sample sizes were 1,500 and 2,000, respectively. Calculate a 95% confidence interval to estimate the proportion of 7- to 11-year-olds who rode their bike at least six times in 2004. A) (0.4481, 0.4919) B) (0.4324, 0.4676) C) (0.4021, 0.5179) D) (0.4712, 0.4888).

Q: The following samples are observations taken from the same elements at two different times: Unit Sample 1 Sample 2 1 15.1 4.8 2 12.3 5.7 3 14.9 6.2 4 17.5 9.4 5 18.1 2.3 6 18.4 4.7 Perform a test of hypothesis to determine if the difference in the means of the distribution at the first time period is 10 units larger than at the second time period. Use a level of significance equal to 0.10. A) Because t = 1.98 > 2.0150, the null hypothesis must be rejected. B) Because t = 1.67 > 2.0150, the null hypothesis must be rejected C) Because t = 1.02 < 2.0150, the null hypothesis cannot be rejected. D) Because t = 0.37 < 2.0150, the null hypothesis cannot be rejected.

Q: The following samples are observations taken from the same elements at two different times: Unit Sample 1 Sample 2 1 15.1 4.8 2 12.3 5.7 3 14.9 6.2 4 17.5 9.4 5 18.1 2.3 6 18.4 4.7 Assume that the populations are normally distributed and construct a 90% confidence interval for the difference in the means of the distribution at the times in which the samples were taken. A) (7.6232, 13.4434) B) (5.2825, 15.8127) C) (6.8212, 14.7821) D) (7.4122, 14.6801)

Q: A paired sample study has been conducted to determine whether two populations have equal means. Twenty paired samples were obtained with the following sample results: Based on these sample data and a significance level of 0.05, what conclusion should be made about the population means? A) Because t = 5.06 > 2.0930, reject the null hypothesis. B) Because t = 3.41 > 2.0930, reject the null hypothesis. C) Because t = 1.82 < 2.0930, do not reject the null hypothesis. D) Because t = 2.02 < 2.0930, do not reject the null hypothesis.

Q: You are given the following results of a paired-difference test: = -4.6 sd = 0.25 n = 16 Construct a 90% confidence interval estimate for the paired difference in mean values. A) -2.86 ------- -2.53 B) -5.72 ------- -5.18 C) -3.81 ------- -3.71 D) -4.71 ------- -4.49

Q: You are given the following results of a paired-difference test: = -4.6 sd = 0.25 n = 16 Construct a 99% confidence interval estimate for the paired difference in mean values. A) -2.912 -------- -2.718 B) -4.784 -------- -4.416 C) -5.241 -------- -4.971 D) -3.141 -------- -2.812

Q: The following paired samples have been obtained from normally distributed populations. Construct a 90% confidence interval estimate for the mean paired difference between the two population means. Sample #Population 1Population 213,6934,63523,6794,26233,9214,29344,1064,19753,8084,53664,3944,49473,8784,094A) -571.92 -172.51B) -487.41 -283.89C) -812.21 -72.61D) -674.41 -191.87

Q: The management of the Seaside Golf Club regularly monitors the golfers on its course for speed of play. Suppose a random sample of golfers was taken in 2011 and another random sample of golfers was selected in 2006. The results of the two samples are as follows: 201120121 = 2252 = 219s1 = 20.25s2 = 21.70n1 = 36n2 = 31Based on the sample results, can the management of the Seaside Golf Club conclude that average speed of play was different in 2012 than in 2011? Conduct the appropriate hypothesis test at the 0.10 level of significance. Assume that the management of the club is willing to accept the assumption that the populations of playing times for each year are approximately normally distributed with equal variances.A) Because the calculated value of t = -2.03 is less than the lower tail critical value of t = - 1.6686, reject the null hypothesis. Based on these sample data, at the = 0.10 level of significance there is sufficient evidence to conclude that the average speed of play is different in 2012 than in 2011.B) Because the calculated value of t = 1.84 is greater than the upper tail critical value of t = 1.6686, reject the null hypothesis. Based on these sample data, at the = 0.10 level of significance there is sufficient evidence to conclude that the average speed of play is different in 2012 than in 2011.C) Because the calculated value of t = 0.89 is neither less than the lower tail critical value of t = - 1.6686, nor greater than the upper tail critical value of t = 1.6686, do not reject the null hypothesis. Based on these sample data, at the = 0.10 level of significance there is not sufficient evidence to conclude that the average speed of play is different in 2012 than in 2011.D) Because the calculated value of t = 1.17 is neither less than the lower tail critical value of t = - 1.6686, nor greater than the upper tail critical value of t = 1.6686, do not reject the null hypothesis. Based on these sample data, at the = 0.10 level of significance there is not sufficient evidence to conclude that the average speed of play is different in 2012 than in 2011.

Q: Descent, Inc., produces a variety of climbing and mountaineering equipment. One of its products is a traditional three-strand climbing rope. An important characteristic of any climbing rope is its tensile strength. Descent produces the three-strand rope on two separate production lines: one in Bozeman and the other in Challis. The Bozeman line has recently installed new production equipment. Descent regularly tests the tensile strength of its ropes by randomly selecting ropes from production and subjecting them to various tests. The most recent random sample of ropes, taken after the new equipment was installed at the Bozeman plant, revealed the following: BozemanChallis1 = 7,200 lb2 = 7,087 lbs1 = 425s2 = 415n1 = 25n2 = 20Descent's production managers are willing to assume that the population of tensile strengths for each plant is approximately normally distributed with equal variances. Based on the sample results, can Descent's managers conclude that there is a difference between the mean tensile strengths of ropes produced in Bozeman and Challis? Conduct the appropriate hypothesis test at the 0.05 level of significance.A) Because the calculated value of t = 0.896 is neither less than the lower tail critical value of t = -2.0167, nor greater than the upper tail critical value of t = 2.0167, do not reject the null hypothesis. Based on these sample data, at the = 0.05 level of significance there is not sufficient evidence to conclude that the average tensile strength of ropes produced at the two plants is different.B) Because the calculated value of t = 0.451 is neither less than the lower tail critical value of t = -2.0167, nor greater than the upper tail critical value of t = 2.0167, do not reject the null hypothesis. Based on these sample data, at the = 0.05 level of significance there is not sufficient evidence to conclude that the average tensile strength of ropes produced at the two plants is different.C) Because the calculated value of t = -2.8126 is less than the lower tail critical value of t = -2.0167, reject the null hypothesis. Based on these sample data, at the = 0.05 level of significance there is sufficient evidence to conclude that the average tensile strength of ropes produced at the two plants is different.D) Because the calculated value of t = 2.8126 is greater than the lower tail critical value of t = -2.0167, reject the null hypothesis. Based on these sample data, at the = 0.05 level of significance there is sufficient evidence to conclude that the average tensile strength of ropes produced at the two plants is different.

Q: Given the following null and alternative hypothesesH0 : 1 - 2 = 0HA : 1 - 0Together with the following sample information Sample 1Sample 2n1 = 125n2 = 120s1 = 31s2 = 381 = 1302 = 105Test the null hypothesis and indicate whether the sample information leads you to reject or fail to reject the null hypothesis, assuming a significance level of 0.05 is to be used. Use the test statistic approach.A) Since 0.812 < 1.9698 reject H0B) Since 1.041 < 1.9698 reject H0C) Since 5.652 > 1.9698 reject H0D) Since 4.418 > 1.9698 reject H0

Q: Given the following null and alternative hypothesesH0 : 1 - 2 = 0HA : 1 - 2 0Together with the following sample information Sample 1Sample 2n1 = 125n2 = 120s1 = 31s2 = 381 = 1302 = 105Develop the appropriate decision rule, assuming a significance level of 0.05 is to be used.A) If t > 1.9698 or t < -1.9698 reject H0, otherwise do not reject H0B) If t > 1.5412 or t < -1.5412 reject H0, otherwise do not reject H0C) If t > 1.8157 or t < -1.8157 reject H0, otherwise do not reject H0D) If t > 2.0124 or t < -2.0124 reject H0, otherwise do not reject H0

Q: Given the following null and alternative hypothesesH0 : 1 2HA : 1 < 2Together with the following sample information Sample 1Sample 2n1 = 14n2 = 181 = 5652 = 578s1 = 28.9s2 = 26.3Assuming that the populations are normally distributed with equal variances, test at the 0.05 level of significance whether you would reject the null hypothesis based on the sample information. Use the test statistic approach.A) Because the calculated value of t = -2.145 is less than the critical value of t = -1.6973, reject the null hypothesis. Based on these sample data, at the = 0.05 level of significance there is sufficient evidence to conclude that the mean for population 1 is less than the mean for population 2.B) Because the calculated value of t = -1.814 is less than the critical value of t = -1.6973, reject the null hypothesis. Based on these sample data, at the = 0.05 level of significance there is sufficient evidence to conclude that the mean for population 1 is less than the mean for population 2.C) Because the calculated value of t = -1.329 is not less than the critical value of t = -1.6973, do not reject the null hypothesis. Based on these sample data, at the = 0.05 level of significance there is not sufficient evidence to conclude that the mean for population 1 is less than the mean for population 2.D) Because the calculated value of t = -1.415 is not less than the critical value of t = -1.6973, do not reject the null hypothesis. Based on these sample data, at the = 0.05 level of significance there is not sufficient evidence to conclude that the mean for population 1 is less than the mean for population 2.

Q: Given the following null and alternative hypothesesH0 : 1 2HA : 1 < 2Together with the following sample information Sample 1Sample 2n1 = 14n2 = 181 = 5652 = 578s1 = 28.9s2 = 26.3Assuming that the populations are normally distributed with equal variances, test at the 0.10 level of significance whether you would reject the null hypothesis based on the sample information. Use the test statistic approach.A) Because the calculated value of t = -1.415 is less than the critical value of t=-1.3104, reject the null hypothesis. Based on these sample data, at the = 0.10 level of significance there is sufficient evidence to conclude that the mean for population 1 is less than the mean for population 2.B) Because the calculated value of t = -1.329 is less than the critical value of t=-1.3104, reject the null hypothesis. Based on these sample data, at the = 0.10 level of significance there is sufficient evidence to conclude that the mean for population 1 is less than the mean for population 2.C) Because the calculated value of t = -0.429 is not less than the critical value of t=-1.3104, do not reject the null hypothesis. Based on these sample data, at the = 0.10 level of significance there is not sufficient evidence to conclude that the mean for population 1 is less than the mean for population 2.D) Because the calculated value of t = -0.021 is not less than the critical value of t=-1.3104, do not reject the null hypothesis. Based on these sample data, at the = 0.10 level of significance there is not sufficient evidence to conclude that the mean for population 1 is less than the mean for population 2.

Q: A decision maker wishes to test the following null and alternative hypotheses using an alpha level equal to 0.05:H0 : 1 - 2 = 0HA : 1 - 2 0The population standard deviations are assumed to be known. After collecting the sample data, the test statistic is computed to be z = 1.78Using the p-value approach, what decision should be reached about the null hypothesis?A) Since p-value = 0.0018 < /2 = 0.025, reject the null hypothesisB) Since p-value = 0.0415 > /2 = 0.025, do not reject the null hypothesis.C) Since p-value = 0.0033 < /2 = 0.025, reject the null hypothesis.D) Since p-value = 0.0375 > /2 = 0.025, do not reject the null hypothesis.

Q: A decision maker wishes to test the following null and alternative hypotheses using an alpha level equal to 0.05:H0 : 1 - 2 = 0HA : 1 - 2 0The population standard deviations are assumed to be known. After collecting the sample data, the test statistic is computed to be z = 1.78Using the test statistic approach, what conclusion should be reached about the null hypothesis?A) Because z = 2.40 > 1.96, we reject the null hypothesis.B) Because z = 2.03 > 1.96, we reject the null hypothesis.C) Because z = 1.78 < 1.96, we do not reject the null hypothesis.D) Because z = 1.45 < 1.645, we do not reject the null hypothesis.

Q: A pet food producer manufactures and then fills 25-pound bags of dog food on two different production lines located in separate cities. In an effort to determine whether differences exist between the average fill rates for the two lines, a random sample of 19 bags from line 1 and a random sample of 23 bags from line 2 were recently selected. Each bag's weight was measured and the following summary measures from the samples are reported: Production Line 1Production Line2Sample Size, n1923Sample Mean, 24.9625.01Sample Standard Deviation, s0.070.08Management believes that the fill rates of the two lines are normally distributed with equal variances.Develop a 95% confidence interval estimate of the true mean difference between the two lines.A) -0.1412 (1 - 2) -0.0912B) -0.0974 (1 - 2) -0.0026C) -0.0231 (1 - 2) -0.0069D) -0.0812 (1 - 2) -0.0188

Q: A pet food producer manufactures and then fills 25-pound bags of dog food on two different production lines located in separate cities. In an effort to determine whether differences exist between the average fill rates for the two lines, a random sample of 19 bags from line 1 and a random sample of 23 bags from line 2 were recently selected. Each bag's weight was measured and the following summary measures from the samples are reported: Production Line 1 Production Line2 Sample Size, n 19 23 Sample Mean, 24.96 25.01 Sample Standard Deviation, s 0.07 0.08 Management believes that the fill rates of the two lines are normally distributed with equal variances. Calculate the point estimate for the difference between the population means of the two lines. A) 0.040 B) 0.034 C) -0.050 D) -0.042