Q&A Hero

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: σ1= σ2 Ha: σ1≠ σ2

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: 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 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 : σ2= 25 The 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 χ2= 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. 12.29 11.88 12.03 12.22 11.76 11.98 Based on these sample data, the test statistic is approximately χ2= 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 χ2= 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: The U.S. Golf Association provides a number of services for its members. One of these is the evaluation of golf equipment to make sure that the equipment satisfies the rules of golf. For example, they regularly test the golf balls made by the various companies that sell balls in the United States. Recently, they undertook a study of two brands of golf balls with the objective to see whether there is a difference in the mean distance that the two golf ball brands will fly off the tee. To conduct the test, the U.S.G.A. uses a robot named "Iron Byron," which swings the club at the same speed and with the same swing pattern each time it is used. The following data reflect sample data for a random sample of balls of each brand. Brand A: 234 236 230 227 234 233 228 229 230 238 Brand B: 240 236 241 236 239 243 230 239 243 240 Given this information, what is the test statistic for testing whether the two population variances are equal? A) Approximately F = 1.145 B) t = 1.96 C) t = -4.04 D) None of the above

Q: A small business owner has two fast food restaurants. The owner wants to determine if there is any difference in the variability of service times at the drive-thru window of each restaurant. A sample of size n = 9 is taken from each restaurant's drive-thru window. To perform a hypothesis test using the 0.05 level of significance the critical value is: A) 3.438 B) 3.197 C) 4.026 D) 4.433

Q: Two airlines are being compared with respect to the time it takes them to turn a plane around from the time it lands until it takes off again. The study is interested in determining whether there is a difference in the variability between the two airlines. They wish to conduct the hypothesis test using an alpha = 0.02. If random samples of 20 flights are selected from each airline, what is the appropriate F critical value? A) 3.027 B) 2.938 C) 2.168 D) 2.124

Q: In performing a one-tailed test for the difference between two population variances, which of the following statements is true? A) The level of alpha needs to be doubled before finding the F-critical value in the table. B) The sample variance that is predicted to be larger in the alternative hypothesis goes in the numerator when forming the F-test statistic. C) You always place the larger of the two sample variances in the numerator. D) The alternative hypothesis must contain the equality.

Q: The Russet Potato Company has been working on the development of a new potato seed that is hoped to be an improvement over the existing seed that is being used. Specifically, the company hopes that the new seed will result in less variability in individual potato length than the existing seed without reducing the mean length. To test whether this is the case, a sample of each seed is used to grow potatoes to maturity. The following information is given: Old Seed New Seed Number of Seeds = 11 Number of Seeds = 16 Average length = 6.25 inches Average length = 5.95 inches Standard Deviation = 1.0 inches Standard Deviation = 0.80 inches The on these data, if the hypothesis test is conducted using a 0.05 level of significance, the calculated test statistic is: A) = 1.25 B) = 0.80 C) = 0.64 D) = 1.56

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: 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.A) 0.04156B) 0.00121C) 0.03418D) 0.00597

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: 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: 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 1 Population 2 1 3,693 4,635 2 3,679 4,262 3 3,921 4,293 4 4,106 4,197 5 3,808 4,536 6 4,394 4,494 7 3,878 4,094 A) -571.92 ≤ μ ≤ -172.51 B) -487.41 ≤ μ ≤ -283.89 C) -812.21 ≤ μ ≤ -72.61 D) -674.41 ≤ μ ≤ -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: 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 α = 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 α = 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 α = 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 α = 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: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 α = 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 α = 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 α = 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 α = 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: μ1- μ2 = 0HA: μ1- μ2 ≠ 0Together with the following sample information Test 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: μ1≥ μ2HA: μ1< μ2Together with the following sample informationA) 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 α = 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 α = 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 α = 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 α = 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: A decision maker wishes to test the following null and alternative hypotheses using an alpha level equal to 0.05: H0: μ1- μ2= 0 HA: μ1- μ2≠ 0 The population standard deviations are assumed to be known. After collecting the sample data, the test statistic is computed to be z = 1.78 Using the p-value approach, what decision should be reached about the null hypothesis? A) Since p-value = 0.0018 < α/2 = 0.025, reject the null hypothesis B) Since p-value = 0.0415 > α/2 = 0.025, do not reject the null hypothesis. C) Since p-value = 0.0033 < α/2 = 0.025, reject the null hypothesis. D) Since p-value = 0.0375 > α/2 = 0.025, 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: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: BoiseRichmondSample Size120135Sample Mean(seconds)195216Sample St. Dev. (seconds)35.1037.80 Using the sample results, develop a 90% confidence interval estimate for the difference between the two population means.A) -29.3124 ≤ (1- 2) ≤ -18.6876B) -24.2412 ≤ (1- 2) ≤ -17.7588C) -26.2941 ≤ (1- 2) ≤ -11.8059D) -28.5709 ≤ (1- 2) ≤ -13.4291

Q: Construct a 95% confidence interval estimate for the difference between two population means based on the following information:A) 25.41 ≤ ( 1- 2) ≤ 44.59B) 35.41 ≤ ( 1- 2) ≤ 40.59C) 15.741 ≤ ( 1- 2) ≤ 54.16D) 22.13 ≤ ( 1- 2) ≤ 47.87

Q: The following information is based on independent random samples taken from two normally distributed populations having equal variances:Based on the sample information, determine the 95% confidence interval estimate for the difference between the two population means.A) -6.23 < (1- 2) < 14.23B) -4.81 < (1- 2) < 16.81C) -5.17 < (1- 2) < 15.17D) -3.25 < (1- 2) < 17.25

Q: A study was recently conducted at a major university to estimate the difference in the proportion of business school graduates who go on to graduate school within five years after graduation and the proportion of non-business school graduates who attend graduate school. A random sample of 400 business school graduates showed that 75 had gone to graduate school while in a random sample of 500 non-business graduates, 137 had gone on to graduate school. Based on a 95 percent confidence level, what is the upper limit of the confidence interval estimate? A) 0.2340 B) 0.1034 C) -0.031 D) -0.018

Q: A study was recently conducted at a major university to determine whether there is a difference in the proportion of business school graduates who go on to graduate school within five years after graduation and the proportion of non-business school graduates who attend graduate school. A random sample of 400 business school graduates showed that 75 had gone to graduate school while in a random sample of 500 non-business graduates, 137 had gone on to graduate school. Based on these sample data, and testing at the 0.10 level of significance, what is the value of the test statistic? A) Approximately z = 1.645 B) About z = -3.04 C) Approximately z = 3.45 D) About z = 1.96

Q: Suppose that two population proportions are being compared to test whether there is any difference between them. Assume that the test statistic has been calculated to be z = 2.21. Find the p-value for this situation. A) p-value = 0.0136 B) p-value = 0.4864 C) p-value = 0.0272 D) p-value = 0.9728

Q: There are a number of highly touted search engines for finding things of interest on the Internet. Recently, a consumer rating system ranked two search engines ahead of the others. Now, a computer user's magazine wishes to make the final determination regarding which one is actually better at finding particular information. To do this, each search engine was used in an attempt to locate specific information using specified keywords. Both search engines were subjected to 100 queries. Search engine 1 successfully located the information 88 times and search engine 2 located the information 80 times. Using a significance level equal to 0.05, which of the following is true? A) Based the sample data, the null hypothesis of equal population proportions is rejected since the test statistic exceeds the critical value. B) Based on the sample data, the null hypothesis should not be rejected since the test statistic of z = 2.04 exceeds the critical value of z = 1.96. C) According to the test results, the hypothesis should be rejected since the test statistic value, z = 1.54, falls in the rejection region. D) Based on the sample data, there is not sufficient evidence to conclude that a difference exists between the proportion of search hits since the test statistic, z = 1.54, does not fall in the rejection region.

Q: Suppose a survey is taken of two groups of people where each person is asked a yes/no question and the proportion of people who answer yes is calculated for each group. Which of the following is true about a hypothesis test of the difference in the two proportions? A) Normality can be assumed if the sample size for each population is at least 30. B) The t-distribution should be used if the standard deviations are unknown. C) The standard deviation must be assumed equal. D) Normality can be assumed if, in each group, at least 5 people say yes, and at 5 people say no.

Q: An advertising company has developed a new ad for one of the national car manufacturing companies. The ad agency is interested in testing whether the proportion of favorable response to the ad is the same between male adults versus female adults. It plans on conducting the test using an alpha level equal to 0.05. A sample of 100 adults of each gender will be used in the study. Each person will be asked to view the ad and indicate whether they find the ad to be "pleasing" or not. The samples resulted in 57 males that liked the ad and 47 females that liked the ad. Based on this information, what is the value of the test statistic? A) z = 1.645 B) z = 1.42 C) t = 2.234 D) z = 1.024

Q: To increase productivity, workers went through a training program. The management wanted to know the effectiveness of the program. A sample of seven workers was taken and their daily production rates before and after the training are shown below. Worker Before After 1 18 22 2 23 25 3 25 27 4 22 25 5 20 24 6 21 19 7 19 20 Based on the data, the training program is: A) effective. B) ineffective. C) neither effective nor ineffective. D) None of the above

Q: Assume that 10 people join a weight loss program for 3 months. Each person's weight both before and after the program is recorded and the number of pounds each person lost is found. The following summarizes the results for the 10 people: Mean weight lost = 9 pounds Sample standard deviation of weight lost = 4.6 pounds Assume that the hypothesis test will be conducted to determine whether or not the weight loss program is effective using a 0.05 level of significance. What is the value of the test statistic? A) t = 6.19 B) t = 1.96 C) z = 1.96 D) z = 6.19

Q: Most companies that make golf balls and golf clubs use a one-armed robot named "Iron Mike" to test their balls for length and accuracy, but because of swing variations by real golfers, these test robots don't always indicate how the clubs will perform in actual use. One company in the golfing industry is interested in testing its new driver to see if it has greater length off the tee than the best-selling driver. To do this, it has selected a group of golfers of differing abilities and ages. Its plan is to have each player use each of the two clubs and hit five balls. It will record the average length of the drives with each club for each player. The resulting data for a sample of 10 players are: Player New Club Leading Club 1 236.4 237.2 2 202.5 200.4 3 245.6 240.8 4 257.4 259.3 5 223.5 218.9 6 205.3 200.6 7 266.7 258.9 8 240 236.5 9 278.9 280.5 10 211.4 206.5 What is the critical value for the appropriate hypothesis test if the test is conducted using a 0.05 level of significance? A) z = 1.645 B) t = 1.7341 C) t = 1.8331 D) t = 2.2622

Q: In testing for differences between the means of two paired populations, an appropriate null hypothesis would be: A) H0: μD= 2 B) H0: μD= 0 C) H0: μD< 0 D) H0: μD> 0

Q: If we are testing for the difference between the means of two paired populations with samples of n1= 20 and n2= 20, the number of degrees of freedom is equal to: A) 39 B) 38 C) 19 D) 18

Q: The t-test for the mean difference between 2 related populations assumes that the respective: A) sample sizes are equal. B) sample variances are equal. C) populations are approximately normal or sample sizes are large. D) All of the above

Q: Suppose that a group of 10 people join a weight loss program for 3 months. Each person's weight is recorded at the beginning and at the end of the 3-month program. To test whether the weight loss program is effective, the data should be treated as: A) independent samples using the normal distribution. B) paired samples using the t-distribution. C) independent samples using the t-distribution. D) independent proportions.

Q: Assume that you are testing the difference in the means of two independent populations at the 0.05 level of significance. The null hypothesis is: H0: μA- μβ ≥ 0 and you have found the test statistic is z = -1.92What should you conclude?A) The mean of pop. A is greater than the mean of pop. B because p-value < α.B) The mean of pop. A is greater than the mean of pop. B because p-value > α.C) There is no significant difference in the two means because p-value > α.D) The mean of pop. B is greater than the mean of pop. A because p-value < α.

Q: There have been complaints recently from homeowners in the north end claiming that their homes have been assessed at values that are too high compare with other parts of town. They say that the mean increase from last year to this year has been higher in their part of town than elsewhere. To test this, the assessor's office staff plans to select a random sample of north end properties (group 1) and a random sample of properties from other areas within the city (group 2) and perform a hypothesis test. The following sample information is available: North End Other Sample Size 20 10 Sample Mean Increase $4,010 $3,845 Sample St. Deviations $1,800 $1,750 Assuming that the null hypothesis will be tested using an alpha level equal to 0.05, what is the critical value? A) z = 1.578 B) t = 1.7011 C) t = 0.2388 D) t = 2.0484

Q: A hypothesis test for the difference between two means is considered a two-tailed test when: A) the population variances are equal. B) the null hypothesis states that the population means are equal. C) the alpha level is 0.10 or higher. D) the standard deviations are unknown.

Q: Under what conditions can the t-distribution be correctly employed to test the difference between two population means? A) When the samples from the two populations are small and the population variances are unknown B) When the two populations of interest are assumed to be normally distributed C) When the population variances are assumed to be equal D) All of the above

Q: In conducting a hypothesis test for the difference between two population means where the standard deviations are known and the null hypothesis is: H0: μA- μβ≥ 0 What is the p-value assuming that the test statistic has been found to be z = 2.52? A) 0.0059 B) 0.9882 C) 0.0118 D) 0.4941

Q: A commuter has two different routes available to drive to work. She wants to test whether route A is faster than route B. The best hypotheses are: A) H0: μA- μβ≥ 0 HA: μA- μβ< 0 B) H0: μA- μβ≤ 0 HA: μA- μβ> 0 C) H0: μA- μβ= 0 HA: μA- μβ≠ 0 D) H0: μA- μβ< 0 HA: μA- μβ≥ 0

Q: A recent study posed the question about whether Japanese managers are more motivated than American managers. A randomly selected sample of each was administered the Sarnoff Survey of Attitudes Toward Life (SSATL), which measures motivation for upward mobility. The SSATL scores are summarized below. AmericanJapaneseSample Size211100Mean SSATL Score65.7579.83Population Std. Dev11.076.41 Which of the following is the correct the null and alternative hypotheses to determine if the average SSATL score of Japanese managers differs from the average SSATL score of American managers?A) H0: μA- μJ≥ 0 versus H1: μA- μJ< 0B) H0: μA- μJ≤ 0 versus H1: μA- μJ> 0C) H0: μA- μJ= 0 versus H1: μA- μJ≠ 0

Q: Given the following information, calculate the degrees of freedom that should be used in the pooled-standard deviation t-test.A) df = 41B) df = 39C) df = 16D) df = 25

Q: A major retail clothing store is interested in estimating the difference in mean monthly purchases by customers who use the store's in-house credit card versus using a Visa, Mastercard, or one of the other major credit cards. To do this, it has randomly selected a sample of customers who have made one or more purchases with each of the types of credit cards. The following represents the results of the sampling: In-House Credit Card National Credit Card Sample Size: 86 113 Mean Monthly Purchases: $45.67 $39.87 Standard Deviation: $10.90 $12.47 Suppose that the managers wished to test whether there is a statistical difference in the mean monthly purchases by customers using the two types of credit cards, using a significance level of .05, what is the value of the test statistic assuming the standard deviations are known? A) t = 3.49 B) z = 11.91 C) z = 2.86 D) z = 3.49

Q: The management of a department store is interested to estimate the difference in the amount of money spent by female and male shoppers. You are given the following information. Female Shoppers Male Shoppers Sample size 64 49 Sample mean $140 $125 Population standard deviation $10 $8 A 95 percent confidence interval estimate for the difference between the average purchases of the customers using the two different credit cards is: A) 49 to 64 B) 11.68 to 18.32 C) 125 to 140 D) 8 to 10

Q: When estimating a confidence interval for the difference between 2 means using the method where sample variances are pooled, which of the following assumptions is not needed? A) The populations are normally distributed. B) The populations have equal variances. C) The samples are independent. D) The sample sizes are equal.

Q: A company in Maryland has developed a device that can be attached to car engines, which it believes will increase the miles per gallon that cars will get. The owners are interested in estimating the difference between mean mpg for cars using the device versus those that are not using the device. The following data represent the mpg for random independent samples of cars from each population. The variances are assumed equal and the populations normally distributed. With Device Without Device 22.6 26.9 23.4 24.4 28.4 20.8 29.0 20.8 29.3 20.2 20.0 26.0 28.1 25.6 Given this data, what is the upper limit for a 95 percent confidence interval estimate for the difference in mean mpg? A) Approximately 3.88 mpg B) About 5.44 mpg C) Just under 25.0 D) None of the above

Q: If the population variances are assumed to be known in an application where a manager wishes to estimate the difference between two population means, the 95 percent confidence interval estimate can be developed using which of the following critical values? A) z = 1.645 B) z = 1.96 C) t value that depends on the sample sizes from the two populations D) z = 2.575

Q: If a manager wishes to develop a confidence interval estimate for estimating the difference between two population means, an increase in the size of the samples used will result in: A) an increase in the size of the critical value. B) a wider confidence interval. C) a more precise confidence interval. D) a less precise confidence interval

Q: A direct retailer that sells clothing on the Internet has two distribution centers and wants to determine if there is a difference between the proportion of customer order shipments that contain errors (wrong color, wrong size, etc.). It calculates a 95 percent confidence interval on the difference in the sample proportions to be -0.012 to 0.037. Based on this, it can conclude that the distribution centers differ significantly for the proportion of orders with errors.

Q: An Internet service provider is interested in estimating the proportion of homes in a particular community that have computers but do not already have Internet access. To do this, the company has selected a random sample of n = 200 homes and made calls. A total of 188 homes responded to the survey question with 38 saying that they had a computer with no Internet access. The 95 percent confidence interval estimate for the true population proportion is approximately 0.1447 - 0.2595.

Q: A major manufacturer of home electronics is interested in determining whether customers have a preference between two new speaker designs for their home entertainment centers. To test this, the design department manager has selected a random sample of customers and shown them the first design. A second sample of customers is shown design 2. The manager then asks each customer whether they prefer the new design they were shown over the one they currently own. The following results were observed: Design 1 Design 2 Sample size n1= 150 n2= 80 Number preferring new x1= 65 x2= 58 Based on these data and a significance level equal to 0.05, the test statistic is approximately -4.22 and thus the null hypothesis should be rejected.

Q: An accounting firm has been hired by a large computer company to determine whether the proportion of accounts receivables with errors in one division (Division 1) exceeds that of the second division (Division 2). The managers believe that such a difference may exist because of the lax standards employed by the first division. To conduct the test, the accounting firm has selected random samples of accounts from each division with the following results. Division 1 Division 2 Sample Size n1= 100 n2= 100 Errors found x1= 13 x2= 8 Based on this information and using a significance level equal to 0.05, the test statistic for the hypothesis test is approximately 1.153 and, therefore, the null hypothesis is not rejected.

Q: The test statistic that is used when testing hypotheses about the difference between two population proportions is the t-value from the t-distribution.

Q: To test whether Model A and Model B cars have the same MPG, the first step is to select two independent random samples of drivers and assign one of them to drive Model A and the other Model B.

Q: Two placement exams are available that students can take to determine which math class they should begin with in their freshman year. It is believed that there is no difference in the mean scores that would be received for the two tests. To test this using a 0.05 level of significance, a randomly selected group of students took both tests and had their scores recorded. The following data were obtained: Student Test A Test B 1 78 82 2 86 74 3 74 79 4 72 93 5 75 80 6 68 82 7 77 99 Based on these data, the test statistic is approximately t = -1.892.

Q: When conducting a hypothesis test to determine whether or not two groups differ, using paired samples rather than independent samples has the advantage of controlling for sources of variation that might distort the conclusions of the study.

Q: If the point estimate in a paired difference estimation example does not fall in the resulting confidence interval, the decision maker can conclude that the two populations likely have different means.

Q: In comparing two populations using paired differences, after the difference is found for each pair, the method for testing whether the mean difference is equal to 0 becomes the same as was used for a one-sample hypothesis test with unknown standard deviation.