Q&A Hero

Q: If the probability of a Type I error is set at 0.05, then the probability of a Type II error will be 0.95.

Q: The significance level in a hypothesis test corresponds to the maximum probability that a Type I error will be committed.

Q: A report recently submitted to the managing partner for a market research company stated "the hypothesis test may have resulted in either a Type I or a Type II error. We won't know which one occurred until later." This statement is one that we might correctly make for any hypothesis that we have conducted.

Q: When a battery company claims that their batteries last longer than 100 hours and a consumer group wants to test this claim, the hypotheses should be: H0: μ ≤ 100 HA: μ > 100

Q: The police chief in a local city claims that the average speed for cars and trucks on a stretch of road near a school is at least 45 mph. If this claim is to be tested, the null and alternative hypotheses are: H0: μ < 45 mph Ha: μ ≥ 45 mph

Q: If a hypothesis test leads to incorrectly rejecting the null hypothesis, a Type II statistical error has been made.

Q: Of the two types of statistical errors, the one that decision makers have most control over is Type I error.

Q: When someone has been accused of a crime the null hypothesis is: H0: innocent. In this case, a Type I error would be convicting an innocent person.

Q: A report recently published in a major business periodical stated that the average salary for female managers is less than $50,000. If we were interested in testing this, the following null and alternative hypotheses would be established: H0: μ ≥ 50,000 Hα: μ < 50,000

Q: A large tire manufacturing company has claimed that its top line tire will average more than 80,000 miles. If a consumer group wished to test this claim, the research hypothesis would be: Ha: μ > 80,000 miles.

Q: In conducting a hypothesis test where the conclusion is to reject the null hypothesis, then either a correct decision has been made or else a Type I error.

Q: The Adams Shoe Company believes that the mean size for men's shoes is now more than 10 inches. To test this, it has selected a random sample of n = 100 men. Assuming that the test is to be conducted using a .05 level of significance, a p-value of .07 would lead the company to conclude that its belief is correct.

Q: In a hypothesis test, the p-value measures the probability that the alternative hypothesis is true.

Q: When using the t-distribution in a hypothesis test, the population does not need to be assumed normally distributed.

Q: A local medical center has advertised that the mean wait for services will be less than 15 minutes. In an effort to test whether this claim can be substantiated, a random sample of 100 customers was selected and their wait times were recorded. The mean wait time was 17.0 minutes. Based on this sample result, there is sufficient evidence to reject the medical center's claim.

Q: In a two-tailed hypothesis test the area in each tail of the rejection region is equal to α.

Q: In a one-tailed hypothesis test, the larger the significance level, the greater the critical value will be.

Q: The following is an appropriate statement of the null and alternate hypotheses for a test of a population mean: H0: μ < 50 HA: μ > 50

Q: If the sample data lead the decision maker to reject the null hypothesis, the alpha level is the maximum probability of committing a Type II error.

Q: If a hypothesis test is conducted for a population mean, a null and alternative hypothesis of the form: H0: μ = 100 HA: μ ≠ 100 will result in a one-tailed hypothesis test since the sample result can fall in only one tail.

Q: Whenever possible, in establishing the null and alternative hypotheses, the research hypothesis should be made the alternative hypothesis.

Q: A one-tailed hypothesis for a population mean with a significance level equal to .05 will have a critical value equal to z = .45.

Q: A conclusion to "not reject" the null hypothesis is the same as the decision to "accept the null hypothesis".

Q: A sample is used to obtain a 95 percent confidence interval for the mean of a population. The confidence interval goes from 15 to 19. If the same sample had been used to test the null hypothesis that the mean of the population is equal to 20 versus the alternative hypothesis that the mean of the population differs from 20, the null hypothesis could be rejected at a level of significance of 0.05.

Q: The null and alternate hypotheses must be opposites of each other.

Q: In testing a hypothesis, statements for the null and alternative hypotheses as well as the selection of the level of significance should precede the collection and examination of the data.

Q: When the decision maker has control over the null and alternative hypotheses, the alternative hypotheses should be the "research" hypothesis.

Q: In hypothesis testing, the null hypothesis should contain the equality sign.

Q: Hypothesis testing and confidence interval estimation are essentially two totally different statistical procedures and share little in common with each other.

Q: Explain what is meant by a p-value.

Q: The produce manager for a large retail grocery store believes that an average head of lettuce weighs more than 1.7 pounds. If she were to test this statistically, what would the null and alternative hypotheses be and what is the research hypothesis?

Q: What is meant by the terms Type I and Type II statistical error?

Q: According to CNN business partner Careerbuilder.com, the average starting salary for accounting graduates in 2008 was at least $47,413. Suppose that the American Society for Certified Public Accountants planned to test this claim by randomly sampling 200 accountants who graduated in 2008. Compute the power of the hypothesis test to reject the null hypothesis if the true average starting salary is only $47,000. Assume that the population standard deviation is known to be $4,600 and the test is to be conducted using an alpha level equal to 0.01. A) 0.0872 B) 0.1323 C) 0.8554 D) 0.9812

Q: Nationwide Mutual Insurance, based in Columbus, Ohio, is one of the largest diversified insurance and financial services organizations in the world, with more than $157 billion in assets. Nationwide ranked 108th on the Fortune 100 list in 2008. The company provides a full range of insurance and financial services. In a recent news release Nationwide reported the results of a new survey of 1,097 identity theft victims. The survey shows victims spend an average of 81 hours trying to resolve their cases. If the true average time spent was 81 hours, determine the probability that a test of hypothesis designed to test that the average was less than 85 hours would reject the research hypothesis. Use α= 0.05 and a standard deviation of 50. A) 0.0123 B) 0.5182 C) 0.1241 D) 0.1587

Q: Waiters at Finegold's Restaurant and Lounge earn most of their income from tips. Each waiter is required to "tip-out" a portion of tips to the table bussers and hostesses. The manager has based the "tip-out" rate on the assumption that the mean tip is at least 15% of the customer bill. To make sure that this is the correct assumption, he has decided to conduct a test by randomly sampling 60 bills and recording the actual tips. Calculate the probability of a Type II error if the true mean is 14%. Assume that the population standard deviation is known to be 2% and that a significance level equal to 0.01 will be used to conduct the hypothesis test. A) 0.0041 B) 0.1251 C) 0.0606 D) 0.4123

Q: Swift is the holding company for Swift Transportation Co., Inc., a truckload carrier headquartered in Phoenix, Arizona. Swift operates the largest truckload fleet in the United States. Before Swift switched to its current computer-based billing system, the average payment time from customers was approximately 40 days. Suppose before purchasing the present billing system, it performed a test by examining a random sample of 24 invoices to see if the system would reduce the average billing time. The sample indicates that the average payment time is 38.7 days. The company that created the billing system indicates that the system would reduce the average billing time to less than 40 days. Conduct a hypothesis test to determine if the new computer-based billing system would reduce the average billing time to less than 40 days. Assume the standard deviation is known to be 6 days. Use a significance level of 0.025. A) Since z = -0.423 > -1.96, we will not reject H0, there is not sufficient evidence to conclude that the new computer-based billing system would reduce the average billing time to less than 40 days. B) Since z = -1.0614 > -1.96, we will not reject H0, there is not sufficient evidence to conclude that the new computer-based billing system would reduce the average billing time to less than 40 days. C) z = -1.0231 > -1.96, we will reject H0, there is sufficient evidence to conclude that the new computer-based billing system would reduce the average billing time to less than 40 days. D) z = 0.341 > -1.96, we will reject H0, there is sufficient evidence to conclude that the new computer-based billing system would reduce the average billing time to less than 40 days.

Q: Calculate the probability of committing a Type II error if the true population mean is 230 gallons. Assume that the population standard deviation is known to be 40 gallons.A) 0.0331B) 0.0712C) 0.0537D) 0.1412

Q: State the appropriate null and alternative hypotheses.A) H0 : μ > 243 Ha : μ ≤ 243B) H0 : μ < 243 Ha : μ ≥ 243C) H0 : μ ≤ 243 Ha : μ > 243D) H0 : μ ≥ 243 Ha : μ < 243

Q: You are given the following null and alternative hypotheses:Calculate the probability of committing a Type II error when the population mean is 505, the sample size is 64, and the population standard deviation is known to be 36A) 0.1562B) 0.5997C) 0.3426D) 0.8888

Q: You are given the following null and alternative hypotheses:If the true population mean is 4,345, calculate the power of the test. Assume the population standard deviation is known to be 200 and the sample size is 100.A) 0.1766B) 0.3876C) 0.0808D) 0.9686

Q: Because of the complex nature of the U.S. income tax system, many people have questions for the Internal Revenue Service (IRS). Yet, an article published by the Detroit Free Press entitled "Assistance: IRS Help Centers Give the Wrong Information" discusses the propensity of IRS staff employees to give incorrect tax information to tax-payers who call with questions. Then IRS Inspector General Pamela Gardiner told a Senate subcommittee that "the IRS employees at 400 taxpayer assistance centers nationwide encountered 8.5 million taxpayers face-to-face last year. The problem: When inspector general auditors posing as taxpayers asked them to answer tax questions, the answers were right 69% of the time." Suppose an independent commission was formed to test whether the 0.69 accuracy rate is correct or whether it is actually higher or lower. The commission has randomly selected n = 180 tax returns that were completed by IRS assistance employees and found that 105 of the returns were accurately completed. Using an α= 0.05 level, based on the sample data, what conclusion should be reached about the IRS rate of correct tax returns? A) The z-critical values from the standard normal table for a two-tailed test with alpha = 0.05 are and z = -1.96. Since z = -0.96 > -1.96, we do not reject the null hypothesis. Thus, based on these sample data, we believe that the accuracy rate is actually higher than the 0.69 rate quoted in the Detroit Free Press article B) The z-critical values from the standard normal table for a two-tailed test with alpha = 0.05 are and z = -1.96. Since z = -0.96 > -1.96, we do not reject the null hypothesis. Thus, based on these sample data, we believe that the accuracy rate is actually higher than the 0.58 rate quoted in the Detroit Free Press article C) The z-critical values from the standard normal table for a two-tailed test with alpha = 0.05 are and z = -1.96. Since z= -3.19 < -1.96, we reject the null hypothesis. Thus, based on these sample data, we believe that the accuracy rate is actually lower than the 0.69 rate quoted in the Detroit Free Press article. D) The z-critical values from the standard normal table for a two-tailed test with alpha = 0.05 are and z = -1.96. Since z = -3.19 < -1.96, we reject the null hypothesis. Thus, based on these sample data, we believe that the accuracy rate is actually lower than the 0.58 rate quoted in the Detroit Free Press article.

Q: A recent article in The Wall Street Journal entitled "As Identity Theft Moves Online, Crime Rings Mimic Big Business" states that 39% of the consumer scam complaints by American consumers are about identity theft. Suppose a random sample of 90 complaints is obtained. Of these complaints, 40 were regarding identity theft. Based on these sample data, what conclusion should be reached about the statement made in The Wall Street Journal? (Test using α= 0.10.) A) Since z = 1.947 > 1.645, we reject the null hypothesis. There is sufficient evidence to conclude that the 0.39 rate quoted in the WSJ article is wrong. B) Since z = 2.033 > 1.96, we reject the null hypothesis. There is sufficient evidence to conclude that the 0.39 rate quoted in the WSJ article is wrong. C) Since z = 1.341 < 1.645, we do not reject the null hypothesis. There is insufficient evidence to conclude that the 0.39 rate quoted in the WSJ article is wrong. D) Since z = 0.97 < 1.645, we do not reject the null hypothesis. There is insufficient evidence to conclude that the 0.39 rate quoted in the WSJ article is wrong.

Q: A major issue facing many states is whether to legalize casino gambling. Suppose the governor of one state believes that more than 55% of the state's registered voters would favor some form of legal casino gambling. However, before backing a proposal to allow such gambling, the governor has instructed his aides to conduct a statistical test on the issue. To do this, the aides have hired a consulting firm to survey a simple random sample of 300 voters in the state. Of these 300 voters, 175 actually favored legalized gambling. Assuming that a significance level of 0.05 is used, what conclusion should the governor reach based on these sample data? A) Since z = 1.1594 < 1.645, do not reject the null hypothesis. The sample data do not provide sufficient evidence to conclude that more than 55 percent of the population favor legalized gambling. B) Since z = 2.1316 > 1.645, reject the null hypothesis. The sample data provide sufficient evidence to conclude that more than 55 percent of the population favor legalized gambling. C) Since z = 1.1594 < 1.645, do not reject the null hypothesis. The sample data do not provide sufficient evidence to conclude that more than 58 percent of the population favor legalized gambling. D) Since z = 2.1316 > 1.645, reject the null hypothesis. The sample data provide sufficient evidence to conclude that more than 58 percent of the population favor legalized gambling.

Q: An issue that faces individuals investing for retirement is allocating assets among different investment choices. Suppose a study conducted 10 years ago showed that 65% of investors preferred stocks to real estate as an investment. In a recent random sample of 900 investors, 540 preferred real estate to stocks. Is this new data sufficient to allow you to conclude that the proportion of investors preferring stocks to real estate has declined from 10 years ago? Conduct your analysis at the α = 0.02 level of significance. A) Because z = -1.915 is not less than -2.055, do not reject H0. A higher proportion of investors prefer stocks today than 10 years ago. B) Because z = -1.915 is not less than -2.055, do not reject H0. A lower proportion of investors prefer stocks today than 10 years ago. C) Because z = -3.145 is less than -2.055, reject H0. A lower proportion of investors prefer stocks today than 10 years ago. D) Because z = -3.145 is less than -2.055, reject H0. A higher proportion of investors prefer stocks today than 10 years ago.

Q: Suppose a recent random sample of employees nationwide that have a 401(k) retirement plan found that 18% of them had borrowed against it in the last year. A random sample of 100 employees from a local company who have a 401(k) retirement plan found that 14 had borrowed from their plan. Based on the sample results, is it possible to conclude, at the α = 0.025 level of significance, that the local company had a lower proportion of borrowers from its 401(k) retirement plan than the 18% reported nationwide? A) The z-critical value for this lower tailed test is z = -1.96. Because -1.5430 is greater than the z-critical value we do not reject the null hypothesis and conclude that the proportion of employees at the local company who borrowed from their 401(k) retirement plan is not less than the national average. B) The z-critical value for this lower tailed test is z = -1.96. Because -1.0412 is greater than the z-critical value we do not reject the null hypothesis and conclude that the proportion of employees at the local company who borrowed from their 401(k) retirement plan is not less than the national average. C) The z-critical value for this lower tailed test is z = 1.96. Because 1.5430 is less than the z-critical value we do not reject the null hypothesis and conclude that the proportion of employees at the local company who borrowed from their 401(k) retirement plan is not less than the national average. D) The z-critical value for this lower tailed test is z = 1.96. Because 1.0412 is less than the z-critical value we do not reject the null hypothesis and conclude that the proportion of employees at the local company who borrowed from their 401(k) retirement plan is not less than the national average.

Q: For the following hypothesis testWith n = 100 and p = 0.66, state the conclusion.A) Because the computed value of z = -2.0785 is less than the critical value of z = -1.96, reject the null hypothesis and conclude that the population proportion is less than 0.75.B) Because the computed value of z = -0.3412 is less than the critical value of z = -1.645, reject the null hypothesis and conclude that the population proportion is less than 0.75.C) Because the computed value of z = 1.4919 is greater than the critical value of z = -1.96, accept the null hypothesis and conclude that the population proportion is greater than 0.75.D) Because the computed value of z = -0.3412 is greater than the critical value of z = -1.645, accept the null hypothesis and conclude that the population proportion is greater than 0.75.

Q: For the following hypothesis test:With n= 0.42 and p = 0.42, state the conclusionA) Because the calculated value of the test statistic, t=0.4122, is neither greater than 2.013 nor less than -2.013, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40.B) Because the calculated value of the test statistic, t=1.7291, is neither greater than 2.013 nor less than -2.013, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40.C) Because the calculated value of the test statistic, z = 1.2412, is neither greater than 2.575 nor less than -2.575, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40.D) Because the calculated value of the test statistic, z = 0.3266, is neither greater than 2.575 nor less than -2.575, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40.

Q: For the following hypothesis test:With n= 64 and p= 0.42, state the decision rule in terms of the critical value of the test statisticA) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is greater than 2.013 or less than -2.013. Otherwise, do not reject.B) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is less than 2.013 or greater than -2.013. Otherwise, do not reject.C) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is greater than 2.575 or less than -2.575. Otherwise, do not reject.D) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is less than 2.575 or greater than -2.575. Otherwise, do not reject.

Q: Given the following null and alternativeTest the hypothesis using α = 0.01 assuming that a sample of n = 200 yielded x = 105 items with the desired attribute.A) Since -2.17 > -2.33, the null hypothesis is not rejected.B) Since -1.86 > -1.02, the null hypothesis is not rejected.C) Since -2.17 > -2.33, the null hypothesis is rejected.D) Since -1.86 > -1.02, the null hypothesis is rejected.

Q: Hono Golf is a manufacturer of golf products in Taiwan and China. One of the golf accessories it produces at its plant in Tainan Hsing, Taiwan, is plastic golf tees. The injector molder produces golf tees that are designed to have an average height of 66 mm. To determine if this specification is met, random samples are taken from the production floor. One sample is contained in the file labeled THeight. If the hypothesis test determines the specification is not being met, the production process will be shut down while causes and remedies are determined. At times this occurs even though the process is functioning to specification. What type of statistical error would this be? A) The null hypothesis, the specification not being met, was not rejected when in fact it was not being met, this is a Type II error. B) The null hypothesis, the specification not being met, was not rejected when in fact it was not being met, this is a Type I error. C) The null hypothesis, the specification is being met, was rejected when in fact it was being met, this is a Type II error. D) The null hypothesis, the specification is being met, was rejected when in fact it was being met, this is a Type I error.

Q: The Center on Budget and Policy Priorities (www.cbpp.org) reported that average out-of-pocket medical expenses for prescription drugs for privately insured adults with incomes over 200% of the poverty level was $173 in 2002. Suppose an investigation was conducted in 2012 to determine whether the increased availability of generic drugs, Internet prescription drug purchases, and cost controls have reduced out-of-pocket drug expenses. The investigation randomly sampled 196 privately insured adults with incomes over 200% of the poverty level, and the respondents' 2012 out-of-pocket medical expenses for prescription drugs were recorded. These data are in the file Drug Expenses. Based on the sample data, can it be concluded that 2012 out-of-pocket prescription drug expenses are lower than the 2002 average reported by the Center on Budget and Policy Priorities? Use a level of significance of 0.01 to conduct the hypothesis test. A) Because t = -2.69 is less than -2.3456, do not reject H0 Conclude that 2012 average out-of-pocket prescription drug expenses are not lower than the 2002 average. B) Because t = -2.69 is less than -2.3456, reject H0 Conclude that 2012 average out-of-pocket prescription drug expenses are lower than the 2002 average. C) Because t = -1.69 is less than -0.8712, reject H0 Conclude that 2012 average out-of-pocket prescription drug expenses are lower than the 2002 average. D) Because t = -1.69 is less than -0.8712, do not reject H0 Conclude that 2012 average out-of-pocket prescription drug expenses are not lower than the 2002 average.

Q: The U.S. Bureau of Labor Statistics (www.bls.gov) released its Consumer Expenditures report in October 2008. Among its findings is that average annual household spending on food at home was $3,624. Suppose a random sample of 137 households in Detroit was taken to determine whether the average annual expenditure on food at home was less for consumer units in Detroit than in the nation as a whole. The sample results are in the file Detroit Eats. Based on the sample results, can it be concluded at the α = 0.02 level of significance that average consumer-unit spending for food at home in Detroit is less than the national average? A) Because t = -13.2314 is less than the critical t value of -1.4126, do not reject H0. The annual average consumer unit spending for food at home in Detroit is not less than the 2006 national consumer unit average B) Because t = -13.2314 is less than the critical t value of -1.4126, reject H0. The annual average consumer unit spending for food at home in Detroit is less than the 2006 national consumer unit average C) Because t = -15.7648 is less than the critical t value of -2.0736, do not reject H0. The annual average consumer unit spending for food at home in Detroit is not less than the 2006 national consumer unit average. D) Because t = -15.7648 is less than the critical t value of -2.0736, reject H0. The annual average consumer unit spending for food at home in Detroit is less than the 2006 national consumer unit average.

Q: At a recent meeting, the manager of a national call center for a major Internet bank made the statement that the average past-due amount for customers who have been called previously about their bills is now no larger than $20.00. Other bank managers at the meeting suggested that this statement may be in error and that it might be worthwhile to conduct a test to see if there is statistical support for the call center manager's statement. The file called Bank Call Center contains data for a random sample of 67 customers from the call center population. Assuming that the population standard deviation for past due amounts is known to be $60.00, what should be concluded based on the sample data? Test using α = 0.10. A) Because p-value = 0.4121 > alpha = 0.10, we do not reject the null hypothesis. The sample data do not provide sufficient evidence to reject the call center manager's statement that the mean past due amount is $20.00 or less. B) Because p-value = 0.4121 > alpha = 0.10, we reject the null hypothesis. The sample data provide sufficient evidence to reject the call center manager's statement that the mean past due amount is $20.00 or less. C) Because p-value = 0.2546 > alpha = 0.10, we do not reject the null hypothesis. The sample data do not provide sufficient evidence to reject the call center manager's statement that the mean past due amount is $20.00 or less. D) Because p-value = 0.2546 > alpha = 0.10, we reject the null hypothesis. The sample data provide sufficient evidence to reject the call center manager's statement that the mean past due amount is $20.00 or less.

Q: The makers of Mini-Oats Cereal have an automated packaging machine that can be set at any targeted fill level between 12 and 32 ounces. Every box of cereal is not expected to contain exactly the targeted weight, but the average of all boxes filled should. At the end of every shift (eight hours), 16 boxes are selected at random and the mean and standard deviation of the sample are computed. Based on these sample results, the production control manager determines whether the filling machine needs to be readjusted or whether it remains all right to operate. At the end of a particular shift during which the machine was filling 24-ounce boxes of Mini-Oats, the sample mean of 16 boxes was 24.32 ounces, with a standard deviation of 0.70 ounce. Assist the production control manager in determining if the machine is achieving its targeted average using test statistic and critical value t. Conduct the test using a p-value. A) p-value = 0.0872 > 0.025; therefore do not reject H0 B) p-value = 0.0422 > 0.005; therefore do not reject H0 C) p-value = 0.0314 < 0.105; therefore reject H0 D) p-value = 0.0121< 0. 0805; therefore reject H0

Q: A mail-order business prides itself in its ability to fill customers' orders in six calendar days or less on the average. Periodically, the operations manager selects a random sample of customer orders and determines the number of days required to fill the orders. Based on this sample information, he decides if the desired standard is not being met. He will assume that the average number of days to fill customers' orders is six or less unless the data suggest strongly otherwise. On one occasion where a sample of 40 customers was selected, the average number of days was 6.65, with a sample standard deviation of 1.5 days. Can the operations manager conclude that his mail-order business is achieving its goal? Use a significance level of 0.025 to answer this question. Conduct the test using this p-value. A) Since 0.024 > 0.0041, reject the null hypothesis. B) Since 0.046 > 0.0025, reject the null hypothesis. C) Since 0.0046 < 0.025, reject the null hypothesis. D) Since 0.0024 < 0.041, reject the null hypothesis.

Q: The director of a state agency believes that the average starting salary for clerical employees in the state is less than $30,000 per year. To test her hypothesis, she has collected a simple random sample of 100 starting clerical salaries from across the state and found that the sample mean is $29,750. Assuming the population standard deviation is known to be $2,500 and the significance level for the test is to be 0.05, what is the critical value (stated in dollars)? A) For alpha = .05 and a one tailed, lower tail test, the critical value is z = -1.645. Solving for the critical x-bar: -1.645 = (x-bar - 30,000)/250, x-bar = $29,588.75 B) For alpha = .05 and a one tailed, lower tail test, the critical value is z = -1.96. Solving for the critical x-bar: -1.96 = (x-bar - 30,000)/250, x-bar = $34,211.14 C) For alpha = .05 and a one tailed, lower tail test, the critical value is z = -1.645. Solving for the critical x-bar: -1.645 = (x-bar - 30,000)/250, x-bar = $34,211.14 D) For alpha = .05 and a one tailed, lower tail test, the critical value is z = -1.96. Solving for the critical x-bar: -1.96 = (x-bar - 30,000)/250, x-bar = $30,411.25

Q: The National Club Association does periodic studies on issues important to its membership. The 2012 Executive Summary of the Club Managers Association of America reported that the average country club initiation fee was $31,912. Suppose a random sample taken in 2009 of 12 country clubs produced the following initiation fees: $29,121 $31,472 $28,054 $31,005 $36,295 $32,771 $26,205 $33,299 $25,602 $33,726 $39,731 $27,816 Based on the sample information, can you conclude at the α = 0.05 level of significance that the average 2009 country club initiation fees are lower than the 2008 average? Conduct your test at the level of significance. A) Because t = -0.4324 is not less than t critical = -1.4512, do not reject Ho. The 2009 average country club initiation fee is not less than the 2008 average. B) Because t = -0.4324 is not less than t critical = -1.4512, reject Ho. The 2009 average country club initiation fee is less than the 2008 average. C) Because t = -0.5394 is not less than t critical = -1.7959, do not reject Ho. The 2009 average country club initiation fee is not less than the 2008 average. D) Because t = -0.5394 is not less than t critical = -1.7959, reject Ho. The 2009 average country club initiation fee is less than the 2008 average.

Q: For the following hypothesis:A) Because the computed value of t = 0.78 is not greater than 2.1727, reject the null hypothesis.B) Because the computed value of t = 0.78 is not greater than 2.1727, do not reject the null hypothesis.C) Because the computed value of t = 0.78 is not greater than 1.3277, reject the null hypothesis.D) Because the computed value of t = 0.78 is not greater than 1.3277, do not reject the null hypothesis.

Q: For the following hypothesis:A) 1.58B) 0.78C) 1.14D) 0.41

Q: For the following hypothesis:A) This is a one-tailed test of the population mean with σ unknown. Therefore, the decision rule is: accept the null hypothesis if the calculated value of the test statistic, t, is greater than 1.3277. Otherwise, reject.B) This is a one-tailed test of the population mean with σ unknown. Therefore, the decision rule is: accept the null hypothesis if the calculated value of the test statistic, t, is greater than 2.1727. Otherwise, reject.C) This is a one-tailed test of the population mean with σ unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is greater than 1.3277. Otherwise, do not reject.D) This is a one-tailed test of the population mean with σ unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is greater than 2.1727. Otherwise, do not reject.

Q: For the following hypothesis test:A) Because the computed value of t = 0.878 is not less than -2.1448 and not greater than 2.1448, do not reject the null hypothesis.B) Because the computed value of t = 1.312 is not less than -2.1448 and not greater than 2.1448, do not reject the null hypothesis.C) Because the computed value of t = 0.878 is not less than -2.1448 and not greater than 2.1448, reject the null hypothesisD) Because the computed value of t = 1.312 is not less than -2.1448 and not greater than 2.1448, reject the null hypothesis

Q: For the following hypothesis test:A) 1.014B) 0.012C) 0.878D) 1.312

Q: For the following hypothesis test:A) This is a two-tailed test of the population mean with σ unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is less than -2.1448 or greater than 2.1448. Otherwise, do not reject.B) This is a two-tailed test of the population mean with σ unknown. Therefore, the decision rule is: accept the null hypothesis if the calculated value of the test statistic, t, is less than -2.1448 or greater than 2.1448. Otherwise, do not accept.C) This is a two-tailed test of the population mean with σ unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is less than -1.1828 or greater than 1.1828. Otherwise, do not reject.D) This is a two-tailed test of the population mean with σ unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is less than -1.1828 or greater than 1.1828. Otherwise, do not accept.

Q: For the following hypothesis test:A) Because the computed value of z = 3.012 is greater than 2.05, accept the null hypothesis and conclude the mean is greater than 45. Also because the p-value is less than 0.02B) Because the computed value of z = 3.012 is greater than 2.05, reject the null hypothesis and conclude the mean is greater than 45. Also because the p-value is less than 0.02C) Because the computed value of z = 2.087 is greater than 2.05, accept the null hypothesis and conclude the mean is greater than 45. Also because the p-value is less than 0.02D) Because the computed value of z = 2.087 is greater than 2.05, reject the null hypothesis and conclude the mean is greater than 45. Also because the p-value is less than 0.02

Q: For the following hypothesis test:A) 3.151B) -2.141C) 2.087D) -3.121

Q: For the following hypothesis test:A) Reject the null hypothesis if the calculated value of the test statistic, z, is greater than the critical value of the test statistic, 1.645. Otherwise, do not reject.B) Reject the null hypothesis if the calculated value of the test statistic, z, is greater than the critical value of the test statistic, 2.05. Otherwise, do not reject.C) Accept the null hypothesis if the calculated value of the test statistic, z, is greater than the critical value of the test statistic, 1.645. Otherwise, do not accept.D) Accept the null hypothesis if the calculated value of the test statistic, z, is greater than the critical value of the test statistic, 2.05. Otherwise, do not accept.

Q: For the following z-test statistic, compute the p-value assuming that the hypothesis test is a one-tailed test: z = -1.55. A) 0.0606 B) 0.1512 C) 0.0901 D) 0.0172

Q: If the hypothesis test you are conducting is a two-tailed test, which of the following is a possible step that you could take to increase the power of the test? A) Reduce the sample size B) Increase alpha C) Increase beta D) Use the t-distribution

Q: A recent report in which a major pharmaceutical company released the results of testing that had been done on the cholesterol reduction that people could expect if they use the company's new drug indicated that the Type II error probability for a given "true" mean was 0.1250 based on the sample size of n = 64 subjects. Given this, what was the power of the test under these same conditions? The alpha level used in the test was 0.05. A) 0.95 B) 0.875 C) Essentially zero D) Power would be undefined in this case since the hypothesis would be rejected.

Q: A company that sells an online course aimed at helping high-school students improve their SAT scores has claimed that SAT scores will improve by more than 90 points on average if students successfully complete the course. To test this, a national school counseling organization plans to select a random sample of n = 100 students who have previously taken the SAT test. These students will take the company's course and then retake the SAT test. Assuming that the population standard deviation for improvement in test scores is thought to be 30 points and the level of significance for the hypothesis test is 0.05, find the critical value in terms of improvement in SAT points, which would be needed prior to finding a beta. A) Reject the null if SAT improvement is > 95 points. B) Reject the null if SAT improvement is < 85.065 points. C) Reject the null if SAT improvement is > 95.88 points. D) Reject the null if SAT improvement is > 94.935 points.

Q: A contract calls for the mean diameter of a cylinder to be 1.50 inches. As a quality check, each day a random sample of n = 36 cylinders is selected and the diameters are measured. Assuming that the population standard deviation is thought to be 0.10 inch and that the test will be conducted using an alpha equal to 0.025, what would the probability of a Type II error be? A) Approximately 0.1267 B) About 0.6789 C) 0.975 D) Can't be determined without knowing the "true" population mean.

Q: Suppose we want to test H0 : μ ≥ 30 versus H1 : μ < 30. Which of the following possible sample results based on a sample of size 36 gives the strongest evidence to reject H0in favor of H1? A) = 28, s = 6 B) = 27, s = 4 C) = 32, s = 2 D) = 26, s = 9

Q: Confidence intervals constructed with small samples tend to have greater margins of error than those constructed from larger samples, all else being constant.

Q: When calculating a confidence interval, the reason for using the t-distribution rather than the normal distribution for the critical value is that the population standard deviation is unknown.

Q: If we are interested in estimating the population mean based on a sample from a population for which we know neither the mean nor the standard deviation, the critical value will be a t value from the t-distribution.

Q: The standard deviation for the checking account balances is assumed known to be $357.50. Recently, a bank manager was interested in estimating the mean balance. To do this, she selected a random sample of 81 accounts and found a mean balance of $1,347.20. At a 95 percent confidence level, the lower limit for the confidence interval is $646.50.

Q: One way to reduce the margin of error in a confidence interval estimate is to lower the level of confidence.

Q: The margin of error is one-half the width of the confidence interval.

Q: In developing a confidence interval estimate, the margin of error is directly dependent on the value of the point estimate.

Q: In a recent report to the supply-chain manager in a major electronics company, the report writer stated that with 90 percent confidence, the manufacturing lead time for a critical part is between 3.34 hours and 4.14 hours. Based on this information, the margin of error for this estimate is .80 hours.