Q&A Hero

Q: When deciding the null and alternative hypotheses, the rule of thumb is that if the claim contains the equality (e.g., at least, at most, no different from, etc.), the claim becomes the null hypothesis. If the claim does not contain the equality (e.g., less than, more than, different from), the claim is the alternative hypothesis.

Q: For testing a research hypothesis, the burden of proof that a new product is no better than the original is placed on the new product, and the research hypothesis is formulated as the null hypothesis.

Q: Lube-Tech is a major chain whose primary business is performing lube and oil changes for passenger vehicles. The national operations manager has stated in an industry newsletter that the mean number of miles between oil changes for all passenger cars exceeds 4,200 miles. To test this, an industry group has selected a random sample of 100 vehicles that have come into a lube shop and determined the number of miles since the last oil change and lube. The sample mean was 4,278 and the standard deviation was known to be 780 miles. Based on this information, the p-value for the hypothesis test is less than 0.10.

Q: Lube-Tech is a major chain whose primary business is performing lube and oil changes for passenger vehicles. The national operations manager has stated in an industry newsletter that the mean number of miles between oil changes for all passenger cars exceeds 4,200 miles. To test this, an industry group has selected a random sample of 100 vehicles that have come into a lube shop and determined the number of miles since the last oil change and lube. The sample mean was 4,278 and the sample standard deviation was 780 miles. Based on this information, the test statistic is approximately t = 1.000.

Q: Lube-Tech is a major chain whose primary business is performing lube and oil changes for passenger vehicles. The national operations manager has stated in an industry newsletter that the mean number of miles between oil changes for all passenger cars exceeds 4,200 miles. To test this, an industry group has selected a random sample of 100 vehicles that have come into a lube shop and determined the number of miles since the last oil change and lube. The sample mean was 4,278 and the standard deviation was known to be 780 miles. Based on a significance level of 0.10, the critical value for the test is approximately z = 1.28.

Q: When using the p-value method for a two-tailed hypothesis, the p-value is found by finding the area in the tail beyond the test statistic, then doubling it.

Q: A two-tailed hypothesis test is used when the null hypothesis looks like the following: H0: = 100.

Q: Generally, it is possible to appropriately test a null and alternative hypotheses using the test statistic approach and reach a different conclusion than would be reached if the p-value approach were used.

Q: When using the p-value method, the null hypothesis is rejected when the calculated p-value >

Q: The state insurance commissioner believes that the mean automobile insurance claim filed in her state exceeds $1,700. To test this claim, the agency has selected a random sample of 20 claims and found a sample mean equal to $1,733 and a sample standard deviation equal to $400. They plan to conduct the test using a 0.05 significance level. Based on this, the null hypothesis should be rejected if > $1,854.66 approximately.

Q: The state insurance commissioner believes that the mean automobile insurance claim filed in her state exceeds $1,700. To test this claim, the agency has selected a random sample of 20 claims and found a sample mean equal to $1,733 and a sample standard deviation equal to $400. They plan to conduct the test using a 0.05 significance level. Given this, the appropriate null and alternative hypotheses areH0 : $1,700HA : > $1,700

Q: A two-tailed hypothesis test with = 0.05 is similar to a 95 percent confidence interval.

Q: The director of the city Park and Recreation Department claims that the mean distance people travel to the city's greenbelt is more than 5.0 miles. Assume that the population standard deviation is known to be 1.2 miles and the significance level to be used to test the hypothesis is 0.05 when a sample size of n = 64 people are surveyed. Given this information, if the sample mean is 15.90 miles, the null hypothesis should be rejected.

Q: The director of the city Park and Recreation Department claims that the mean distance people travel to the city's greenbelt is more than 5.0 miles. Assuming that the population standard deviation is known to be 1.2 miles and the significance level to be used to test the hypothesis is 0.05 when a sample size of n = 64 people are surveyed, the critical value is approximately 4.75 miles.

Q: The loan manager for State Bank and Trust has claimed that the mean loan balance on outstanding loans at the bank is over $14,500. To test this at a significance level of 0.05, a random sample of n = 100 loan accounts is selected. Assuming that the population standard deviation is known to be $3,000, the null and alternative hypotheses to be tested are:H0 : $14,500HA : > $14,500

Q: The loan manager for State Bank and Trust has claimed that the mean loan balance on outstanding loans at the bank is over $14,500. To test this at a significance level of 0.05, a random sample of n = 100 loan accounts is selected. Assuming that the population standard deviation is known to be $3,000, the value of that corresponds to the critical value is approximately $14,993.50.

Q: The critical value in a null hypothesis test is called alpha.

Q: Type II error is failing to reject the null hypothesis when the null is actually false.

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: 100HA : > 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: < 45 mphHa : 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 : 50,000Hα : < 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 : > 80,000 miles.

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, they would formulate the following null and alternative hypotheses:H0 : 80,000Hα : 80,000

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: A local medical center has advertised that the mean wait for services will be less than 15 minutes. Given this claim, the hypothesis test for the population mean should be a one-tailed test with the rejection region in the lower (left-hand) tail of the sampling distribution.

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: < 50HA : > 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 : = 100HA : 100will 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: A contract with a parts supplier calls for no more than .04 defects in the large shipment of parts. To test whether the shipment meets the contract, the receiving company has selected a random sample of n = 100 parts and found 6 defects. If the hypothesis test is to be conducted using a significance level equal to 0.05, what is the test statistic and what conclusion should the company reach based on the sample data?

Q: A national car rental chain believes that more than 80 percent of its customers are satisfied with the check-in process that the company is using. To test this, a random sample of n = 200 customers are surveyed. These sample results show 168 that say they were satisfied. If the test is to be conducted using a .05 level of significance, what is the critical value?

Q: The Gordon Beverage Company bottles soft drinks using an automatic filling machine. When the process is running properly, the mean fill is 12 ounces per can. The machine has a known standard deviation of 0.20 ounces. Each day, the company selects a random sample of 36 cans and measures the volume in each can. They then test to determine whether the filling process is working properly. The test is conducted using a 0.05 significance level. Using the test statistic approach, what conclusion should the company reach if the sample mean is 12.02 ounces? What type of statistical error may have been committed?

Q: The Gordon Beverage Company bottles soft drinks using an automatic filling machine. When the process is running properly, the mean fill is 12 ounces per can. The machine has a known standard deviation of 0.20 ounces. Each day, the company selects a random sample of 36 cans and measures the volume in each can. They then test to determine whether the filling process is working properly. The test is conducted using a 0.05 significance level. What is the critical value in ounces?

Q: A small city is considering breaking away from the county school system and starting its own city school system. City leaders believe that more than 60 percent of residents support the idea. A poll of n = 215 residents is taken and 134 people say they support starting a city school district. Using a 0.10 level of significance, conduct a hypothesis test to determine whether this poll supports the belief of city leaders.

Q: Explain why an increase in sample size will reduce the probability of a Type II error but will not impact the probability of a Type I error.

Q: A company makes a device that can be fitted to automobile engines to improve the mileage. The company claims that if the device is installed, owners will observe a mean increase of more than 3.0 mpg. Assuming that the population standard deviation of increase is known to be 0.75 mpg, and a sample of size 64 cars is selected, what is the probability of "accepting" the null hypothesis if the true population mean is 3.10 mpg increase? Assume that the test will be performed using a 0.05 level of significance.

Q: A company makes a device that can be fitted to automobile engines to improve the mileage. The company claims that if the device is installed, owners will observe a mean increase of more than 3.0 mpg. Assuming that the population standard deviation of increase is known to be 0.75 mpg, and a sample of size 64 cars is selected with an = 3.25 mpg, use the p-value approach to test the null hypothesis using a significance level of 0.05.

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

Q: If a real estate market is strong there will be a close relationship between the asking price for homes and the selling price. Suppose that one analyst believes that the mean difference between asking price and selling price for homes in a particular market area is less than $2,000. To test this using an alpha level equal to 0.05, a random sample of n = 15 homes that have sold recently was selected. The sample differences between asking price and selling price are in the following table: $2,053 $1,693 $1,854 $1,747 $869 $1,396 $2,473 $1,931 $2,303 $1,502 $1,038 $2,755 $2,084 $1,664 $2,104 Based on these sample data, what is the critical value expressed in dollars?

Q: A cell phone manufacturer claims that its phone will last for more than 8 hours of continuous talk time when the battery is fully charged. To test this claim a sample of n = 18 phones were tested. The results showed a sample mean of 8.2 hours and a sample standard deviation of 0.4 hour. Conduct the hypothesis test using a 0.5 level of significance and determine whether or not the company's claim is supported.

Q: One of the factors that a company will use in determining whether it will locate a new facility in a community is the status of the real estate market. The managers believe that an important measure of the real estate market is the average length of time that homes stay on the market before selling. They believe that if the mean time on the market is less than 45 days, the real estate market is favorable. To test this in a particular area, a random sample of n = 100 homes that sold during the past six months was selected. The mean for this sample was 40 days. It is believed that the population standard deviation is 15 days. If the test is conducted using a 0.05 level of significance, what conclusion should be reached?

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: 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.State the appropriate null and alternative hypotheses.A) H0 : $47,413 HA : < $47,413B) H0 : < $47,413 HA : $47,413C) H0 : $47,413 HA : > $47,413D) H0 : > $47,413 HA : $47,413

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 = 0.05 and a standard deviation of 50.A) 0.0123B) 0.5182C) 0.1241D) 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: 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.State the appropriate null and alternative hypotheses.A) H0 : 15 Ha : < 15B) H0 : 15 Ha : > 15C) H0 : 9 Ha : < 9D) H0 : 9 Ha : > 9

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: According to data from the Environmental Protection Agency, the average daily water consumption for a household of four people in the United States is approximately at least 243 gallons. (Source: http://www.catskillcenter.org/programs/csp/H20/Lesson3/house3.htm) Suppose a state agency plans to test this claim using an alpha level equal to 0.05 and a random sample of 100 households with four people. 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) 0.0331 B) 0.0712 C) 0.0537 D) 0.1412

Q: According to data from the Environmental Protection Agency, the average daily water consumption for a household of four people in the United States is approximately at least 243 gallons. (Source: http://www.catskillcenter.org/programs/csp/H20/Lesson3/house3.htm) Suppose a state agency plans to test this claim using an alpha level equal to 0.05 and a random sample of 100 households with four people.State the appropriate null and alternative hypotheses.A) H0 : > 243 Ha : 243B) H0 : < 243 Ha : 243C) H0 : 243 Ha : > 243D) H0 : 243 Ha : < 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 36 A) 0.1562 B) 0.5997 C) 0.3426 D) 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) 0.1766 B) 0.3876 C) 0.0808 D) 0.9686

Q: You are given the following null and alternative hypotheses: If the true population mean is 4,345, determine the value of beta. Assume the population standard deviation is known to be 200 and the sample size is 100. A) 0.9192 B) 0.8233 C) 0.6124 D) 0.0314

Q: You are given the following null and alternative hypotheses: If the true population mean is 1.25, calculate the power of the test. Assume the population standard deviation is known to be 0.50 and the sample size is 60. A) 0.49 B) 0.20 C) 0.96 D) 0.60

Q: You are given the following null and alternative hypotheses: If the true population mean is 1.25, determine the value of beta. Assume the population standard deviation is known to be 0.50 and the sample size is 60. A) 0.40 B) 0.04 C) 0.51 D) 0.80

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 = 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 articleB) 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 articleC) 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: 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.State the appropriate null and alternative hypotheses.A) H0: p = 0.69 Ha : p 0.69B) H0 : p = 0.58 Ha : p 0.58C) H0 : p > 0.69 Ha : p 0.69D) H0 : p > 0.58 Ha : p 0.58

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 = 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.