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Question
A major car magazine has recently collected data on 30 leading cars in the U.S. market. It is interested in building a multiple regression model to explain the variation in highway miles. The following correlation matrix has been computed from the data collected:| mileage, highway | mileage, city | Curb Weight | cylinders | Horsepower | |
| mileage, highway | 1 | ||||
| mileage, city | 0.857550598 | 1 | |||
| Curb Weight | -0.739110566 | -0.70765104 | 1 | ||
| cylinders | -0.694837149 | -0.866135056 | 0.596475711 | 1 | |
| Horsepower | -0.549172956 | -0.684199197 | 0.293202385 | 0.840347219 | 1 |
The analysts also produced the following multiple regression output using curb weight, cylinders, and horsepower as the three independent variables. Note, a number of the output fields are missing, but can be determined from the information provided.
Based on the information provided, using a 0.05 level of statistical significance, both curb weight and horsepower are statistically significant variables in explaining the variation in the dependent variable when they are included in the model along with cylinders.
Answer
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Related questions
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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.
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In a two-tailed hypothesis test the area in each tail of the rejection region is equal to α.
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The null and alternate hypotheses must be opposites of each other.
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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:
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 = 0.05 and a standard deviation of 50.A) 0.0123B) 0.5182C) 0.1241D) 0.1587
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:
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:
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:
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:
For the following hypothesis test With n = 100 and p = 0.66, state the calculated value of the test statistic.
A) 2.7299
B) -2.0785
C) 1.4919
D) -0.3421
Q:
For the following hypothesis test: With n= 0.42 and p = 0.42, state the conclusion
A) 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:
Given the following null and alternativeTest the hypothesis using = 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 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:
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. Use = 0.05. 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.A) Since -1.2445 < 1.013 < 1.2445, do not reject H0 and conclude that the filling machine remains all right to operate.B) Since -1.2445 < 1.013 < 1.2445, reject H0 and conclude that the filling machine needs to be moderated.C) Since -2.1315 < 1.83 < 2.1315, do not reject H0 and conclude that the filling machine remains all right to operate.D) Since -2.1315 < 1.83 < 2.1315, reject H0 and conclude that the filling machine needs to be moderated.
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. Use = 0.05. Establish the appropriate null and alternative hypotheses to be tested for boxes that are supposed to have an average of 24 ounces.A) H0 : = 32 ounces Ha: 32 ouncesB) H0: = 16 ounces Ha: 16 ouncesC) H0: = 22 ounces Ha: 22 ouncesD) H0: = 24 ounces Ha: 24 ounces
Q:
The bottlers of a new fruit juice daily select a random sample of 12 bottles of the drink to estimate the mean quantity of juice in the bottles filled that day. On one such day, the following results were observed: = 12.03; s = 0.12. Based on this information, the upper limit for a 95 percent confidence interval estimate is approximately 12.106 ounces.
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, the sample mean that generated the confidence interval was 3.60.
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A random sample of 100 boxes of cereal had a sample mean weight of 396 grams. The standard deviation is known to be 5 grams. The upper end of the confidence interval for the mean is 405.8 grams.
Q:
When using a 95 percent confidence interval for a mean, the area in the upper tail of the distribution that is outside the interval is 5 percent.
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A statement in the newspaper attributed to the leader of a local union stated that the average hourly wage for union members in the region is $13.35. He indicated that this number came from a survey of union members. If an estimate was developed with 95 percent confidence, we can safely conclude that this value is within 95 percent of the true population mean hourly wage.
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The product manager for a large retail store has recently stated that she estimates that the average purchase per visit for the store's customers is between $33.00 and $65.00. The $33.00 and the $65.00 are considered point estimates for the true population mean.
Q:
Sampling error is the difference between a statistic computed from a sample and the corresponding parameter computed from the population.
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A human resources manager wishes to estimate the proportion of employees in her large company who have supplemental health insurance. What is the largest size sample she should select if she wants 95 percent confidence and a margin of error of 0.01?
Q:
Under what circumstances would you wish to select a pilot sample?
Q:
In a recent audit report, an accounting firm stated that the mean sale per customer for the client was estimated to be between $14.50 and $28.50. Further, this was based on a random sample of 100 customers and was computed using 95 percent confidence. Provide a correct interpretation of this confidence interval estimate.
Q:
Assume that a standard deck of 52 playing cards is randomly shuffled and the first 2 cards are dealt to you. What is the probability that you have a blackjack? A blackjack is where one card is an ace and the other card is worth 10 points. The 10-point cards are kings, queens, jacks and 10's.