2. The mean annual premium for automobile insurance in the United States is $1503 (Insure.com website, March 6, 2014). Being from Pennsylvania, you believe automobile insurance is cheaper there and wish to develop statistical support for your opinion. A sample of 25 automobile insurance policies from the state of Pennsylvania showed a mean annual premium of $1440 with a standard deviation of s = $165.
1. Consider the following hypothesis test:
H_{0}: μ ≥ 45 H_{a}: μ < 45 Save your time  order a paper!Get your paper written from scratch within the tight deadline. Our service is a reliable solution to all your troubles. Place an order on any task and we will take care of it. You won’t have to worry about the quality and deadlines Order Paper NowA sample of 36 is used. Identify the pvalue and state your conclusion for each of the following sample results. Use α = .01. a. With x = 44 and s = 5.2, the pvalue is Can it be concluded that the popluation mean is less than 45? b. With x = 43 and s = 4.6, the pvalue is Can it be concluded that the population mean is less than 45? c. With x = 46 and s = 5.0, the pvalue is Can it be concluded that the popluation mean is less than 45?

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2. The mean annual premium for automobile insurance in the United States is $1503 (Insure.com website, March 6, 2014). Being from Pennsylvania, you believe automobile insurance is cheaper there and wish to develop statistical support for your opinion. A sample of 25 automobile insurance policies from the state of Pennsylvania showed a mean annual premium of $1440 with a standard deviation of s = $165.
If required, enter negative values as negative numbers. a. Develop a hypothesis test that can be used to determine whether the mean annual premium in Pennsylvania is lower than the national mean annual premium. H_{0}: μ – Select your answer – greater than 1503 greater than or equal to 1503 equal to 1503 less than or equal to 1503 less than 1503 not equal to 1503 Item 1 b. What is a point estimate of the difference between the mean annual premium in Pennsylvania and the national mean (to the nearest dollar)? c. At α = .05, test for a significant difference by completing the following. Calculate the value of the test statistic (to 2 decimals). The pvalue is – Select your answer – less than .005 between .005 and .01 between .01 and .025 between .025 and .05 between .05 and .10 between .10 and .20 greater than .20 Item 5 What is your conclusion?

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3.What percentage of the population live in their state of birth? According to the U. S. Census Bureau’s American Community Survey, it ranges from 25% in Nevada to 78.7 percent in Louisiana (AARP Bulletin, March, 2014). The average percentage across all states and the District of Columbia is 57.7%. The data in the WEBfile Homestate are consistent with the findings in the American Community Survey. The data are for a random sample of 120 Arkansas residents and for a random sample of 180 Virginia residents. If required, enter negative values as negative numbers. a. Formulate hypotheses that can be used to determine whether the percentage of stay at home residents in the two states differs from the overall average of 57.7%. H_{0}: p – Select your answer – greater than 0.577 greater than or equal to 0.577 equal to 0.577 less than or equal to 0.577 less than 0.577 not equal to 0.577 Item 1 b. Estimate the proportion of stay at home residents in Arkansas (to 4 decimals). Does this proportion differ significantly from the mean proportion for all states? Use α = .05. c. Estimate the proportion of stay at home residents in Virginia (to 3 decimals). Does this proportion differ significantly from the mean proportion for all states? Use α = .05. d. Would you expect the proportion of stay at home residents to be higher in Virginia than in Arkansas? Support your conclusion with the results obtained in parts (b) and (c). The input in the box below will not be graded, but may be reviewed and considered by your instructor.

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4. Consider the following hypothesis test:
H_{0}: μ = 18 A sample of 48 provided a sample mean x = 17 and a sample standard deviation s = 4.9.
a. Compute the value of the test statistic (to three decimal places.)
b. Use the t distribution table (Table 2 in Appendix B) to compute a range for the pvalue. (to two decimal places) pvalue is between is
c. At α = .05, what is your conclusion? pvalue is Select greater than or equal to 0.05, reject greater than 0.05, do not reject less than or equal to 0.05, do not reject less than 0.05, reject equal to 0.05, do not reject not equal to 0.05, do not reject Item 4 H_{0}
d. What is the rejection rule using the critical value? Reject H_{0} if t is Select greater than or equal to 2.012 greater than 2.012 less than or equal to 2.012 less than 2.012 equal to 2.012 not equal to 2.012 Item 5 or t is Select greater than or equal to 2.012 greater than 2.012 less than or equal to 2.012 less than 2.012 equal to 2.012 not equal to 2.012 Item 6
What is your conclusion? t = ; Select do not reject reject Item 8 H_{0}

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