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Part (a) According to the National Association of Insurance Commissioners, the mean expenditure for auto insurance in North Carolina was $705 in 2017. An insurance salesperson in NC believes that the mean expenditure for auto insurance is less today. He obtains a simple random sample of 36 auto insurance policies and determines the mean expenditure to be $681 with a standard deviation of $40. Is there enough evidence to support the salesperson's claim that the mean expenditure for auto insurance is less than the 2017 amount at the a= 0.05 level of significance? Answer Step O: Verify Assumptions Is the sample n obtained by simple random sampling (SRS)? Is the sample size n ≥ 30 ? Step 1: Null and alternative hypotheses Ho : μ = Is the sample size n no more than 5% of the population size N (i.e. n < 0.05 N)? Step 2: Test statistic to= Step 3: Classical and p-value approaches -ta= p-value = (yes, no) Step 4: Conclusion versus H₁ μ Does the test statistic fall (lie) in the rejection (critical) region? (Round the critical value to 2 decimal places) The confidence interval is ( (Round the p-value to 4 decimal places) (Reject, Fail to reject) Ho. There (yes, no) H₁. Part (b) Use part (a) to construct a 99% confidence interval about the population mean μ. (yes, no) (is, is not) sufficient evidence at the 0.05 significant level to support the alternative hypothesis (yes, no) ). (Round to 2 decimal places).

Part (a) According to the National Association of Insurance Commissioners, the mean expenditure for auto insurance in North Carolina was $705 in 2017. An insurance salesperson in NC believes that the mean expenditure for auto insurance is less today. He obtains a simple random sample of 36 auto insurance policies and determines the mean expenditure to be $681 with a standard deviation of $40. Is there enough evidence to support the salesperson's claim that the mean expenditure for auto insurance is less than the 2017 amount at the a= 0.05 level of significance? Answer Step O: Verify Assumptions Is the sample n obtained by simple random sampling (SRS)? Is the sample size n ≥ 30 ? Step 1: Null and alternative hypotheses Ho : μ = Is the sample size n no more than 5% of the population size N (i.e. n < 0.05 N)? Step 2: Test statistic to= Step 3: Classical and p-value approaches -ta= p-value = (yes, no) Step 4: Conclusion versus H₁ μ Does the test statistic fall (lie) in the rejection (critical) region? (Round the critical value to 2 decimal places) The confidence interval is ( (Round the p-value to 4 decimal places) (Reject, Fail to reject) Ho. There (yes, no) H₁. Part (b) Use part (a) to construct a 99% confidence interval about the population mean μ. (yes, no) (is, is not) sufficient evidence at the 0.05 significant level to support the alternative hypothesis (yes, no) ). (Round to 2 decimal places).

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Part (a) According to the National Association of Insurance Commissioners, the mean expenditure for auto insurance in North Carolina was $705 in 2017. An insurance salesperson in NC believes that the mean expenditure for auto insurance is less today. He obtains a simple random sample of 36 auto insurance policies and determines the mean expenditure to be $681 with a standard deviation of $40. Is there enough evidence to support the salesperson's claim that the mean expenditure for auto insurance is less than the 2017 amount at the a = 0.05 level of significance? Answer Step 0: Verify Assumptions Is the sample n obtained by simple random sampling (SRS)? Is the sample size n > 30 ? Is the sample size n no more than 5% of the population size N (i.e. n Step 1: Null and alternative hypotheses Ho : μ = Step 2: Test statistic to = Step 3: Classical and p-value approaches -ta= p-value = Step 4: Conclusion (yes, no) Does the test statistic fall (lie) in the rejection (critical) region? H₁. versus H1 : μ (Round the critical value to 2 decimal places) The confidence interval is ( (Round the p-value to 4 decimal places) (Reject, Fail to reject) Ho. There (yes, no) 0.05 N)? Part (b) Use part (a) to construct a 99% confidence interval about the population mean u. (yes, no) (is, is not) sufficient evidence at the 0.05 significant level to support the alternative hypothesis (yes, no) ). (Round to 2 decimal places).
Transcribed Image Text:Part (a) According to the National Association of Insurance Commissioners, the mean expenditure for auto insurance in North Carolina was $705 in 2017. An insurance salesperson in NC believes that the mean expenditure for auto insurance is less today. He obtains a simple random sample of 36 auto insurance policies and determines the mean expenditure to be $681 with a standard deviation of $40. Is there enough evidence to support the salesperson's claim that the mean expenditure for auto insurance is less than the 2017 amount at the a = 0.05 level of significance? Answer Step 0: Verify Assumptions Is the sample n obtained by simple random sampling (SRS)? Is the sample size n > 30 ? Is the sample size n no more than 5% of the population size N (i.e. n Step 1: Null and alternative hypotheses Ho : μ = Step 2: Test statistic to = Step 3: Classical and p-value approaches -ta= p-value = Step 4: Conclusion (yes, no) Does the test statistic fall (lie) in the rejection (critical) region? H₁. versus H1 : μ (Round the critical value to 2 decimal places) The confidence interval is ( (Round the p-value to 4 decimal places) (Reject, Fail to reject) Ho. There (yes, no) 0.05 N)? Part (b) Use part (a) to construct a 99% confidence interval about the population mean u. (yes, no) (is, is not) sufficient evidence at the 0.05 significant level to support the alternative hypothesis (yes, no) ). (Round to 2 decimal places).
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