Question

12-18. To investigate whether online automobile insurance he cheaper than the policies the offer, a local insurance agent compared the mobile insurance cmiums for five of this customers with price quotations for the same coverage from o n e panies. Results are shown in table. Test the null hypothesis that the mean of the difference all t h e policies is zero versus the relevant alternative, as indicated below Us 01 Asume a normal distribuce of differences. (dependent) Policy Local (1) Online (2) Difference (pl-2) 568 191 451 488 605 677 783 703 907 1008 12. What is the mean of the difference 13. State the null and alternative hypotheses for the test. What is the alternative hypothesis a. b. HS0 c0 14. State the correct formula for the test statistic. Justify your answer a. 2 sample T test because sample sire is small and normally distributed b. Paired Test because sample size is small and normally distributed, source is dependent c. One sample 2 test because sample size is small but standard deviation is known and normally distributed d. 2 sample proportion test because n is small enough 15. Construct the rejection region a. (3.747, infinity) b. (-infinity. -3.747] c. (3.365, infinity) d. (- infinity. -3.365] c. (-infinity. -4.604]0[4.604, Infinity) 16. Compute the value of the test statistic, 17. Make a decision a. Fail to reject Ho, because test statistie in the Rejection region b. Fail to reject Ho, because test statistic is not in the Rejection region c. Reject Ho, because test statistic in the Rejection region d. Reject Ho, because test statistic is not in the Rejection region 18 State a conclusion in the context of the problem, about the mean of the difference in premiums for all policies, based on the test you performed Online is cheaper than local insurance policy at percent level of significance b. Online is not cheaper than local insurance policy at I percent level of significance c. There is enough evidence support the claim al percent level of significance

Answer 1

Answer Date: 08/04/2020 The difference table is, Local 568 451 605 783 907 Online Difference 391 177 488 -37 677 -72 703 80 1008 -101 12) The mean difference is, d=20 _177+(-37)+(-72)+80 +(-101) = 9.40

The mean difference is 9.40). 13) To test the hypothesis is that online automobile insurance policies might be cheaper than the local insurance at 1% significance level. The null and alternative hypothesis is, H:lla = 0 H : Ha> 0 The answer is (a). 14) Paired-T test because sample size is less than 30 which is 5, normally distributed, source is dependent and each sample local and online is taken from single population is correct formula for statistics.

The answer is b. 15) The t critical value is, The sample size is small and right-tailed test. Look in the column headed a = 0.01 and the row headed in the t distribution table by using degree of freedom is, d.f.=n-1 = 5-1 = 4 The t critical value is 3.747, infinity|- 16) The t-test statistics is, First, compute the standard deviation for computing the t-test statistics.

The standard deviation is, (d, -d)? Sa= n -1 (177 -9.40)2 + (-37-9.40)2 + ... +(80 -9.40)²+(-101-9.402 5-1 = 116.234 The t-test statistics is, 9.40 |(116.234)?/5 = 0.181 The t-test statistics is 0.181.

17) Decision The conclusion is that the t value corresponds to sample statistics is not fall in the critical region, so the null hypothesis is fail to reject at 1% level of significance. The answer is b. 18) There is insufficient evidence to indicate that online automobile insurance policies might be cheaper than the local insurance. The result is not statistically significant. The answer is b