Question

ty updates, fixes, and improvements, choose Check for Updates. BPA 232, Business Statistics Fall 2017 Section 4: Make and interpret your decision using p-value 15. Last year, an insurance company found that the average claim was $2,000 with a standard devíation of $500. If the average claim has increased, the company must raise its rates. This year, mean of a random sample of 40 claims is $2,150. At the .05 level, has the mean claim increased? . Hosu s 2,000; H: μ > 2.000 Level of significance: 0.05 Test statistic: z . we calculate the value of z for the data using the formula L2/X-μ) / (o/ vn)-190 a. Calculate p-value b. Compare p-value to alpha. State the decision and the strength of the evidence. c. Interpret the decision. 16. An article on the Google News website said that nonresidential college students have an average commute of 30 minutes with a standard deviation of 6 minutes. A survey of 30 AACC students showed a mean commute of 29 minutes. At the 10 level of significance, does the commute time of AACC students differ from the time in the article? Test statistic: z Level of significance: 0.10 we calculate thrvalue of z for the data using the formula L-(X-μ) / (o/ vnj:-0 91 Calculate p-value Compare p-value to alpha. State the decision and the strength of the evidence. Interpret the decision. . a. b. c. 17. In the June 2011, the average time to find a job was found to be 39.7 weeks. A survey of 100 individuals in June 2012 found an average time of 35 weeks with a standard deviation of 6.3 weeks At the.05 level of significance, has the time needed to find a job decreased?

need answer for 15 and 16

Answer 1

15)

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ < 2000 Ha: μ > 2000 This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation will be used (2) Rejection Region Based on the information provided, the significance level is a 0.05, and the critical value for a right-tailed test is 1.64 The rejection region for this right-tailed test is R = {z : z > 1.64) (3) Test Statistics The z-statistic is computed as follows: X 2150 - 2000 σ/Vn 500/V40_T.OY' (4) Decision about the null hypothesis Since it is observed that 1.897 zc1.64, it is then concluded that the null hypothesis is rejected 0.0289, and since p 0.0289 < 0.05, it is Using the P-value approach. The p-value is p concluded that the null hypothesis is rejected (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is greater than 2000, at the 0.05 significance level

16)

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ 30 Ha: 30 This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used (2) Rejection Region Based on the information provided, the significance level is α 0.1, and the critical value for a two- tailed test is z1.64 The rejection region for this two-tailed test is R (3) Test Statistics The z-statistic is computed as follows {z : |z| > 1.64 X-29- 30 6//30 0.913 (4) Decision about the null hypothesis Since it is observed that0.913 ze 1.64, it is then concluded that the null hypothesis is not rejected Using the P-value approach: The p-value is p 0.3613, and since p - 0.3613 0.1, it is concluded that the null hypothesis is not rejected (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is different than 30, at the 0.1 significance level