I need answers ASAP 6. A simple random sample of 60 test scores on a statewide exam, given to allgraduating high school students, produced a sample mean of 52. The stateTeacher,s Association believes the mean test score for the population of studentswho took the test is greater than 50. Assume the population standard deviation is15 and the significance level for the test is a =0.01a) State the appropriate null and alternative hypothesesb) Compute the value of the test statisticc) State the conclusion

6. a) Let μ denotes the population mean test score for the students who took the test. The claim is that the mean test score for the population of students who took the test is greater than 50. The hypothesis is, Null hypothesis: H0:μ≤50 Alternative hypothesis: Ha:μ>50b) The value of the test statistic is, z=x¯-μσn=52-501560=21.9365=1.03 Thus, the value of the test statistic is 1.03. c) Computation of critical value: The hypothesis is right tailed. The probability value is, α=0.011-α=0.99 The critical value of z-distribution can be obtained using the excel formula “=NORM.S.INV (0.99)”. The critical value is 2.33. Decision rule: If test statistic > critical value, then reject the null hypothesis. Otherwise, do not reject the null hypothesis. Conclusion: The test statistic (1.03) is less than the critical value (2.33). Based on decision rule, fail to reject the null hypothesis. Thus, there is no sufficient evidence that the mean test score for the population of students who took the test is greater than 50.