Question

PROJECT: The national mean annual salary for a school administrator is $90,000 a year (the Cincinnati Enquirer, April 7, 2012). A school official took a sample of 25 school administrators in the state of Ohio to learn about salaries in that state to see if they differed from the national average. The sample data for 25 Ohio administrators is contained in the excel file named Administrator. Use these data for the following questions: Q1.) Formulate hypotheses that can be used to determine whether the population mean annual administrator salary in Ohio differs from the national mean of $90,000. (5 marks) Q2.) Compute the value of the test statistics? What is the p-value for your hypothesis test in part (Q1)? (5 marks) Q3.) At alpha=0.05, can your null hypothesis be rejected? What is your conclusion? Repeat the preceding hypothesis test using the critical value approach. (3 marks) Q4.) At alpha=0.01, can your null hypothesis be rejected? What is the rejection rule using the critical value approach? What is your conclusion? (2 marks)

Answer 1

Let u denotes the population mean. Hypotheses : To test null hypothesis H. :u =90000 against alternative hypothesis H1 :p # 90000 Let i denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, s 11217.34 T = 85045.8333, s2 = 125828677.5362,8 = 11217.3383, n=24, standard error of mean = = 2289.7296 /n V24 The test statistic can be written as: Vnī - 90000) s which under Ho follows at distribution with n-1 df. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tobs > t0.025,n–1 = 2.068658 0.4- 0.3 density 0.2- 0.1 0.0 Now, Value of the test statistic: The value of the test statistic tobs = V24(85045.8333 – 90000) 11217.3383 = -2.163647 -2.16 associated degrees of freedom = 23 The critical value =Łt 0.05 = = 0.025,23 = +2.068658 +2.0687 2,1-1 P-value = P(tn-1|> tobs) = 2 * P(t23 <-2.163647) = 2 * 0.020562 = 0.041123 0.0411 Conclusion : Since p-value < 0.05 and tobs > tcritical = 2.068658, so we reject H, at 0.05 level of significance Hence, we can conclude that the the population mean is significantly different from 90000.

4.

Decision rule / Rejection region : We reject H, at 0.01 level of significance if P-value < 0.01 or if tobs > to.005,n–1 = 2.807336 0.4 - 0.3 0.2 ^ 0.1- 0.0- 0 Now, Value of the test statistic: The value of the test statistic tobs = 24 (85045.8333 - 90000) 11217.3383 -2.163647 -2.16 associated degrees of freedom = 23 The critical value =Łt 0.01 = £70.005,23 = +2.807336 +2.8073 2.n-1 P-value = P(|tn-1| > tobs ) = 2 * P(t23 < -2.163647) = 2 * 0.020562 = 0.041123 0.0411 Conclusion : Since p-value > 0.01 and tobs tcritical = 2.807336, so we fail to reject H, at 0.01 level of significance Hence, we can conclude that the population mean is not significantly different from 90000.