Question

The development office and the registrar of your university have provided you with anonymous matches of starting salaries and GPAs for 108 graduating economics majors. Your sample contains a variety of jobs, from church pastor to stockbroker. a) The average starting salary for the 108 students was $38,584 with a sample standard deviation of $7,500. The mean starting salary for economics majors from other universities is $36,000. At a 5% significance level, is there statistical evidence showing that the mean for your university is different from the other universities' mean? b) Compute the p-value from the previous test. What does it mean? c) Construct a 95% confidence interval for the starting salary of all economics majors at your university. How do you interpret this interval? d) You wonder if it pays to get good grades by calculating the average salary for economics majors who graduated with a cumulative GPA of B+ or better, and those who had a Bor worse. The data is as shown in the accompanying table. Cumulative GPA Average Earning Y Sample Standard Dev. s, B+ or better $39,900 $8,300 59 B or worse $37,000 $6,100 49 Using a 5% significance level test, is there statistical evidence showing that students with high GPA have a higher starting salary than students with low GPA? e) Construct a 95% confidence interval for the difference between the starting salaries of high and low GPA students. How do you interpret this interval?

Answer 1

a)

Ho: u = 36000 Ha: u = 36000 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown 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 two- tailed test is te = 1.982. The rejection region for this two-tailed test is R={t: t > 1.982} (3) Test Statistics The t-statistic is computed as follows: t = X - MO s/n 38584 – 36000 7500/7108 = 3.58 (4) Decision about the null hypothesis = 1.982, it is then concluded that the null hypothesis is Since it is observed that t = 3.58 > t rejected. Using the P-value approach: The p-value is p= 0.0005, and since p = 0.0005 < 0.05, it is 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 u is different than 36000, at the 0.05 significance level.

There is statistical difference

b)

p-value = 0.0005

it means that there is 0.0005 probability of observing this sample or extreme than this when null hypothesis is true

c)

Confidence Interval The 95% confidence interval is 37153.338 <u < 40014.662.

Please post rest parts again