Question

The file salaries.mtw contains the starting salaries of a sample of new commerce graduates.

Salary from file:
47508
52802
54906
62425
44996
41279
54308
56199
74659
53188
58387
62125
56557
44685
30962
53814
64389
57446
57694
51371
52474
47797
51889
45710
46557
52960
70545
50568
34922
64528
40670
53530
65170

(a) Test at the 0.05 level of significance whether there is sufficient evidence to conclude that the average starting salary of all new commerce grads exceeds $50,000. Use the critical value approach and show manually how the t-statistic is calculated.

(b) Now show how you would calculate the p-value for the result in part (a).

(c) Calculate manually a 95% 1-sided confidence interval for the average salary.

(d) Explain why the p-value and confidence interval do or do not support your conclusion in part (a).

(e) Show a boxplot of the data and explain whether the assumptions of the test and confidence interval are reasonable.

Answer 1

Calculate the mean and standard deviation of the salaries data. Mean(X 47508+52802 +...+ 65170 1757020 -53243.03 Standard deviation(s) n-1 (47508 53243.03) +... (65170-53243.03) 33-1 9445.46744305453 9445.47 9445.47 (Round to 2 decimal 1pl Null hypothesis, Ho : μ-50000 Alternative hypothesis, H1:μ>50000

Level of significance, a 0.05 Calculate the test statistic value. sIvn 53243.03-50000 9445.47 /V33 Hence, the required test statistic is 1.97 Calculate the degrees of freedom. df -n-1 -33-1 - 32 Calculate t - critical value. t-critical TINV(0.95,32)) (Use MS Excel) 1.693889 Conclusion The tatis ie Value is greater than the tCrisical value, so we reject the null hypothesis and conclude that the average starting salary of all new commerce grads exceeds 50000.

b) Calculate the P-value of the Hypothesis test. P-value (TDIST(1.97,32,1)) (Use MS Excel) = 0.028772 Conclusion The P-value is less than the given significance level 0.05, so we reject the null hypothesis and conclude that the average starting salary of all new commerce grads exceeds 50000 c) Calculate the 95% 1-sided confidence interval for the average. 95%CI 1 sided-1-tcritical 3445A1 /33 -53243.03-1.693889 = 53243.03-2785.169 50457.86 Therefore, the 95% confident that the population mean for the average salary is 5045786 d) The P-value and the confidence interval calculated in the part (a) supports your conclusion that there is sufficient evidence to conclude that the population mean of the average starting salary of all new commerce graduates exceeds $50000.

e) Use MS Excel to draw a Box plot for the salaries data. Step-1: Enter the data into excel sheet Step-2: Click Add-Ins ->Mega Stat ->Descriptive Statistics Step-3: Enter the data into Input range box. Step-4: Click Box plot Step-5: Click OK. The obtained output is shown in below, BoxPlot 20000 30000 40000 50000 60000 70000 80000 90000 100000 data The box plot of salary indicates that the distribution is approximately normally distributed Since, the mean is greater than that of the median of salaries. There exists outlier in the data. The assumption of the t-test conducted and the confidence intervals calculated in the parts (a) and (c) yields the conclusions are reasonable.