Question

Problem 3: A Corporate Perceptions Study surveyed 200 readers and asked them on how they would rate XYZ Corporation on the “Quality of Management” and the “Reputation of the Company”. The two variables of the study were rated on a categorical scale as excellent, good, and fair. The sample data on 200 responses for the study is summarized as follows. Conduct an appropriate hypothesis test for this problem to check if the variables are independent of each other. Reputation of Company Quality of Management Excellent Good Fair Excellent 39 26 5 Good 36 34 10 Fair 25 10 15 Answer the following questions based on your analysis. Problem 3d) Calculate and show the chi-square components in a table. What is the test statistic value? Problem 3e) Type in the Excel function with inputs to be used to determine p-value. What is the p-value? Conduct Hypothesis test using p-value approach (at .004 level of significance). What is the decision on the hypothesis test? Problem 3f) Type in the Excel function with inputs to be used to determine Critical-value. What is the Critical-value? Conduct Hypothesis test using Critical-Value approach (at .008 level of significance). What is the decision on the hypothesis test? Problem 3g) Are the decisions under the p-value and critical-value approaches different? If they are different, why are they different? If they are not different, why are they not different? Problem 3h) Based on test decisions under parts 3e) and 3f), what conclusion would you draw? Does your conclusion sound meaningful?

Answer 1

3d)

Following table shows the row total and column total:

		Quality of managemet
		Good	Fair	Excellent	Total
	Excellent	39	26	5	70
Reputation of company	Good	36	34	10	80
	Fair	25	10	15	50
	Total	100	70	30	200

Expected frequencies will be calculated as follows:

Following table shows the calculations for expected freqeuncies:

		Quality of managemet
		Good	Fair	Excellent	Total
	Excellent	35	24.5	10.5	70
Reputation of company	Good	40	28	12	80
	Fair	25	17.5	7.5	50
	Total	100	70	30	200

Following table shows the calculations for chi square test statistics:

O	E	(O-E)^2/E
39	35	0.457142857
36	40	0.4
25	25	0
26	24.5	0.091836735
34	28	1.285714286
10	17.5	3.214285714
5	10.5	2.880952381
10	12	0.333333333
15	7.5	7.5
Total		16.16326531

The test statistics is:

3e)

Degree of freedom: df =( number of rows -1)*(number of columns-1) = (3-1)*(3-1)=4

The p-value using excel function "=CHIDIST(16.163,4)" is 0.0028.

Since p-value is less than 0.004 so we reject the null hypothesis.

3f)

The critical value for test using excel function "=CHIINV(0.008,4)" is 13.789.

Since test statistics is greater than critical value so we reject the null hypothesis.

3g)

No decisions are same

3h)

The variables “Quality of Management” and the “Reputation of the Company” are not independent of each other.