Question

A national survey of U.S. consumers conducted by the Wall Street Journal asked the question: "In general, how would you rate the service that American businesses provide?" The distribution of responses to this question was as follows:

• Excellent 8%
• Pretty good 47%
• Only fair 34%
• Poor 11%

Suppose a store manager wants to find out whether the results of this consumer survey apply to customers of supermarkets in his city. To do so, he interviews 207 randomly selected customers as they leave the supermarkets in various parts of the city as to their rating of service. Following is the number of responses and distribution of this survey:

• Excellent 21
• Pretty good 109
• Only fair 62
• Poor 15

Please answer the following:

1. Determine whether the observed frequencies of responses from this survey are the same as the frequencies that would be expected on the basis of the national survey. Hint: Think Chi-square goodness-of-fit, and see Ppts and Selected Example Solutions, and of course, the internet.
2. Can you describe a chi-squared application, either goodness-of-fit or test-for-independence, in your profession?

Answer 1

Solution:

Part a

Here, we have to use chi square test for goodness of fit.

Null hypothesis: H₀: Given data follows the frequency distribution of the national survey.

Alternative hypothesis: H_a: Given data do not follow the frequency distribution of the national survey.

We assume level of significance = α = 0.05

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

We are given

N = 4

Degrees of freedom = df = N – 1 = 4 – 1 = 3

α = 0.05

Critical value = 7.814727764

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Category	Exp.Prob.	O	E	(O - E)^2/E
Excellent	0.08	21	16.56	1.190434783
Pretty good	0.47	109	97.29	1.409436736
Only fair	0.34	62	70.38	0.997789145
Poor	0.11	15	22.77	2.651422925
Total	1	207	207	6.249083588

Chi square = ∑[(O – E)^2/E] = 6.249083588

P-value = 0.100100991

(By using Chi square table or excel)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that given data follows the frequency distribution of the national survey.

There is sufficient evidence to conclude that the observed frequencies of responses from this survey are the same as the frequencies that would be expected on the basis of the national survey.

Part b

Yes, we can use the Chi squared application in our profession. We can use Chi square test for goodness of fit for checking whether the distribution of the product line follows the entire population distribution or not. Also, we can use chi square test for independence for checking whether the two categorical variables such as gender and satisfaction of employee are independent or not. There would be several real life situations in the profession whether we can use these chi squared applications.