Question

Chi- Squared Test of Independence

6) Business Non-response questionnaires were mailed to 300 randomly selected businesses in each of three categorical sizes. The following data shows the number of responses.

Size of Company	Response	No Response
Small	175	125
Medium	145	155
Large	120	180

Test the claim that the size of the company is independent of the company`s response

a. State the null and alternative hypothesis
b. Give the p-value
c. Give a conclusion for the hypothesis test

Answer 1

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a. Hypothesis design:

Null hypothesis, Ho:  size of the company is independent of the company`s response

Alternate hypothesis, Ha:  size of the company is NOT independent of the company`s response

b. p-value.

The contingency table below provides the following information: the observed cell totals, (the expected cell totals) and [the chi-square statistic for each cell].

For example for 1st cell:

We calculate Expected value, E = RowSum*ColumnSum/Total

= 300*440/900 = 146.67

Chi-square = (O-E)^2/E = (146.67-175)^2 / 146.67 = 5.47

Results
	response	non response				Row Totals
small	175  (146.67)  [5.47]	125  (153.33)  [5.24]				300
medium	145  (146.67)  [0.02]	155  (153.33)  [0.02]				300
large	120  (146.67)  [4.85]	180  (153.33)  [4.64]				300
Column Totals	440	460				900  (Grand Total)

The chi-square statistic is the sum of individual Chi-Squares in square brackets in the table above = 5.47 + 5.24 + .02 + .02 + 4.85+ 4.64 = 20.2322.

df = (rows-1)*(column -1) = (3-1)*(2-1) = 2

p-value = 1 - CHISQ.DIST(20.2322, 2, TRUE) = .00004

The p-value is .00004.

Answer: p-value = .00004

c.

The result is significant at p < .05. We reject null hypothesis.

Answer: Conclusion: Size of the company is NOT independent of the company`s response