Answer to Question #174617 in Statistics and Probability for kavinuwan

Question

A well-known consulting firm wants to test how it can influence the proportion of questionnaires returns for its surveys. In the belief that the inclusion of an inducement to respond may be influential, the firm sends out 1000 questioners: 200 promise to send respondents a summary of the survey results; 300 indicate that 20 respondents (selected by a lottery) will be awarded gifts; and 500 are accompanied by no inducements. Of these, 80 questionnaires promising a summary, 100 questionnaires offering gifts, and 120 questionnaires offering no inducements are returned. What can you conclude from these results?

Answer 1

To test independence of two attributes we use chi-square test.

The given information is:

number of rows m=2

number of column n=3

Hypothesis:

H₀ - responded and not responded are independent.

H₁ - responded and not responded are dependent.

Test statistic:

$χ^2 = \sum^{n}_{j=1} \sum^{m}_{i=1} \frac{(O_{ij}-E_{ij})^2}{E_{ij}} \\ χ^2 = \sum^{n}_{j=1} \sum^{m}_{i=1} \frac{O_{ij}^2}{E_{ij}} – N$

Is follow χ² distribution with (m-1)(n-1) degrees of freedom.

O​​​​​_ij - observed frequency i​​​​​​^throw and j​​​​​​^th column

E​​​​​​_ij - expected frequency i​​​​​​^th row and j​​​​​​^th column

$E_{ij} = \frac{(i^{th} \; row \; total) \times (j^{th}) \;column \;total}{N} \\ E_{11} = \frac{R_1 \times C_1}{N} \\ = \frac{300 \times 200}{1000} = 60 \\ E_{12} = \frac{R_1 \times C_2}{N} \\ = \frac{300 \times 300}{1000} = 90 \\ E_{13} = \frac{R_1 \times C_3}{N} \\ = \frac{300 \times 500}{1000} = 150 \\ E_{21} = \frac{R_2 \times C_1}{N} \\ = \frac{700 \times 200}{1000} = 140 \\ E_{22} = \frac{R_2 \times C_2}{N} \\ = \frac{700 \times 300}{1000} = 210 \\ E_{23} = \frac{R_2 \times C_3}{N} \\ = \frac{700 \times 500}{1000} = 350 \\ χ^2 = \sum^{n}_{j=1} \sum^{m}_{i=1} \frac{O_{ij}^2}{E_{ij}} -N \\ χ^2 = \frac{80^2}{60} + \frac{100^2}{90} + \frac{120^2}{150} + \frac{120^2}{140} + \frac{200^2}{210} + \frac{380^2}{350} -1000 \\ = 106.66 + 111.11 + 96 + 102.85 + 190.47 + 412.57 -1000 \\ = 1019.66 -1000 \\ = 19.66$

Critical value or table value at

$(m-1)(n-1)=(2-1)(3-1)= 1 \times 2=2$ degrees of freedom and 5 % level of significance.

Tab $χ^2 = 5.991$

$Cal \; χ^2 > Tab \; χ^2$

Reject null hypothesis at 5% level of significance.

There is enough evidence to conclude that the responded and not responded are dependent at 5% level of significance.