Answer to Question #226747 in Statistics and Probability for csk

Question

Chi-Square Testing 

A survey was carried out in a firm of the smoking habits of men and women employees with the following results:

       

Men Women

Smokers 48 27

Non-smokers 58 57

 

It is required to test whether, at the 5% level of significance, the survey reveals any difference in the smoking habits of men and women.

Answer 1

Expected values:

$\begin{matrix} & Men&Women \\ Smokers & 41.84 & 33.16\\ Non-smokers & 64.16 & 50.84 \end{matrix}$

Test statistic: $\chi^2=\frac{(48-41.84)^2}{41.84}+\frac{(27-33.16)^2}{33.16}+\frac{(58-64.16)^2}{64.16}+\frac{(57-50.84)^2}{50.84}=3.387.$

Degrees of freedom: $df=(2-1)(2-1)=1.$

P-value: $p=P(\chi^2>3.387)=0.0657.$

Since the p-value is greater than 0.05, fail to reject the null hypothesis.

The survey did not reveal any difference in the smoking habits of men and women.