Question

A researcher finds that of 1000 people who said that they attend a religious service at least once a week, 31 stopped to help a person with car trouble. Of 1200 people interviewed who had not attended a religious service at least once a month, 22 stopped to help a person with car trouble. At the 0.05 significance level, test the claim that the two proportions are different.

Answer 1

Answer:

Out of 1000 people who said that they attend a religious service at least once a week, 31 stopped to help a person with car trouble

$n_1=1000 , y_1=31$

Of 1200 people interviewed who had not attended a religious service at least once a month, 22 stopped to help a person with car trouble.

$n_2=1200 , y_2=22$

We will use Comparing Two Proportions

$\widehat{p_1}=\frac{y_1}{n_1}$

$\widehat{p_1}=\frac{31}{1000}$

$\widehat{p_1}=0.031$

$\widehat{p_2}=\frac{y_2}{n_2}$

$\widehat{p_2}=\frac{22}{1200}$

$\widehat{p_2}=0.0183$

$H_0:p_1=p_2$i.e. religion service makes no difference

$H_a:p_1 \neq p_2$  i.e. religion service makes difference

$\widehat{p}=\frac{y_1+y_2}{n_1+n_2}=\frac{31+22}{1000+1200} =0.024$

Formula of test statistic : $\frac{\widehat{p_1}-\widehat{p_2}}{\sqrt{\widehat{p}(1-\widehat{p})(\frac{1}{n_1}+\frac{1}{n_2})}}$

Substitute the values  

test statistic : $\frac{0.031-0.0183}{\sqrt{0.024(1-0.024)(\frac{1}{1000}+\frac{1}{1200})}}= 1.937$

Refer the z table for p value

p value = 0.9726

α=0.05

p value >α

So, we failed to reject null hypothesis

Hence religion service makes no difference