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

Respuesta :

Answer:

There is enough evidence to support the claim that the proportions are not equal. (P-value: 0.048).

Step-by-step explanation:

The question is incomplete:

"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."

This is a hypothesis test for the difference between proportions.

The claim is that the proportions are not equal.

Then, the null and alternative hypothesis are:

[tex]H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq 0[/tex]

The significance level is 0.05.

The sample 1, of size n1=1000 has a proportion of p1=0.031.

[tex]p_1=X_1/n_1=31/1000=0.031[/tex]

The sample 2, of size n2=1200 has a proportion of p2=0.018.

[tex]p_2=X_2/n_2=22/1200=0.018[/tex]

The difference between proportions is (p1-p2)=0.013.

[tex]p_d=p_1-p_2=0.031-0.018=0.013[/tex]

The pooled proportion, needed to calculate the standard error, is:

[tex]p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{31+22}{1000+1200}=\dfrac{53}{2200}=0.024[/tex]

The estimated standard error of the difference between means is computed using the formula:

[tex]s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.024*0.976}{1000}+\dfrac{0.024*0.976}{1200}}\\\\\\s_{p1-p2}=\sqrt{0+0}=\sqrt{0}=0.007[/tex]

Then, we can calculate the z-statistic as:

[tex]z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.013-0}{0.007}=\dfrac{0.013}{0.007}=1.98[/tex]

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

[tex]P-value=2\cdot P(z>1.98)=0.048[/tex]

As the P-value (0.048) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the proportions are not equal.