Question

In a recent survey of 500 grocery shoppers, 29% of males and 40% of females make an impulse purchase when they shop.

If the survey consists of 250 males and 250 females, at .05 significance is there evidence of a difference in proportion between the two populations?

Answer 1

solution:

the given information as follows:

let sample 1 denotes male survey and sampel 2 denotes the details for female survey.

so,

for sample 1:

n_1 = 250

proportion of male who make an impulse purchase when they shop = = 29% = 0.29

P1

for sample 2:

proportion of female who make an impulse purchase when they shop = $\hat p_2$ = 40% = 0.40hjn mkmjvggf

value of the pooled proportion is computed as $\bar p=\frac{0.29+0.40}{2}=0.345$

we have to test is there significance difference in two proportion.

so, null and alternative hypothesis:

ad id: "H

it is a two tailed test:

test statistics:

$z=\frac{\hat p_1-\hat p_2}{\sqrt{\bar p(1-\bar p)\left ( \frac{1}{n_1}+\frac{1}{n_2} \right )}}=\frac{0.29-0.40}{\sqrt{0.345(1-0.345)\left ( \frac{1}{250}+\frac{1}{250} \right )}}=-2.59$

significance level = $\alpha$ = 0.05

critical value = $z_c=z_{\alpha/2}=z_{0.05/2}=\pm1.96$

(critical value is found from the z table)

since , z= -2.59 < -1.96 , so rejecting the null hypothesis

conclusion;

there is significance evidence at 0.05 significance level that there is difference in proportion between the two populations.