Expert Answer to "Customers randomly selected at a grocery store included..."

UkusakazaL

Answered question

2021-08-09

Customers randomly selected at a grocery store included 172 women and 45 men. 74 of the women and 12 of the men used coupons. Find the

95 %

confidence interval.

Answer & Explanation

Ian Adams

Skilled2021-08-16Added 163 answers

Step 1
Given:
Number of women $(n_{1}) = 172$
Number of men $(n_{2}) = 45$
Number of women used coupons $(x_{1}) = 74$
Number of men used coupons $(x_{2}) = 12$
Confidence level $= 0.95$
So, $α = 1 - 0.95 = 0.05$
Step 2
Sample proportions
${\hat{p}}_{1} = \frac{x_{1}}{n_{1}} = \frac{74}{172}$
${\hat{p}}_{2} = \frac{x_{2}}{n_{2}} = \frac{12}{45}$
From z tables, $P (z > 1.96) = 0.025$
Step 3
Finding confidence interval
$({\hat{p}}_{1} - {\hat{p}}_{2}) - z_{\frac{α}{2}} \times \sqrt{\frac{{\hat{p}}_{1} \times (1 - {\hat{p}}_{1})}{n_{1}} + \frac{{\hat{p}}_{2} \times (1 - {\hat{p}}_{2})}{n_{2}}}, ({\hat{p}}_{1} - {\hat{p}}_{2}) + z_{\frac{α}{2}} \times \sqrt{\frac{{\hat{p}}_{1} \times (1 - {\hat{p}}_{1})}{n_{1}} + \frac{{\hat{p}}_{2} \times (1 - {\hat{p}}_{2})}{n_{2}}}$
$(\frac{74}{172} - \frac{12}{45}) - 1.96 \times \sqrt{\frac{\frac{74}{172} \times (1 - \frac{74}{172})}{172} + \frac{\frac{12}{45} \times (1 - \frac{12}{45})}{45}}, (\frac{74}{172} - \frac{12}{45}) + 1.96$
$\times \sqrt{\frac{\frac{74}{172} \times (1 - \frac{74}{172})}{172} + \frac{\frac{12}{45} \times (1 - \frac{12}{45})}{45}}$
0.0147, 0.3125
The confidence interval for difference in women and men proportions of using coupons is 0.0147, 0.3125.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in College Statistics

Didn't find what you were looking for?