# 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.

UkusakazaL 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 $$\displaystyle{95}\%$$ confidence interval.

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

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