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.

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Answered 2021-08-16 Author has 29527 answers

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.

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