Question

A survey found that 31 of 60 randomly selected women and 38 of 73 randomly selected men follow a regular exercise program. Find a 95% confidence interval for the difference between the proportions of women and men who follow a regular exercise program.

Answer 1

Answer:

95% confidence interval for the difference between the proportions of women and men who follow a regular exercise program

(-0.1740 , 0.1672)

Step-by-step explanation:

Step(i):-

Given A survey found that 31 of 60 randomly selected women

First sample proportion

                $p_{1} = \frac{x_{1} }{n_{1} } = \frac{31}{60} = 0.5166$

Second sample proportion

               $p_{2} = \frac{x_{2} }{n_{2} } = \frac{38}{73} = 0.520$

Level of significance = 0.05

Z₀.₀₅ = 1.96

Step(ii):-

95% confidence interval for the difference between the proportions of women and men who follow a regular exercise program

$((p_{1} ^{-}-p^{-} _{2} - Z_{0.05} S.E(p_{1} ^{-} - p^{-} _{2} ), (p_{1} ^{-}-p^{-} _{2} +Z_{0.05} S.E(p_{1} ^{-} - p^{-} _{2} ))$

where

$Se(p_{1} - p_{2} ) = \sqrt{\frac{p_{1} (1-p_{1} )}{n_{1} }+\frac{p_{2}(1-p_{2} }{n_{2} } }$

$Se(p_{1} - p_{2} ) = \sqrt{\frac{0.5166 (1-0.5166 )}{60 }+\frac{0.520(1-0.520 }{73 } } = 0.08706$

95% confidence interval for the difference between the proportions of women and men who follow a regular exercise program

$((p_{1} ^{-}-p^{-} _{2} - Z_{0.05} S.E(p_{1} ^{-} - p^{-} _{2} ), (p_{1} ^{-}-p^{-} _{2} +Z_{0.05} S.E(p_{1} ^{-} - p^{-} _{2} ))$

((0.5166-0.520-1.96(0.08706) , 0.5166-0.520+1.96(0.08706))

(-0.1740 , 0.1672)

Final answer:-

95% confidence interval for the difference between the proportions of women and men who follow a regular exercise program

(-0.1740 , 0.1672)