Question

The U.S. Commission on Crime randomly selects 600 files of recently committed crimes in an area and finds 380 in which a firearm was reportedly used. Find a 95% confidence interval for p, the true fraction of crimes in the area in which some type of firearm was reportedly used.

Answer 1

Answer:

95% confidence interval for p, the true fraction of crimes in the area in which some type of firearm was reportedly used.

(0.59445 , 0.67155)

Step-by-step explanation:

Explanation:-

Given random sample size 'n' = 600

sample proportion

                            $p^{-} = \frac{x}{n} = \frac{380}{600} = 0.633$

95% of confidence interval for Population proportion is determined by

$(p^{-} - Z_{\frac{\alpha }{2} } \sqrt{\frac{p^{-} (1-p^{-} )}{n} } , p^{-} +Z_{\frac{\alpha }{2} } \sqrt{\frac{p^{-} (1-p^{-} )}{n} })$

Level of significance : α = 0.05

   $z_{\frac{0.05}{2} } = Z_{0.025} =1.96$

$(0.633 - 1.96 \sqrt{\frac{0.633 (1-0.633 )}{600 }}, (0.633 + 1.96 \sqrt{\frac{0.633 (1-0.633 )}{600 }$

On calculation , we get

(0.633 - 0.03855 , (0.633 + 0.03855)

(0.59445 , 0.67155)

Conclusion:-

95% confidence interval for p, the true fraction of crimes in the area in which some type of firearm was reportedly used.

(0.59445 , 0.67155)