Question

A survey of an urban university (population of 25,450) showed that 750 of 1,100 students sampled attended a home football game during the season. Using the 90% level of confidence, what is the confidence interval for the proportion of students attending a football game

Answer 1

Answer:

The 90% level of confidence interval For the Population proportion

(0.6569 ,0.7031)

Step-by-step explanation:

Explanation:-

Given sample size 'n' = 1100

Given sample proportion

        $p^{-} =\frac{x}{n} = \frac{750}{1100} = 0.68$

The 90% level of confidence interval For the Population proportion is determined by

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

90% of level of significance Z-value

$Z_{\frac{0.10}{2} } = Z_{0.05} =1.645$

$(0.68 - 1.645 \sqrt{\frac{0.68(1-0.68)}{1100} } ,(0.68 + 1.645 \sqrt{\frac{0.68(1-0.68)}{1100}$

On calculation , we get

(0.68 -0.0231 ,0.68 +0.0231)

(0.6569 ,0.7031)

Final answer:-

The 90% level of confidence interval For the Population proportion

(0.6569 ,0.7031)