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fiasKO [112]
3 years ago
12

A survey of an urban university (population of 25,450) showed that 750 of 1,100 students sampled attended a home football game d

uring the season. Using the 90% level of confidence, what is the confidence interval for the proportion of students attending a football game
Mathematics
1 answer:
JulsSmile [24]3 years ago
6 0

Answer:

The 90% level of confidence interval For the Population proportion

(0.6569 ,0.7031)

Step-by-step explanation:

<u><em>Explanation</em></u>:-

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)

<u><em>Final answer</em></u>:-

The 90% level of confidence interval For the Population proportion

(0.6569 ,0.7031)

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