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igor_vitrenko [27]
3 years ago
5

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex

amine the records for all​ students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. Use a 99​% confidence interval to estimate the true proportion of students on financial aid. A. 0.59plus or minus0.003 B. 0.59plus or minus0.090 C. 0.59plus or minus0.007 D. 0.59plus or minus0.623
Mathematics
1 answer:
Pachacha [2.7K]3 years ago
8 0

Answer: B. 0.59 plus or minus 0.090

Step-by-step explanation:

The confidence interval of population proportion is given by :-

\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

Given : Sample size : n=200

The proportion of students are receiving financial aid :hat{p}=\dfrac{118}{200}=0.59

Significance level : \alpha: 1-0.99=0.01

Critical value : z_{\alpha/2}=2.576

Then, the confidence interval of population proportion is given by :-

0.59\pm (2.576)\sqrt{\dfrac{0.59(1-0.59)}{200}}\\\\=0.59\pm0.0895877841673\\\\\approx0.59\pm0.090

Hence,  a 99​% confidence interval to estimate the true proportion of students on financial aid = 0.59 plus or minus 0.090

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