Question

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

Answer 1

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