Question

A university dean is interested in determining the proportion of students who receive some financial aid. The dean randomly selects 200 students and finds that 118 of them are receiving financial aid. Use a 95% confidence interval to estimate the proportion of students on financial aid.

Answer 1

Answer: $(0.522,\ 0.658)[$

Step-by-step explanation:

The formula to calculate the confidence interval is given by :-

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where n= sample size .

$\hat{p}$ = sample proportion.

z* = cRitical value .

Let p be the population proportion of students who receive some financial aid.

The dean randomly selects 200 students and finds that 118 of them are receiving financial aid.

i.e. n= 200

$\hat{p}=\dfrac{118}{200}=0.59$

We know that the critical value for 95% confidence interval : z*= 1.96

Then, the 95% confidence interval for population proportion will be :-

$(0.59)\pm (1.96)\sqrt{\dfrac{0.59(1-0.59)}{200}}$

$0.59\pm (1.96)\sqrt{0.0012095}$

$0.59\pm (1.96)(0.0347778665246)$

$0.59\pm0.0681646183882\approx0.59\pm0.068\\\\=(0.59-0.068,\ 0.59+0.068)\\\\=(0.522,\ 0.658)$

Hence, 95% confidence interval to estimate the proportion of students on financial aid = $(0.522,\ 0.658)[$