Answer to Question #219307 in Statistics and Probability for Akoo

Question

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examining the records for all students, the dean randomly selected 200 students and finds that 118 of them are receiving financial aid. Use a 90% confidence interval to estimate the true proportion of students who receive financial aid.

Answer 1

The sample proportion is computed as follows, based on the sample size $N=200$ and the number of favorable cases $X=118:$

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

The critical value for $\alpha=0.01$  is $z_c=z_{1-\alpha/2}=1.6449.$ The corresponding confidence interval is computed as shown below:

$$CI(Proportion) ​$$

$$=(\hat{p}-z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})$$

$$=(0.59-1.6449\sqrt{\dfrac{0.59(1-0.59)}{200}},$$

$$0.59+1.6449\sqrt{\dfrac{0.59(1-0.59)}{200}})$$

$$=(0.5328, 0.6472)$$

Therefore, based on the data provided, the 90% confidence interval for the population proportion is $0.5328<p<0.6472,$ which indicates that we are 90% confident that the true population proportion $p$ is contained by the interval $(0.5328, 0.6472).$