Answer to Question #346700 in Statistics and Probability for Pearl

Question

A sample of 390 senior high school students applied to different state universities, wherein 117 are female. Find the 99% confidence interval of the true population proportion who applies for different state universities

Answer 1

The sample proportion is computed as follows, based on the sample size $N = 390$ and the number of favorable cases $X = 117$

$$\hat{p}=\dfrac{X}{N}=\dfrac{117}{390}=0.3$$

The critical value for $\alpha = 0.01$ is $z_c = z_{1-\alpha/2} = 2.5758.$

The corresponding confidence interval is computed as shown below:

$$CI(Proportion)=(\hat{p}-z_c\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{N}},$$

$$\hat{p}+z_c\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{N}})$$

$$=(0.3-2.5758\times \sqrt{\dfrac{0.3(1-0.3)}{390}},$$

$$0.3+2.5758\times \sqrt{\dfrac{0.3(1-0.3)}{390}})$$

$$=(0.24023, 0.35977)$$

Therefore, based on the data provided, the 99% confidence interval for the population proportion is $0.24023 < p < 0.35977,$ which indicates that we are 99% confident that the true population proportion $p$ is contained by the interval $(0.24023, 0.35977).$