Question

In a recent survey of 1159 students, 968 of them would like to recommend a website to their friends. Construct a 95% confidence interval to estimate the proportion of all students who would recommend the website to their friends.

Give your answers to three decimals:

_____,_____

Answer 1

The sample proportion is computed as follows

$$\hat{p}=\dfrac{X}{N}=\dfrac{968}{1159}$$

The critical value for $\alpha = 0.05$ is $z_c = z_{1-\alpha/2} = 1.96.$

The corresponding confidence interval is computed as shown below:

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

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

$$=(\dfrac{968}{1159}-1.96\sqrt{\dfrac{\dfrac{968}{1159}(1-\dfrac{968}{1159})}{1159}},$$

$$\dfrac{968}{1159}+1.96\sqrt{\dfrac{\dfrac{968}{1159}(1-\dfrac{968}{1159})}{1159}})$$

$$=(0.814,0.857)$$

Therefore, based on the data provided, the 95% confidence interval for the population proportion is $0.814 < p < 0.857,$ which indicates that we are 95% confident that the true population proportion $p$ is contained by the interval $(0.814, 0.857).$