Question

A university surveyed recent graduates of the English department about their starting salaries. Four hundred graduates returned the survey. The average salary was $25,000. Assume for purposes of this problem that the population standard deviation is known, and it is $2,500. What is the 95% confidence interval for the mean salary of all graduates from the English department?

Answer 1

Answer:   (24755, 25245)

Step-by-step explanation:

Given : Sample size : n= 400

Sample mean : $\overline{x}= \ 25,000$

Standard deviation : $\sigma=\ 2,500$

Significance level : $\alpha: 1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

The confidence interval for population mean is given by :-

$\overline{x}\pm\ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=25000\pm(1.96)\dfrac{2500}{\sqrt{400}}\\\\=25000\pm245\\\\=(25000-245,\ 25000+245)=(24755,\ 25245)$

Hence, the 95% confidence interval for the mean salary of all graduates from the English department is (24755, 25245)