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)

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)(\sigma)/(√(n))\\\\=25000\pm(1.96)(2500)/(√(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)