Answer:
Explanation:
We want to determine a 95% confidence interval for the mean salary of all graduates from the English department.
Number of sample, n = 400
Mean, u = $25,000
Standard deviation, s = $2,500
For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.
We will apply the formula
Confidence interval
= mean ± z × standard deviation/√n
It becomes
25000 ± 1.96 × 2500/√400
= 25000 ± 1.96 × 125
= 25000 ± 245
The lower end of the confidence interval is 25000 - 245 =24755
The upper end of the confidence interval is 25000 + 245 = 25245
Therefore, with 95% confidence interval, the mean salary of all graduates from the English department is between $24755 and $25245