Recently, a university surveyed recent graduates of the English Department for their starting salaries. Four hundred graduates returned the survey. The average salary was $25,577. The population standard deviation is $2,566. What is the 95% confidence interval for the mean salary of all graduates from the English Department?
[$25,326, $25,828]
[$25,365, $25,789]
[$25,246, $25,908]
[$25,336, $27,101]
[$25,326, $25,828]
Thank you!
To find the 95% confidence interval for the mean salary of all graduates from the English Department, we can use the formula:
Confidence Interval = Sample Mean ± (Z-score * Standard Error)
Where:
- Sample Mean is the average starting salary of the graduates ($25,577)
- Z-score is the critical value that corresponds to the desired confidence level (95% confidence level corresponds to a Z-score of 1.96)
- Standard Error is the standard deviation of the sample mean, which is calculated as the population standard deviation divided by the square root of the sample size (σ / √n)
Given that the population standard deviation is $2,566 and the sample size is 400, we can calculate the standard error as:
Standard Error = 2,566 / √400 = $128.30
Now, we can substitute the values into the formula to find the confidence interval:
Confidence Interval = 25,577 ± (1.96 * 128.30)
Confidence Interval = [25,577 - (1.96 * 128.30), 25,577 + (1.96 * 128.30)]
Confidence Interval = [25,577 - 251.87, 25,577 + 251.87]
Confidence Interval = [25,325.13, 25,828.87]
Therefore, the 95% confidence interval for the mean salary of all graduates from the English Department is [$25,326, $25,829]. Hence, the correct option is [$25,326, $25,828].