SOLUTION: (I'm stuck on this question, can someone please help?) Recently, a university surveyed recent graduates of the English Department for their starting salaries. Four hundred graduat

Question 865293: (I'm stuck on this question, can someone please help?)
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]

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Use this table to get the critical value of (look in the row that starts with . In this row, you'll find a value that's above the 95% confidence level, which is 1.960)

The sample mean is xbar = 25577 (given)
The population standard deviation is sigma = 2566 (given)
The sample size is n = 400 (given)
From above, the critical value for a 95% confidence interval is z = 1.960 (this is approximate)

----------------------------------------------------------------------------

We will use these values to plug them into the lower bound L and upper bound U formulas below

L = xbar - z*sigma/sqrt(n)
U = xbar + z*sigma/sqrt(n)

where the confidence interval is [L, U]

----------------------------------------------------------------------------

Now let's find the lower bound L of the confidence interval

L = xbar - z*sigma/sqrt(n)

L = 25577 - 1.960*2566/sqrt(400)

L = 25577 - 251.468

L = 25325.532

L = 25326

----------------------------------------------------------------------------

Now let's find the upper bound U of the confidence interval

U = xbar + z*sigma/sqrt(n)

U = 25577 + 1.960*2566/sqrt(400)

U = 25577 + 251.468

U = 25828.468

U = 25828

So the confidence interval [L,U] is [25326, 25828]

This is choice A) [$25,326, $25,828]

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