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]
|
|
|
