Question

Starting salaries of 110 college graduates who have taken a statistics course have a mean of $44,836. Suppose the distribution of this population is approximately normal and has a standard deviation of $10,796.
Using a 75% confidence level, find both of the following:

(a) The margin of error:

(b) The confidence interval for the mean μ: ____ < μ < _____

Answer 1

solution:

From the given information

Samle size (n) = 110

Sample Mean () = $44,836

Standard deviation () = $10796

For 75% confidence level , = 1 - CL = 1 - 0.75 = 0.25

Critical value : Z(/2) = Z(0.125) = 1.15 [ since ,use z-distribution table ]

a) The margin of error (ME) = Z*(/ ) = 1.15*(10796/ ) = 1183.762

b) The confidence interval for mean is given by

CI : ( -ME , + ME)

: (44836-1183.762 , 44836+1183.762)

: ( 43652.238 , 46019.762)

Therefore,The confidence interval for mean is  43652.24 < < 46019.76