Question

How to calculate margin of error for this question?

Starting salaries of 80 college graduates who have taken a statistics course have a mean of $42,893. Suppose the distribution of this population is approximately normal and has a standard deviation of $10,748.
Use a 93% confidence level.

Can't get the answer right :(

Answer 1

Solution :

Given that,

Population standard deviation =     = $10748

Sample size n =80

At 93% confidence level the z is ,

= 1 - 93% = 1 - 0.93= 0.07

/ 2 = 0.035

Z / 2   = Z0.035 = 1.81 ( Using z table )

Margin of error = E =   Z / 2     * ( /n)
= 1.81 * ( 10748 / 80 )

E= 2175.0099