Question

A survey of nonprofit Universities showed that online fundraising has increased in the past year. Based on a sample of 80 Universities, the mean quantity per donation in the past year was $82, with a standard deviation of $9. You are required to create a 95% confidence interval with the data.

What is the margin of error?

What is the point estimate for this for μ using the data above?

Answer 1

Solution :

Given that,

Point estimate = sample mean =     = $82

Population standard deviation =     =$9

Sample size n =80

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z / 2   = Z_0.025 = 1.96   ( Using z table )

Margin of error = E =   Z / 2     * ( /n)
= 1.96 * (9 / 80 )

= 1.9722
At 95% confidence interval estimate of the population mean
is,

- E <   < + E

82-1.9722 <   <82 + 1.9722

80.0278 <   < 83.9722

( 80.0278 , 83.9722)