Question

The president of a University wishes to find the average age of students presently enrolled. From past studies, the standard deviation is known to be 2 years. A random sample of 50 students is selected and the mean is found to be 23.2 years.

A. Find the 95% confidence interval for the population mean.

B. Find the 99% confidence interval for the population mean?

Answer 1

Solution :

Given that,

Point estimate = sample mean =     =23.2

Population standard deviation =    = 2

Sample size n =50

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 * (2 / 50 )

E= 0.5544
At 95% confidence interval estimate of the population mean
is,

- E < < + E

23.2- 0.5544 <   <23.2 + 0.5544

22.6456 <   < 23.7544

( 22.6456, 23.7544 )

(B)

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576   ( Using z table )

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

= 2.576 * (2 / 50 )

= 0.7286

At 99% confidence interval estimate of the population mean is,

- E < < + E

23.2- 0.7286< < 23.2+0.7286

22.4714< < 23.9286

(22.4714 , 23.9286)