Question

The mean age of all students at a local community college for a recent Fall term was 32.4. The population standard deviation has been pretty consistent at 12. Suppose that 25 students from the college were randomly selected. The mean age for the sample was 29.7. The true mean age of the community college students, given a confidence level of 95% lies between [x] and [y].  Please enter the lower limit of the interval first and the upper limit second. Please round to two decimal places.

I answered with (25.00, 34.40) and received partial credit... though I don't know what was right and what was wrong.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 29.7

Population standard deviation =    = 12

Sample size = n = 25

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96

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

= 1.96 * ( 12 /  25 )

= 4.70

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

  ± E

29.7 ± 4.70

( 25.00, 34.40 )