Question

A group of 76 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test. Construct and interpret a 99% confidence interval for the mean score, μ, of all students taking the test.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 29.5

Population standard deviation =    = 5.2
Sample size = n =76

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* (5.2 / 76)

= 1.5365

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

- E < < + E

29.5- 1.5365< < 29.5+1.5365

27.9635< < 31.0365

(27.9635, 31.0365 )