Question

Thirty randomly selected students took the calculus final. If the sample mean score was 79 and the standard deviation was 7 , find the 99% confidence interval for the mean score of all students. Assume the population is normally distributed and round to two decimal places.

Answer 1

Solution :

Given that,

= 79

s = 7

n = 30

Degrees of freedom = df = n - 1 = 30 - 1 = 29

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

_{t /2  df} = t_0.005,29 = 2.756

Margin of error = E = t_/2,df * (s /n)   

= 2.756 * (7 / 30) = 3.52

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

- E < < + E

79 - 3.52 < < 79 + 3.52

75.48 < < 82.52

(75.48, 72.52 )