Question

Suppose scores on exams in statistics are normally distributed. A random sample of 28 scores is taken and gives a sample mean of 72 and sample standard deviation of 4. What is the 99% confidence interval for the true (population) mean of statistics exam scores?

Answer 1

)solution

Given that,

= 72

s =4

n = 28

Degrees of freedom = df = n - 1 =28 - 1 = 27

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2  df = t0.005,27 = 2.771 ( using student t table)

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

= 2.771 * ( 4/ 28) = 2.09

The 99% confidence interval is,

- E < < + E

72 - 2.09 < < 72+ 2.09

69.91 < < 74.09

(69.91 , 74.09)