Question

If n=20, ¯ x (x-bar)=36, and s=13, construct a confidence interval at a 99% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place.

Answer 1

Given that,

= 36

s =13

n = 20

Degrees of freedom = df = n - 1 =20 - 1 = 19

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, 19=2.861 ( using student t table)

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

=2.861 * ( 13/ 20)

E= 8.3

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

- E < < + E

36 -8.3 < <36 + 8.3

27.7 < < 44.3

(27.7 , 44.3 )

The 99% confidence interval estimate of the population mean is,(27.7 , 44.3 )