Question

If a sample of size n = 16 is chosen from a normal population, which one of the following is a 99% confidence interval for the population mean when the sample mean is 30 and the sample standard deviation is 2.4?

A 30 ± 2.602(2.4)

B 30 ± 2.583(2.4)

C 30 ± 2.326(2.4)

D 30 ± 2.602(0.6)

E 30 ± 2.947(0.6)

Answer 1

Given that,

= 30

s =2.4

n = 16

Degrees of freedom = df = n - 1 = 16- 1 = 15

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

_{t /2  df} = t_0.005,15= 2.947

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

= 2.947 * (2.4 / 16) = 2.947(0.6)

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

-+ E

30+- 2.947(0.6)

CORRECT OPTION E