Question

A simple random sample of size

n equals n=40

is drawn from a population. The sample mean is found to be

x overbar equals 120.3

and the sample standard deviation is found to be

s equals 12.3

Construct a​ 99% confidence interval for the population mean.

Answer 1

Solution :

Given that,

= 120.3

s = 12.3

n = 40

Degrees of freedom = df = n - 1 = 40 - 1 = 39

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

_{t /2  df} = t_0.005,39 = 2.708

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

= 2.708 * (12.3 /40 ) = 5.2665

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

- E < < + E

120.3 - 5.2665 < < 120.3 + 5.2665

115.0337 < < 125.5665

(115.0337, 125.5665 )