Question

A simple random sample of size n equals 40 is drawn from a population. The sample mean is found to be x overbar equals 121.9 and the sample standard deviation is found to be s equals 12.9. Construct a​ 99% confidence interval for the population mean.

Lower bound:

Upper bound:

Answer 1

solution

Given that,

= 121.9

s =12.9

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 = t0.005, 39= 2.708 ( using student t table)

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

= 2.708* ( 12.9/ 40) = 5.5234

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

- E < < + E

121.9  - 5.5234 < < 121.9 + 5.5234

116.3766 < < 127.4234

(116.3766,127.4234 )

Lower bound:116.3766

Upper bound:127.4234