Question

A simple random sample of size

n equals 40n=40

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

x overbar equals 121.1x=121.1

and the sample standard deviation is found to be

s equals 12.7s=12.7.

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

The lower bound is..

the upper bound is..

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 121.1

sample standard deviation = s = 12.7

sample size = 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 * (12.7 / 40)

= 5.4

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

- E < < + E

121.1 - 5.4 < < 121.1 + 5.4

115.7 < < 126.5

Lower bound = 115.7

Upper bound = 126.5