Question

A simple random sample of size n = 40 is drawn from a population. The sample mean is found to be x= 120.9 and the sample standard deviation is found to be s = 12.1. Construct a 99% confidence interval for the population mean. The lower bound is (Round to two decimal places as needed.)

Answer 1

Solution :

Given that,

$\bar x$ = 120.9

s = 12.1

n = 40

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

At 99% confidence level the t is ,

$\alpha$ = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

t_{$\alpha$ /2,df} = t_0.005,39 =2.708

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 2.708 * (12.1 / $\sqrt$ 40)

= 5.18

Margin of error = 5.18

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

$\bar x$ - E <$\mu$  < $\bar x$ + E

120.9 - 5.18 < $\mu$ < 120.9 + 5.18

115.72 < $\mu$ < 126.08

(115.72, 126.08 )