Question

A simple random sample of size n=40 is drawn from a population. The sample mean is found to be x overbar equals 120.7 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.) The upper bound is ________ (Round to two decimal places as needed.)

Answer 1

Solution :

Given that,

$\bar x$ = 120.7

s =12.1

n =40

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

At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

  $\alpha$/ 2$\alpha$ = 0.05 / 2 = 0.025

t_{$\alpha$ /2,df} = t_0.025,39 = 2.708

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

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

= 5.18

The 95% confidence interval is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

120.7-5.18- < $\mu$ < 120.7+ 5.18

lower bound 115.52

upper bound 125.88