Question

The upper bound is ___

A simple random sample of size n = 40 is drawn from a population. The sample mean is found to be x = 120.2 and the sample standard deviation is found to be s = 12.7. 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,

$\bar x$ = 120.2

s =12.7

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

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

= 2.708* ( 12.7/ $\sqrt$ 40) = 5.44

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

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

120.2 - 5.44 < $\mu$ <120.2 + 5.44

114.76 < $\mu$ < 125.64

(lower bound= 114.76 , upper bound=125.64)