Question

Assume that a sample is used to estimate a population mean μ . Find the 99.9% confidence interval for a sample of size 52 with a mean of 80.8 and a standard deviation of 17.2. Enter your answer as an open-interval (i.e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

99.9% C.I. =

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 80.8

sample standard deviation = s = 17.2

sample size = n = 52

Degrees of freedom = df = n - 1 = 51

At 99.9% confidence level the t is ,

$\alpha$ = 1 - 999% = 1 - 0.999 = 0.001

$\alpha$ / 2 = 0.001 / 2 = 0.0005

t$\alpha$ /2,df = t0.0005,24 = 3.492

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

= 3.492* (17.2 / $\sqrt$ 52)

= 8.3

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

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

80.8 - 8.3 < $\mu$ < 80.8 + 8.3

72.5 < $\mu$ < 89.1

99.9% C.I. = (72.5 , 89.1)