Question

A sample of 8 items has an average fat content of 18.6 grams and a standard deviation of 2.4 grams. Assuming a normal distribution, construct a 99 percent confidence interval for μ.

Answer 1

Solution :

sample size = n = 8

Degrees of freedom = df = n - 1 = 7

t_{$\alpha$ /2,df} = 3.499

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

= 3.499 * (2.4 / $\sqrt$ 8)

Margin of error = E = 3.0

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

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

18.6 - 3.0 < $\mu$ < 18.6+ 3.0

15.6 < $\mu$ < 21.6

(15.6 , 21.6)