Question

A machine fills boxes of cereal in a factory. The average weight of cereal in a random sample of 17 boxes is calculated to be 350 grams and the sample standard deviation is calculated to be 8 grams. Weights of cereal per box are known to follow a normal distribution. We calculate a 90% confidence interval for the true mean weight of cereal per box. The margin of error for the appropriate confidence interval is:

Answer 1

Solution :

Given that,

$\bar x$ = 350

s =8

n =

Degrees of freedom = df = n - 1 =17 - 1 = 16

a )

At 90% confidence level the t is ,

$\alpha$ = 1 - 90% = 1 - 0.90 = 0.1

$\alpha$ / 2 = 0.1 / 2 = 0.05

t$\alpha$ /2,df = t0.05,16= 1.746 ( using student t table)

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

= 1.746 * (8 / $\sqrt$ 17)

Margin of error = E= 3.3877

The 90% confidence interval estimate of the population mean is,

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

350- 3.3877 < $\mu$ < 350+ 3.3877

346.6123 < $\mu$ < 353.3877

(346.6123 , 353.3877 )