Question

Thirty six bearings made by a certain process have a mean diameter of 0.64 cm and standard deviation 0.05 cm.

Construct:

1. 99% confidence interval for the actual average diameter of bearings made by this process
2. 88% confidence interval for the actual average diameter of bearings made by this process.

Answer 1

Solution :

Given that,

i

t_{$\alpha$ /2,df} = 2.724

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

= 2.724 * (0.05 / $\sqrt$ 36)

Margin of error = E = 0.02

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

$\bar x$ - E 9 $\mu$ < $\bar x$ + E

0.64 - 0.02 < $\mu$ < 0.64 + 0.02

0.62 < $\mu$ < 0.66

(0.62 , 0.66)

ii.

t_{$\alpha$ /2,df} = 1.594

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

= 1.594 * (0.05 / $\sqrt$ 36)

Margin of error = E = 0.01

The 88% confidence interval estimate of the population mean is,

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

0.64 - 0.01 < $\mu$ < 0.64 + 0.01

0.63 < $\mu$ < 0.65

(0.63 , 0.65)