Question

The article "Study on the Life Distribution of Microdrills” (Z. Yang, Y. Chen, and Y. Yang, Journal of Engineering Manufacture, 2002:301-305) reports that in a sample of 50 microdrills drilling a low-carbon alloy steel, the average lifetime (expressed as the number of holes drilled before failure) was 12.68 with a standard deviation of 6.83. Find a 95% confidence interval for the mean lifetime of microdrills under these conditions.

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 12.68

Population standard deviation =   $\sigma$ = 6.83

Sample size n =50

At 95% confidence level the z is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z$\alpha$/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E = Z$\alpha$/2 * ( $\sigma$ /$\sqrt$n)

= 1.96* (6.83 /  $\sqrt$50 )

E= 1.8932
At 95% confidence interval estimate of the population mean
is,

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

12.68- 1.8932 <  $\mu$ < 12.68 +1.8932

10.7868 <  $\mu$ < 14.5732

( 10.7868 , 14.5732 )