Question

11. A sample of 62 aircraft “C” checks generated an average of 462 write-ups (squawks), with a standard deviation of 57. The VP of operations wants to know how many squawks to expect on the next “C” check performed. Can you give him an estimate based on a 90% degree of confidence?

Answer 1

solution:

Given that,

Point estimate = sample mean = $\bar x$ = 462

Population standard deviation =   $\sigma$ = 57

Sample size = n =62

At 95% confidence level the z is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.1

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

Z_$\alpha$/2 = Z_0.05 = 1.645

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

= 1.645* (57 /  $\sqrt$62 )

= 11.9081
At 90% confidence interval estimate of the population mean
is,

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

462 -11.9081 <  $\mu$ < 462+ 11.9081

450.0918 <  $\mu$ < 473.9081

(450.0918 , 473.9081)