Question

In: Statistics and Probability

1. A sample of 48 aircraft “C” checks generated an average of 238 write-ups (squawks), with...

1. A sample of 48 aircraft “C” checks generated an average of 238 write-ups (squawks), with a standard deviation of 39. 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 95% degree of confidence?

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 238

Population standard deviation = = 39

Sample size = n =48

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

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

= 1.96 * (39 / $\sqrt$ 48 )

= 11.0332
At 95% confidence interval estimate of the population mean
is,

- E < < + E

238 -11.0332 < < 238+ 11.0332

226.9668 < < 249.0332

( 226.9668 , 249.0332 )

orchestra answered 1 year ago

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