2. The same sample of 48 aircraft revealed that the average time that the aircraft was...

2. The same sample of 48 aircraft revealed that the average time that the aircraft was down for “C” check was 27 days with a standard deviation of 5 days. The VP of operations wants to know how long the next aircraft due inspection will be out of service. Can you give him an estimate based on a 90% degree of confidence?

Solutions

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 27

Population standard deviation = = 5

Sample size = n =48

At 90% confidence level

= 1 - 90%

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

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

= 1.645 * ( 5 / $\sqrt$ 48 )

= 1.1872
At 90% confidence interval estimate of the population mean
is,

- E < < + E

27- 1.1872 <   <27 + 1.1872

25.8128<   < 28.1872

(  25.8128 ,  28.1872)

orchestra answered 1 year ago

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