A simple random sample of 80 widgets is drawn from a manufacturing lot of 1900 w

Question

A simple random sample of 80 widgets is drawn from a manufacturing lot of 1900 widgets. The widgets in the sample are run until they fail. 30 of the widgets in the sample last more than 3,000 hours.

An unbiased estimate of the fraction of widgets in the manufacturing lot that will last more than 3,000 hours is? 0.375
The finite population correction for the standard error of the sample percentage is? 0.9789786297?

The largest the standard error of the sample percentage could be is ________
The bootstrap estimate of the standard error of the sample percentage is?______

Explanation / Answer

Solution:- Given that sample n = 80 , x = 30

a) p = 30/80 = 0.375

b) sqrt((N - n)/(N-1)) = sqrt((1900 - 80)/(1900-1) ) = 0.9789786

c) sqrt(p*q/n) * sqrt((N-n)/(N-1)) = sqrt((0.375*0.625/80)*sqrt((1900 - 80)/(1900-1) ) = 0.05355465

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