Question

Company records show that the proportion of defective parts has historically been 5%, however a quality control engineer believes that due to budget cuts these numbers have increased. To test this claim a random sample of N = 15 parts will be taken and if more than two defective parts are found the inspector will conclude that the proportion of defective parts has increases. If the true probability of a defective part is in fact p = 10% what is the power of the test?

a. 3.6%

b. 82%

c. 96.4%

d. 18%

Answer 1

Let X denote the number of feffecti es found in 15 items.

Then X is a binomial variate with n=15 and p =0.05 under null hypothesis and p=0.1 under alternate hypothesis.

Critical region is then given as X>2

Power of test is defined as P( X falls in rejection region given alternate hypothesis is correct )= P(X>2| p=0.1)=1-P(X<=2|p=0.1)

This can be calculated using binomial probability as

1-[P(X=0)+P(X=1)+P(X=2)] given p=0.1

Where P(X=X)= 15Cx p​​​​​​x (1-p)15-x

We get power of test = 0.18=18% (answer)