Question

What is the probability of making a type ii error when the machine is overfilling by .5 ounces (to 4 decimals)?

Answer 1

Part A

The probability of making a type ii error is equal to 1 minus the power of a hypothesis testing.

The power of a hypothesis test is given by:

$\beta(\mu')=\phi\left[z_{\alpha/2}+ \frac{\mu-\mu'}{\sigma/\sqrt{n}} \right]-\phi\left[-z_{\alpha/2}+ \frac{\mu-\mu'}{\sigma/\sqrt{n}} \right]$

Given that the machine is overfilling by .5 ounces, then $\mu-\mu'=-0.5$, also, we are given that the sample size is 30 and the population standard deviation is  = 0.8 and α = 0.05

Thus,

$\beta(16.5)=\phi\left[z_{0.025}+ \frac{-0.5}{0.8/\sqrt{30}} \right]-\phi\left[-z_{0.025}+ \frac{-0.5}{0.8/\sqrt{30}} \right] \\ \\ =\phi\left[1.96+ \frac{-0.5}{0.1461} \right]-\phi\left[-1.96+ \frac{-0.5}{0.1461} \right] \\ \\ =\phi(1.96-3.4233)-\phi(-1.96-3.4233) \\ \\ =\phi(-1.4633)-\phi(-5.3833)=0.07169$

Therefore, the probability of making a type II error when the machine is overfilling by .5 ounces is 1 - 0.07169 = 0.9283



Part B:

From part A, the power of the statistical test when the machine is overfilling by .5 ounces is 0.0717.