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