If 5 for 25.59 how much for 9
1 answer:
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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]](https://tex.z-dn.net/?f=%5Cbeta%28%5Cmu%27%29%3D%5Cphi%5Cleft%5Bz_%7B%5Calpha%2F2%7D%2B%20%5Cfrac%7B%5Cmu-%5Cmu%27%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%20%5Cright%5D-%5Cphi%5Cleft%5B-z_%7B%5Calpha%2F2%7D%2B%20%5Cfrac%7B%5Cmu-%5Cmu%27%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%20%5Cright%5D)
Given that the machine is overfilling by .5 ounces, then
, 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](https://tex.z-dn.net/?f=%5Cbeta%2816.5%29%3D%5Cphi%5Cleft%5Bz_%7B0.025%7D%2B%20%5Cfrac%7B-0.5%7D%7B0.8%2F%5Csqrt%7B30%7D%7D%20%5Cright%5D-%5Cphi%5Cleft%5B-z_%7B0.025%7D%2B%20%5Cfrac%7B-0.5%7D%7B0.8%2F%5Csqrt%7B30%7D%7D%20%5Cright%5D%20%5C%5C%20%20%5C%5C%20%3D%5Cphi%5Cleft%5B1.96%2B%20%5Cfrac%7B-0.5%7D%7B0.1461%7D%20%5Cright%5D-%5Cphi%5Cleft%5B-1.96%2B%20%5Cfrac%7B-0.5%7D%7B0.1461%7D%20%5Cright%5D%20%5C%5C%20%20%5C%5C%20%3D%5Cphi%281.96-3.4233%29-%5Cphi%28-1.96-3.4233%29%20%5C%5C%20%20%5C%5C%20%3D%5Cphi%28-1.4633%29-%5Cphi%28-5.3833%29%3D0.07169)
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.
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:
Given that the machine is overfilling by .5 ounces, then
Thus,
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.
It looks like your first selection (a) is appropriate.