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The modulus of rupture (MOR) for a particular grade of pencil lead is known to have a standard deviation of 250 psi. Process standards call for a target value of 6500 psi for the true mean MOR. For each batch, an inspector tests a random sample of 16 leads. Management wishes to detect any change in the true mean MOR. (Assume normal distribution.)

QUESTION : Find the probability of type II error of the test when the true mean MOR is 6400.

How do we solve for two tails? i am able to get 0.4821 but with 1 tail method. 6397-6400/62.5 = 0.045 than using normcdf on MATLAB, i got this value

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You have not said what significance level you are using, but it seems that what you did for one tail used $5\%$ and

  • standard error of the mean $\dfrac{250}{\sqrt{16}} = 62.5$
  • If the true population mean is $6500$ then there is a $5\%$ probability that the the sample mean will be below $6500 + 62.5 \Phi^{-1}(0.05) \approx 6397$
  • If the true population mean is $6400$ then there is about a $\Phi\left(\dfrac{6397-6400}{62.5}\right) \approx 48.21\%$ probability that the the sample mean will be below $6397$

You need to do something similar for two tails:

  • If the true population mean is $6500$ then there is a $2.5\%$ probability that the the sample mean will be below $6500 + 62.5 \Phi^{-1}(0.025) \approx 6377.5$
  • If the true population mean is $6500$ then there is a $2.5\%$ probability that the the sample mean will be above $6500 + 62.5 \Phi^{-1}(0.975) \approx 6622.5$
  • If the true population mean is $6400$ then there is about a $\Phi\left(\dfrac{6377.5-6400}{62.5}\right) \approx 35.94\%$ probability that the the sample mean will be below $6377.5$
  • If the true population mean is $6400$ then there is about a $1-\Phi\left(\dfrac{6622.5-6400}{62.5}\right) \approx 0.02\%$ probability that the the sample mean will be above $6622.5$
  • So the probability of a type II error of the test when the true mean is $6400$ is about $35.96\%$
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