If is $\bar{X}$ the mean of a random sample of size $n$ from a normal distribution with mean $\mu$ and unknown standard deviation, then I know $\frac{\bar{X}-\mu}{S\sqrt{N}}$ has a T distribution with n-1 degrees of freedom.

According to my book, it is very difficult to calculate the probability of a type II error for a give $\mu_1$, and so we must use these graphs to estimate $\beta$. I don't see why we cannot follow the same procedure we did with a large sample, or Normal with known-variance.

For instance, suppose $H_0$ is $\mu \geq \mu_0$, and $H_a = \mu < \mu_0$, and I want $\alpha=0.5$.

I first find $C$ so that $P(\bar{X} < C | \mu \geq \mu_0) = \alpha$

Then when I want to calculate $\beta$ for say $\mu_1$, I just use the same exact $C$. I calculate $P(\bar{X} > C | \mu = \mu_1)$

Isn't it the case that $\frac{\bar{X}-\mu_1}{S\sqrt(n)}$ is a T distribution with n-1 df too? It seems then I just need to find $\frac{C-\mu_1}{S\sqrt(n)}$ in my t-Table if for n-1 df to find $\beta$, just like I did in the other cases (normal known variance, or large sample.) (Though admittedly, that entry may be missing...but if I had a more complete table, would this work?)

  • $\begingroup$ I suggest you reduce your question from a long chat ... down to whatever it is you seek ... stated in 3 or 4 lines. $\endgroup$ – wolfies Nov 25 '14 at 15:14
  • $\begingroup$ What you say is correct. Indeed $\frac{\bar{X}-\mu_1}{S\sqrt(n)}$ has a t-distribution with $n-1$ df. So, if for example you use Excel (for the $t$-distribution percentiles) then yes, it would work. Otherwise, what you say is perfectly ok, you proceed for the calculation of Type II exactly as when you have normal distribution. Now, perhaps your book does not want to confuse you, or wants to introduce the power curve, or is too introductory and prefers to say that this is computationally too hard. Otherwise, you have it correctly according to my understanding. $\endgroup$ – Jimmy R. Nov 26 '14 at 23:00

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