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The exercise is:

Assume that the population mean is actually 110 grams and that the distribution is normal with standard deviation 4 grams. In a z test of $H_0$: u = 113 against $H_a$: u < 113 with $\alpha = 0.05$, find the probability of rejecting $H_0$ with six observations.

From part (a) in that exercise i found out:

$$\hat x = 112.967 - s = 4.28 - n = 6- t,0.05,5 = -2.015$$

My approach is: Find the type II error and do $$ 1 - \beta(110) $$

so:

$$P\left( \frac{\hat x - 113}{\frac{4.28}{\sqrt{6}}} > -2.015 | u = 110, \sigma=4\right) = \alpha $$ $$1 - P\left(\hat x < -2.015\cdot{\frac{4.28}{\sqrt{6}}} + 113\right) $$ $$1 - P\left(\frac{\hat x - 110}{\frac{4}{\sqrt{6}}} < \frac{-2.015\cdot{\frac{4.28}{\sqrt{6}}} + 3)}{\frac{4}{\sqrt{6}}}\right) $$

is:

$$1-P\left(\frac{\hat x - 110}{\frac{4}{\sqrt{6}}} < -0.32\right)$$

which gives (1-(1-0.3745)) = 0.3745, while the solution is 0.58.

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1 Answer

up vote 0 down vote accepted

The sample mean doesn't enter into this calculation at all. You're answering a question about the probability that (under repeated sampling) you'd reject the null when you're drawing from a population with mean 110 while the hypothesized mean is 113.

Further note that the question specifies a z test, not a t-test - and specifies $\sigma$ - so your critical value is wrong. You need to pay careful attention to the question.

(By comparison, your sample either leads to rejection or it doesn't.)

Here's a calculation (done in the statistical package R) you may find of some value, by considering where the inputs come from, and what they mean:

> pnorm(-1.645-(110-113)/(4/sqrt(6)))
[1] 0.5761748

Note that used with a single argument, pnorm is just the cdf of the standard normal.

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