# Test hypothesis, probability of rejecting $H_0$

A mechanic wants to test how difficult it is to unscrew the bolts that hold the tires in place in a car. Assume that out of $$8$$ cars tested, the average force was $$259.634$$ newtons. Also assume that the force has a standard deviation $$\sigma = 36$$

Test the hypothesis $$H_0: \mu = 310$$ vs $$H_1: \mu \neq 310$$ at significance level $$0.01$$. If the real value of $$\mu = 250$$, what is the probability of rejecting $$H_0$$?

As an estimator for $$\mu$$, I use $$\hat \mu = \bar X = 259.6$$. Assuming $$H_0$$ correct I get $$Z = \frac{\bar X-\mu}{\sigma/\sqrt{n}}$$. Using $$\bar X = 259.6$$ and $$\mu = 310$$ I get $$Z = -3.96$$.

I want to reject $$H_0$$ if $$Z or $$Z > z_{upper}$$. Since we have a significance level of $$1\%$$, and I assume a two-tailed test (?), from the table I find $$\alpha = 0.01/2 = 0.005$$ and thus $$z_{\alpha} = 2.576$$

Because of symmetry in the normal distribution, $$z_{lower} = -2.576$$ and since $$-3.96 < -2.576$$ we reject $$H_0$$.

As for the second part of the question, I'm not sure what to do.

I tried something like $$\frac{\bar X-310}{36/\sqrt{8}} = 2.326$$

$$\bar X = 339.6$$

$$\frac{339.6-250}{36/\sqrt{8}} = 7.04$$ which is obviously incorrect as the $$z$$ values in standard normal distribution don't go higher than approx. $$3.7$$...

The answer is supposed to be $$0.9838$$ if that is of interest.

You will reject $$H_0$$ if the sample mean $$\overline{X}$$ lies below $$\mu-z\sigma/\sqrt{n}$$ or above $$\mu+z\sigma/\sqrt{n}$$; here $$\sigma=36$$, $$n=8$$, $$\mu=310$$, $$z=2.57583$$ as you obtained above. Call these $$X_L$$ and $$X_U$$, which I obtain as 277.215 and 342.785 resp. Now, if the true mean is 250, $$\overline{X}$$ is Normal with mean 250 and stdev $$36/\sqrt{8}$$. So the probability of it being below $$X_L$$ is $$\Phi\left(\frac{X_L-\mu}{\sigma/\sqrt{n}}\right)$$, and above $$X_U$$ is $$\Phi\left(\frac{\mu-X_U}{\sigma/\sqrt{n}}\right)$$; the second of these is negligible, and the first gives you the desired 0.9838.