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 $Z<z_{lower}$ 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.


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