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.