$ 239$ subjects had their cholesterol measured, and then were put on high-fiber diets. After a month on the high-fiber diet, the cholesterol was measured again. The mean LDL cholesterol level before the experiment was $118.5 $ mg/DL. The mean level after the subjects were on a diet for a month was $113.7$ mg/DL. The sample standard deviation of the amount of cholesterol each patient lost was $ 38.37$ mg/DL. (a) Test whether or not the high-fiber data significantly reduced the LDL cholesterol level at an α = 0.05 level.
For this one I just used a paired to test. The hypothesis are $ H_0: \mu_d\leq 0$ and $ H_a: \mu_d>0$. We see that $\bar{d}=\bar{x}_{before}-\bar{x}_{after}=118.5-113.7=4.8$. It follows that $t_0=\dfrac{\bar{d}}{\dfrac{s_d}{\sqrt{n}}}=\frac{4.8}{\frac{38.37}{\sqrt{239}}}=1.933\implies 1.933>t_{0.05,238}=1.645$ Thus we reject the null hypothesis $ H_0$
(b) If the true reduction in cholesterol was 10 mg/DL, what is the power of this test?
We assume that $ \mu_d=10$. Since [math] n[/math] is large we can use a normal approximation. Thus it follows that $ \bar{d}=N(\mu_d,\dfrac{\sigma^2_{before}}{239}+\dfrac{\sigma^2_{after}}{239}$. Also we shall assume that $\dfrac{s_d}{\sqrt{239}}$ is approximately $\sqrt{\dfrac{\sigma^2_{before}}{239}+\dfrac{\sigma^2_{after}}{239}}$. Thus Power$$=P(t_0\in R|\mu_d=10)=P(\dfrac{\bar{d}}{\dfrac{38.37}{\sqrt{239}}}>1.645|\mu_d=10)= P(\dfrac{\bar{d}}{\dfrac{38.37}{\sqrt{239}}}-\dfrac{10}{\dfrac{38.37}{\sqrt{239}}}>1.645-{\dfrac{38.37}{\sqrt{239}}}) =P(\dfrac{\bar{d}-10}{\dfrac{38.37}{\sqrt{239}}}>1.62814)=P(Z>1.62814)]= 1-P(Z\leq 1.62814 )=1-.94825=.05175$$.
This is what I get for part b. I think it looks right but im not entirely sure. I'm basing the rejection region off of part (a) that is reject the null hypothesis as long as the T statistic is greater that $ T_{0.05,238}$.