TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2023W/HW07.3

Test power in the z-test

Let $X_{1},\dots ,X_{25}$ be i.i.d. random variables with $X_{1}\sim {\mathcal {N}}(2,4),{\text{ and }}H_{0}:\mu =1$.

(a) Compute the test power of the left-sided z-test. Express it through the cumulative distribution function of the standard normal distribution, depending on the significance level $\alpha$.

(b) Comment on the impact of $\alpha$ on the test power.

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Lösungsvorschlag von Lessi

--Lessi 2024-02-07T13:04:11Z

Let $X_{1},\dots ,X_{2}5$ be i.i.d random variables with $X_{1}\sim {\mathcal {N}}(2,4)$, and $H_{0}:\mu =1$.

a) test power of the left-sided z-test.

Expressed as the cdf of a standard normal distribution depending on the significance level $\alpha$ In the context of a left sided z-test we define the alternative Hypothesis is $H_{A}:\mu <1$, $R=(-\infty ,q_{\alpha }]$, $p=\mathbb {P} _{H_{0}}(Z\leq z)$

Written informally the test power is the probability P(reject the hypothesis | the hypothesis is false). In this case we know that the hypothesis is false since we know the underlying distribution $X_{1}\sim {\mathcal {N}}(2,4)$ and therefore actually know $\mu =2$. Now the test power is $\Phi ({\frac {{\bar {X}}-\mu }{\sigma /{\sqrt {n}}}}\leq q_{\alpha })=\Phi ({\frac {{\bar {X}}-2}{2/5}}\leq q_{\alpha })$, so the probability that the standardized sample mean from any sample we draw is in the rejection region.

b) Comment on the impact of $\alpha$ on the test power.

Since this is a left-sided test and the actual $\mu$ is to the right of $\mu _{0}$, we generally would get a relatively small test-power. Since the probability depends on the quantile of $\alpha$, if the significance would be increased the test power would also increase.