TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2023W/HW07.1

z-test (without R)

In the context of a two-sided $z$-test let ${\bar {x}}={\frac {4}{3}}$, $\sigma =1$, $n=9$ and $H_{0}:\mu ={\frac {2}{3}}$.

The $\alpha$-quantiles $q_{\alpha }$ of ${\mathcal {N}}(0,1)$ are as follows (table below).

How do you decide for a significance level of 1%, (b) 3%, (c) 6% and (d) 10% by only using the information above? Justify your answers.

Dieses Beispiel ist als solved markiert. Ist dies falsch oder ungenau? Aktualisiere den Lösungsstatus (Details: Vorlage:Beispiel)

$\alpha$	0.005	0.010	0.015	0.020	0.025	0.030	0.035	0.040
$q_{\alpha }$	-2.576	-2.326	-2.170	-2.054	-1.960	-1.881	-1.812	-1.751

Lösungsvorschlag von Simplex

$z$-statistic: $z={\frac {{\bar {x}}-\mu _{0}}{\sigma /{\sqrt {n}}}}={\frac {{\frac {4}{3}}-{\frac {2}{3}}}{1/{\sqrt {9}}}}={\frac {2/3}{1/3}}=2$

(a)

$\alpha =0.01$: $R=\overbrace {(-\infty ,-2.576]} ^{\alpha /2}\cup \overbrace {[2.576,\infty )} ^{1-\alpha /2}\Rightarrow z\not \in R\rightarrow$ no rejection of $H_{0}$

(b)

$\alpha =0.03$: $R=(-\infty ,-2.170]\cup [2.170,\infty )\Rightarrow z\not \in R\rightarrow$ no rejection of $H_{0}$

(c)

$\alpha =0.06$: $R=(-\infty ,-1.881]\cup [1.881,\infty )\Rightarrow z\in R\rightarrow$ rejection of $H_{0}$

(d)

$\alpha =0.10$: also rejected because of (c) being rejected too

--Simplex 19:20, 3. Feb. 2023 (CET)