TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2023W/HW10.3

One-sample test for proportions (without R)

In the context of the one-sample situation for proportions let the observed relative frequency be ${\textstyle h={\frac {1}{2}}}$. Let the null hypothesis be ${\textstyle H_{0}:p=0.4}$ and further let the (approximate) test be two-sided. Answer the following questions only using the table below, which shows the ${\textstyle \alpha }$-quantiles ${\textstyle q_{\alpha }}$ of the ${\textstyle N(2,1)}$-distribution.

a) What is the value of the question mark in the table?

b) For ${\textstyle n=49}$ the null hypothesis is rejected on the 10%-level?

c) For ${\textstyle n=144}$ the null hypothesis is rejected on the 3%-level?

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

${\textstyle \alpha }$	0.01	0.05	0.1	0.2	0.5
${\textstyle q_{\alpha }}$	-0.32	0.36	0.72	1.16	?

Lösungsvorschlag von Friday

--Friday Sa 30 Jan 2021 18:18:43 CET

a)

The ${\textstyle N(2,1)}$ distribution has a mean of 2 and is symetric. Therefore the ${\textstyle \alpha }$-quantile for ${\textstyle 0.5}$ must be ${\textstyle \mathbf {2} }$.

b)

While the table in the description certently is useful, it is not exactly what we need. First, there are the quantiles for the ${\textstyle N(2,1)}$-distribution while we need the ${\textstyle N(0,1)}$-distribution. Moreover we need the ${\textstyle q_{\alpha /2}}$ quantiles.

The adapted table for the ${\textstyle N(0,1)}$-distribution looks like:

${\textstyle \alpha }$	0.02	0.1	0.2	0.4	1
${\textstyle q_{\alpha /2}}$	-2.32	-1.64	-1.28	-0.84	0

Now we can test the hypothesis:

$${\begin{aligned}se_{h}&={\sqrt {\frac {h(1-h)}{n}}}={\sqrt {\frac {0.5(1-0.5)}{49}}}={\sqrt {\frac {0.25}{49}}}\approx 0.0714\\z&={\frac {h-p_{0}}{se_{h}}}={\frac {0.5-0.4}{0.0714}}\approx 1.4000\\R&=(-\infty ,-1.64]\cup [1.64,\infty )\end{aligned}}$$

As ${\textstyle z\not \in R}$, we do not reject ${\textstyle H_{0}}$ on the 10%-level.

c)

$${\begin{aligned}se_{h}&={\sqrt {\frac {h(1-h)}{n}}}={\sqrt {\frac {0.5(1-0.5)}{144}}}={\sqrt {\frac {0.25}{144}}}\approx 0.0417\\z&={\frac {h-p_{0}}{se_{h}}}={\frac {0.5-0.4}{0.0417}}\approx 2.4\end{aligned}}$$

Although the table hasn't got the exact values for the 3% level, we can say that z is in the rejection region for the 2% level (-.32-2 = -2.32) so the H0 would be already rejected at this level and therfore would also be rejected at the 3% level (edited by M.T.)