TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2023W/HW08.1

One-sample t-test (without R)

In the context of a one-sample $t$-test let ${\bar {x}}=2.05$, $s^{2}=4$, $n=16$ and $H_{0}:\mu =1$. The values of the distribution function $F$ of the $t(n)$-distribution are (table below).

What is your conclusion for the following cases:

(a) at a right-sided test at level 5%

(b) at a right-sided test at level 1%?

(c) at a left left-sided test at level 5%?

(d) at a two-sided test at level 5%?

Briefly explain your choice. (Only use the upper information).

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

$t$	-2.300	-2.200	-2.100	-2.000	-1.900	-1.800
$F(t)$ (for $n=15$)	0.018	0.022	0.027	0.032	0.038	0.046
$F(t)$ (for $n=16$)	0.018	0.021	0.026	0.031	0.038	0.045

Lösungsvorschlag von Lessi

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

This means that the t-value is $t={\frac {{\bar {x}}-\mu _{0}}{s/{\sqrt {n}}}}=2.1$ and $T={\frac {{\bar {X}}-\mu _{0}}{s/{\sqrt {n}}}}{\stackrel {H_{0}}{\sim }}t(15)$

a) right-sided at level $\alpha =5\%$

$p=\mathbb {P} _{H_{0}}(T\geq t)=1-\mathbb {P} _{H_{0}}(T\leq 2.1)=1-F(2.1)=F(-2.1)=0.027$

which means that $p<\alpha =5\%\implies$ we reject $H_{0}$

b) right-sided at level $\alpha =1\%$

which means that $p>\alpha =1\%\implies$ we don't reject $H_{0}$

c) left-sided at level $\alpha =5\%$

$p=\mathbb {P} _{H_{0}}(T\leq t)=F(2.1)=1-F(-2.1)=0.973$

which means that $p>\alpha =5\%\implies$ we don't reject $H_{0}$

c) two-sided at level $\alpha =5\%$

$p=\mathbb {P} _{H_{0}}(|T|\geq |t|)=1-\mathbb {P} _{H_{0}}(-2.1\leq T\leq 2.1)=1-(F(2.1)-F(-2.1))=0.053$

which means that $p>\alpha =5\%\implies$ we don't reject $H_{0}$