TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Bura)/Übungen 2020W/HW13.5

Cigarettes - yummy (part 1)

100 Viennese students were asked of whether they smoke. The possible answers were (1) ’yes, I do’ (2) ’no, I stopped smoking’, or (3) ’no, I never smoked’. The survey resulted in 22 smokers and 55 that never smoked. Everybody answered.

Test the null hypothesis that the proportion of Viennese students that never smoked is double of those that do smoke and also double of those that stopped smoking. Choose a 1%-significance level. Formulate a result.

Lösungsvorschlag von Friday

--Friday Sa 30 Jan 2021 18:43:49 CET

First, lets define the probabilities for the hypothesis:

$$H_{0}:p=p_{0}:=(0.25,0.25,0.5)$$

Now, we can create a table for the occupation and the expected values under the null hypothesis:

${\textstyle k}$	1	2	3	Σ
${\textstyle x_{k}}$	22	23	55	100
${\textstyle \mathbb {E} _{H_{0}}[X_{k}]}$	25	25	50	100

Perform the ${\textstyle \chi ^{2}}$-test, goodness of fit:

$${\begin{aligned}x&=\sum _{k=1}^{d}{\frac {(x_{k}-\mathbb {E} _{H_{0}}[X_{k}])^{2}}{\mathbb {E} _{H_{0}}[X_{k}]}}\\&={\frac {(22-25)^{2}}{25}}+{\frac {(23-25)^{2}}{25}}+{\frac {(55-50)^{2}}{50}}\\&={\frac {18}{50}}+{\frac {8}{50}}+{\frac {25}{50}}={\frac {51}{50}}=1.02\end{aligned}}$$

With R we can calulate qchisq(0.99,2):

$$q_{1-\alpha }\approx 9.21$$

We cannot reject the null hypothesis as ${\textstyle x^{2}=1.02\not \in R}$ with ${\textstyle R=[9.21,\infty )}$.