Question

Homework > Homework 6.3 You are conducting a Goodness of Fit hypothesis test for the claim that all 5 categories are equally likely to be selected. Complete the table. Report all answers correct to three decimal places. Category Observed Expected (obs-exp)2/exp Frequency Frequency A 21 B 15 c 10 D 23 E 20 What is the chi-square test statistic for this data? x At the alpha 0.05 level, what is the conclusion for this test? Fail to reject the null hypothesis Reject the null hypothesis Report all answers accurate to three decimal places. Question Help: Video Video Message instructor Submit Question MacBook F esc 80 F3 DOO 000 F4 FI F2 F5 FO @ # $ % A 1 2 3 4 5 6

Answer 1

We have to perform Chi-square test for goodness of fit.

We have to test for null hypothesis $\tiny H_0:\text{All 5 categories are equally likely to be selected.}$

against the alternative hypothesis

H1 : All 5 categories are not equally likely to be selected

Our Chi-square test statistic is given by

x2 = Σ (fi - e;)2 i=1 ei

Here,

Number of categories

n = 5

Necessary calculations are as follows.

(f-e)^2/e Category Observed Expected frequency (f) frequency (e) A 21 17.80 B 15 17.80 C 10 17.80 D 23 17.80 E 20 17.80 Total 89 89.00 0.5753 0.4404 3.4180 1.5191 0.2719 6.2247

$\tiny \therefore \chi^2_{calculated}=\sum_{i=1}^{5}\frac{\left(f_{i}-e_{i}\right)^2}{e_{i}}=6.2247$

Degrees of freedom $\tiny v=n-1=4$

$\tiny \therefore \text{p-value}=P\left(\chi^2_{4}>6.2247\right)=0.1829842$ [Using R-code '1-pchisq(6.2247,4)']

Level of significance $\tiny \alpha=0.05$

We reject our null hypothesis if $\tiny \text{p-value}<\alpha$

Here, we observe that $\tiny \text{p-value}=0.1829842\nless0.05=\alpha$

Hence, we fail to reject the null hypothesis.