Question

A researcher drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 209 times. The observed frequencies for the outcomes of​ 1, 2,​ 3, 4,​ 5, and 6 respectively are 36​, 30​, 41​, 40​, 23​, and 39. Use a 0.05 significance level to test the claim that the outcomes are not equally likely. Does it appear that the loaded die behaves differently than a fair​ die?

Answer 1

Explanation:

Given that the observed frequencies for the outcomes as follows:

To check this we can use chi square goodness of fit test.

`H_0: equally likely\\H_a: not equally likely`

(Two tailed test at 5% significance level)

Assuming equally likely expected observations are found out and then chi square is calculated as (0-E)^2/E

Df = 6-1 =5

Outcome Frequency Expected frequency (Obs-exp)^2/Exp

1 36 34.83333333 0.03907496

2 30 34.83333333 0.670653907

3 41 34.83333333 1.091706539

4 40 34.83333333 0.766347687

5 23 34.83333333 4.019936204

6 39 34.83333333 0.498405104

209 209 7.086124402

p value =0.214

Since p >alpha, we accept null hypothesis

It appears that the loaded die does not behave differently than a fair​ die at 5% level of significance