Aresearcher 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?

Step-by-step explanation:

Given that the observed frequencies for the outcomes as follows:

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

(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

OutcomeFrequencyExpected frequency(Obs-exp)^2/Exp

13634.833333330.03907496

23034.833333330.670653907

34134.833333331.091706539

44034.833333330.766347687

52334.833333334.019936204

63934.833333330.498405104

209209          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

the correct answer to your question is a. 4^4

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16.80 times .20= 3.36

step-by-step explanation: