The chart shows the number of points Jerry scored on each test this week out of the number of points possible for the test.
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Answer:
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
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