In order to answer this question, we need to calculate the test statistic Tn and compare it to the critical value C.
First, let's calculate the observed value of Tn. Based on the data set given, we can tabulate the frequencies for each outcome:
Outcome 1: 3 occurrences
Outcome 2: 1 occurrence
Outcome 3: 1 occurrence
Outcome 4: 2 occurrences
Outcome 5: 2 occurrences
Outcome 6: 6 occurrences
Next, we calculate Tn using the formula:
Tn = max(3, 1, 1, 2, 2, 6) = 6
Now, we need to determine the critical value C that gives the test a significance level of 5%. This can be done using computational tools or statistical tables. Let's assume that the critical value is C = 2.
Finally, we compare the observed value of Tn to the critical value C. Since Tn = 6 is greater than C = 2, we reject the null hypothesis that the die is fair.
Therefore, the answer is: Reject
We use the same statistical set-up as above. Recall that you use a test of the form
\psi _ n = \mathbf{1}\left( T_ n > C \right),
where C is a constant chosen in the previous problem so that \psi _ n has asymptotic level 5 \%. Suppose you observe that data set
5,6,1,6,4,1,2,4,6,6,1,6,6,3,5.
Do you reject or fail to reject the null hypothesis that the die is fair? (You are encouraged to use computational tools.)
Reject
Fail to reject
1 answer
1 answer