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
To determine whether to reject or fail to reject the null hypothesis that the die is fair, we can follow these steps using computational tools:
Step 1: Calculate the test statistic T_n for the observed data set.
- The test statistic T_n is the sum of the observed dice rolls.
- T_n = 5 + 6 + 1 + 6 + 4 + 1 + 2 + 4 + 6 + 6 + 1 + 6 + 6 + 3 + 5 = 68.
Step 2: Choose the constant C so that the test has an asymptotic level of 5%.
- The asymptotic level refers to the probability of rejecting the null hypothesis when it is true, also known as a Type I error.
- Since the asymptotic level is 5%, we need to choose C such that the probability of T_n being greater than C when the die is fair is 5%.
- We can use a statistical table or computational tools to find the appropriate value of C.
Step 3: Compare the test statistic T_n with the chosen constant C.
- If T_n is greater than C, we reject the null hypothesis.
- If T_n is less than or equal to C, we fail to reject the null hypothesis.
Based on the provided data set and the calculated test statistic, we cannot determine whether to reject or fail to reject the null hypothesis without knowing the chosen constant C.
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