We use the same statistical set-up as above. In particular, our test-statistic is
T_ n = n \sum _{j = 1}^6 \frac{(\widehat{p }_ j - \frac{1}{6} )^2}{\frac{1}{6} }.
You use a test of the form
\psi _ n = \mathbf{1}\left( T_ n > C \right).
What value of C should be chosen so that \psi is a test of asymptotic level 5 \%? Give a numerical value with at least 2 decimals.
Use this table to find the quantiles of a chi-squared distribution.
To find the appropriate value of C for a test of asymptotic level 5%, we need to determine the critical value from the chi-squared distribution.
The test statistic T_n follows a chi-squared distribution with 6 degrees of freedom under the null hypothesis.
Using the chi-squared distribution table, we need to find the value of C such that the probability of observing a test statistic greater than C is equal to 5%.
Looking at the table, the closest probability value we can find is 0.05, which corresponds to the critical value of 12.59.
Therefore, the value of C should be chosen as 12.59 for the test to have an asymptotic level of 5%.
To find the value of C that should be chosen so that the test \(\psi\) has an asymptotic level of 5%, we need to find the 5% quantile of the chi-squared distribution with 5 degrees of freedom.
Here is the table of critical values for the chi-squared distribution with different probabilities (quantiles) and degrees of freedom:
```
Quantile df=5
0.500 0.28
0.900 7.81
0.950 11.07
0.975 15.09
0.990 20.52
```
From the table, we can see that the 5% quantile (0.05 probability) for 5 degrees of freedom is approximately 11.07.
Therefore, the value of C that should be chosen for the test \(\psi\) to have an asymptotic level of 5% is approximately 11.07.