Consider the same statistical set-up as above. In particular, we have the test statistic
T_ n := n \sum _{j =0}^ K \frac{\left( \frac{N_ j}{n} - f_{\widehat{\theta }}(j) \right)^2}{ f_{\widehat{\theta }}(j) }.
where \widehat{\theta } is the MLE for the binomial statistical model (\{ 0,1, \ldots , K\} , \{ \text {Bin}(K, \theta ) \} _{\theta \in (0,1)}).
We define our test to be
\psi _ n = \mathbf{1}( T_ n > \tau ),
where \tau is a threshold that you will specify. For the remainder of this page, we will assume that K = 3 (the sample space is \{ 0,1,2,3\}).
What value of \tau should be chosen so that \psi _ n is a test of asymptotic level 5 \%? Give a numerical value with at least 3 decimals.
(Use this table or software to find the quantiles of a chi-squared distribution.)
\tau =\quad
To determine the value of τ that gives a test of asymptotic level 5%, we need to find the quantile of the chi-squared distribution for a given significance level.
Let's assume that the significance level is α = 0.05. We can use a chi-squared distribution table or software to find the 1 - α quantile.
From the given set-up, we have T_n as the test statistic, which follows a chi-squared distribution with degrees of freedom equal to the number of categories in the sample space minus the number of parameters estimated (in this case, K - 1).
Since K = 3, the degrees of freedom for T_n will be 3 - 1 = 2.
Using a chi-squared distribution table or software, we can find the 1 - α quantile for this distribution. The 1 - α quantile corresponds to the value of τ that gives a test of asymptotic level 5%.
The value of τ that should be chosen is equal to the 1 - α quantile.
Please consult a chi-squared distribution table or use statistical software to find the 1 - α quantile for a chi-squared distribution with 2 degrees of freedom.
In order to determine the value of \(\tau\) that ensures the test \(\psi_n\) has an asymptotic level of 5%, we need to find the quantile of a chi-squared distribution.
The test statistic \(T_n\) follows a chi-squared distribution with \(K\) degrees of freedom under the null hypothesis. In this case, \(K=3\).
Since we want the asymptotic level to be 5%, we need to find the cutoff point where the probability of observing a test statistic greater than \(\tau\) is 5%.
To find the required quantile, we can use a chi-squared distribution table or software. The quantile that corresponds to a 5% level of significance is denoted as \(\chi_{1-\alpha}^2\) with \(\alpha = 0.05\).
Using the table or software, we can look up the value of \(\chi_{1-\alpha}^2\) with \(K=3\) and \(\alpha=0.05\).