To determine if we reject H_0 at asymptotic level 5%, we need to compare the test statistic T_n^Wald to the critical value C.
From the information given, we have n=100, 40 X_i's are negative, and 60 X_i's are non-negative. We can use these values to calculate the test statistic T_n^Wald.
First, we need to find the MLE of theta, hattheta. Since the X_i's are negative when -1 ≤ X < 0, the MLE of theta is the proportion of negative X_i's, which is 40/100 = 0.4.
Next, we can substitute the values into the test statistic formula:
T_n^Wald = ((4*n)/(1-(hattheta)^2))*(hattheta-sqrt(hattheta))^2
T_n^Wald = ((4*100)/(1-(0.4)^2))*(0.4-sqrt(0.4))^2
T_n^Wald = 51.2
Now we compare T_n^Wald to the critical value C. The critical value is determined by the significance level, which in this case is 5%.
Since T_n^Wald = 51.2 > C = 3.84, we reject H_0 at the asymptotic level of 5%.
Therefore, the answer is: We reject H_0.
Recall from the setup that X_1,\ldots ,X_ n\sim X are i.i.d. random variables with density f_\theta, for some unknown \theta \in (0,1):
f_\theta (x)=\left\{ \begin{array}{ll} \theta ^2& \text {if } -1 \le x<0\\ 1- \theta ^2 & \text {if } 0\leq x\leq 1\\ 0 & \text {otherwise.} \end{array}\right.
Consider the following hypotheses:
\displaystyle \displaystyle H_0 \displaystyle : X \sim \textsf{Unif}(-1,1)
\displaystyle H_1 \displaystyle : X \text { not distributed as } \textsf{Unif}(-1,1).
Write down the test statistic \, T_ n^{\text {Wald}}\, (that is quadratic in \hat\theta ^{\text {MLE}}) for Wald's test (involving a chi-square distribution) for the above hypothesis. Use the value of \theta that defines H_0 as the argument of the asymptotic variance \, V(\theta )\,.
Hint: Rewrite the hypothesis in terms of the parameter \theta.
(Enter hattheta for \hat\theta ^{\text {MLE}}.)
(To avoid double jeopardy, you may use V for the asymptotic variance \, V(\theta )\, under H_0.)
\, T_ n^{\text {Wald}}=\quad
((4*n)/(1-(hattheta)^2))*(hattheta-sqrt(hattheta))^2
submitted
((4*n)/(1-(hattheta)^2))*(hattheta-sqrt(hattheta))^2
What is the form of this Wald's test?
\, \mathbf{1}(T_ n^{\text {Wald}}\, {\color{blue}{>}} \, C) for some C{\color{blue}{>}} 0
\, \mathbf{1}(T_ n^{\text {Wald}}\, {\color{blue}{>}} \, C) for some C{\color{blue}{<}} 0
\, \mathbf{1}(T_ n^{\text {Wald}}\, {\color{blue}{<}} \, C) for some C{\color{blue}{>}} 0
\, \mathbf{1}(T_ n^{\text {Wald}}\, {\color{blue}{<}} \, C) for some C{\color{blue}{<}} 0
submitted
Find C such that Wald's test has asymptotic level 5\%.
(Enter a numerical value accurate to at least 2 decimal places.)
C=\quad
3.84
submitted
You obtain a sample of size n=100, of which 40 of the X_ i are negative (X_ i<0) and 60 of the X_ i are non-negative (X_ i\geq 0).
Do we reject H_0 at asymptotic level 5\%?
We reject H_0.
We fail to reject H_0
Cannot be determined without more information.
1 answer