Let Y₁, Y2, Y3,..., Ynd N(μ, o²), where o²1. Find the asymptotic distribution of Z =the asymptotic distribution of μ?n= 25 is known.=2. Suppose now we do not know o² and instead plug in the MLE estimator ²Ĥ-μ-, where is the MLE. Hint: what is√o²/n'Σ1(Yi - μ)² to obtain Z := Is the asymptotic distribution still N(0, 1)? Ifyes, why? (hint: what do we know about consistency of MLEs? Slutsky's theorem?)..=3. Suppose we collect data of size n = 16 and the MLE turns out to be û = 2 and ô² = 25.Using the result from the previous part, conduct an a = .05 level test of Hou 0in favor of H₁ : µ ‡0. Find the critical value, rejection region, and p-value. Interpretthe p-value. Do you reject or fail to reject the null?

Given that Y1,Y2,.........Yn ~iidNμ,σ2 where σ2=25 is known Here we use the central limit theorem and convergence in probability for obtaining the desired asymptotic distribution.Part 1 The MLE μ is Y¯ as θ^=Y¯=∑i=1nYin=μ^ Here, θ^~NY¯,1I(θ) Now considering μ=Y¯, then μ^~Nμ,1I(θ) where I(θ) is the information matrix For σ known I(θ)=1σ2n2 Therefore, μ^-μ1I(θ)=μ^-μσ2n~N(0,1) According to the central limit theorem, as μ^ is asymptotically normal with Nμ,σ2nPart 2 Now σ2 is not known and if we consider σ^2=1n∑i=1nYi-μ2 If we use σ12=1n-1∑i=1nYi-μ¯2 then Z will lead to t-distribution for σ^2=σ12 Here, we use σ^2=1n∑i=1nYi-μ2 Then we use the Slutsky theorem which indicates a consistent estimator for σ2 Now, due to convergence in probability σ^2→Pσ2 Therefore, Z=μ^-μσ2n→Pμ^-μσ2n~N(0,1)Part 3 Given that n=16μ^=2σ^2=25α=0.05 Hypothesis H0:μ=0H1:μ≠0 Test statistic is Z=X¯-μσn =2-0516 =1.6 The critical value for Z0.05=1.96 Now the rejection region is Z>1.96 Hence, the p-value isPZ>1.6=0.1096 So, the p-value is the probability that the value of Z under H0 is true at least as extreme as 1.6 Hence p-value is greater than 0.05, we fail to reject H0