4. Recall the "means ANOVA model Y,-μί +Gj where μί-B( ) is thepopulation mean of the...
Problem 2. Consider the one-way layout ANOVA model, where we assume that Yij = μί-cij,に1, . . . , I and J 1, . . . ,J, where μί's are fixed unknown with zero mean treatment means and eiy's are random errors , al such that Σ-lai -0 and E[Yj-μ + ai,1- Show that there exists unique numbers μ, ai, a. b. Show that the null hypothesis Ho : μ,-...- μι is equivalent to Ho : 01 ,-. . .-a1-0...
Problem 2. Consider the one-way layout ANOVA model, where we assume that Yij = μί-cij,に1, . . . , I and J 1, . . . ,J, where μί's are fixed unknown with zero mean treatment means and eiy's are random errors , al such that Σ-lai -0 and E[Yj-μ + ai,1- Show that there exists unique numbers μ, ai, a. b. Show that the null hypothesis Ho : μ,-...- μι is equivalent to Ho : 01 ,-. . .-a1-0
9. Consider the ANOVA model where Xij ~ N(μί,02). Then show that SSE (a) the random variable-O-~ χ2(1(J-1), and (b) the statistics SSE and SSTr are independent. Further, if the null hypothesis Ho : μ.-μ2-...-μι-, μ is true, then SSTr (c) the random variable-"K2(1-1) MSTr MSE (e) the random variable ~ χ2(U-1). (d) the statistics MsF), and
2. (a) Let us consider a full model of a balanced (all t treatments have equal number of observations r) CRD design with t treatments and r replications of each treatment, hence having n rt observations. . Minimizing sum of square error Δ/u114%)-ΣΊ ΣΊ (Vij-μ-%)2 with respect to μ and Ti find the least square estimators of μ and Ti as μ and T. Hint: Take derivative of the objective function with respect to μ and Ti and equate then...
ITs a stats problem Note that the analysis of variance F test for testing for treatment effects in a CRD with p- 2 treatments is equivalent to testing for equality of treatment means in the pooled two-sample t-test. Recall that the pooled two-sample t test statistic is given by (ni-1)sỈ + (n2-1)s n n2-2 where s2- 2, 2-i is the sample mean of the ith treatment group, s? is the sample variance of the ih treatment group, and n, is...
Please help! (a) Let us consider a full model of a balanced (all t treatments have equal number of observations r) CRD design with t treatments and r replications of each treatment, hence having n-rt observations. 2. i. Minimizing sum of square error Δfull (μ'Ti) -Σι-1 Σ-1 (Vi,- μ-Ti)2 with respect to μ and Ti find the least square estimators of μ and Ti as μ and Ti. Hint: Take derivative of the objective function with respect to μ and...
Problem 4: Complete the ANOVA table based on the following data: Replications Standard Deviation Factor A Level 1 Level 2 Mean 15 ANOVA Table Degree Freedom of Sum of square Mean sum of F-valuc Source error square error Treatment Error Total NA NA NA Recall that for single-factor ANOVA, we have the following -SSTreatments+SSp where ss,-Σ Σ(y)-T)-total sum of squares n Σ(vi-r-treatment sum of squares i- SSE _ Σ Σ(w-5)2-error sum of squares Given a dataset like the following: Treatment...
2. (a) Let us consider a full model of a balanced (all t treatments have equal number of observations r) CRD design with t treatments and r replications of each treatment, hence having n-rt observations i. Minimizing sum of square error Δfull(μ, Tỉ)-Σι-12jai (Vij-l-ri)2 with respect to μ and Ti find the least square estimators of μ and Te as μ and Ti Hint: Take derivative of the objective function with respect to u and Ti and equate then to...
Question 5 /10 points/ MLE of Variance is Biased For each n-1.M, let Xn ~ N(μ, Σ) denote an instance drawn (independently) from a Gaussian distribution with mean μ and convariance Σ. Recall /IML Xm, and Show that EML]-NN Σ Y ou mav want to prove, then use . where àn,m = 1 if m n and = 0 otherwise. Question 5 /10 points/ MLE of Variance is Biased For each n-1.M, let Xn ~ N(μ, Σ) denote an instance...
Find a consistent estimator of µ 2 , where E(Y ) = µ is the population mean and Y¯ n is the sample mean. 2 If E(Y 2 ) = µ 0 2 then prove that 1 n Pn i=1 Y 2 i is an consistent estimator of µ 0 2 3 We define σ 2 = µ 0 2 − µ 2 . Show that S 2 n = 1 n Pn i=1 Y 2 i − Y¯ 2...