The University of Sydney School of Mathematics and Statistics Computer Exercise Week 8 STAT3023: Statistical Inference Semester 2, 2021 Lecturers: Neville Weber and Michael Stewart Prepare your computer report and submit to the appropriate Canvas portal by 11:59pm Sunday 10th October. Please include in your report all the code, plots and any comments required by the questions. The permitted format of the file that you upload will be restricted to PDF, HTML or Word, and should be created in a reproducible fashion (e.g. compiled from an Rmarkdown file). Please do not have your name visible in your report. Two-sided tests for a normal variance Suppose X1, . . . , Xn are iid N(µ, σ 2). In the week 7 Tutorial it was noted that the statistic Y = (n − 1)S2 = ∑ni=1(Xi − X¯)2 (where X¯ = 1n∑ni=1Xi and S2 are the sample mean and variance) has a σ2χ2n−1 distribution (note we are not multiplying by 1 2 as we did in the week 7 Tutorial!). Consider testing H0 : σ 2 = 1 against H1 : σ 2 6= 1. (1) 1. One possible level-α test is the “equal-tailed” test based on Y , where we reject for Y < a or Y > b where P0 {Y < a} = P0 {Y > b} = α 2 . (a) Taking α = 0.04 and n = 5, find appropriate values a and b. (b) Defining sig.sq=(50:150)/100 plot the power of the test against sig.sq. Add a horizontal dotted line to indicate the level. 2. In Tutorial week 7 we also saw that the UMPU test rejects for large values of S2 − log(S2) which is equivalent to rejecting for small values of the statistic T = (n− 1) log Y − Y ; to see this, write log(S2) = log Y − log(n − 1), multiply through by n − 1 and ignore the (n − 1) log(n− 1) term. If the test is to have level α, we reject for Y ≤ c or Y ≥ d where P0(Y ≤ c) + P0(Y ≥ d) = α (2) and (n− 1) log(c)− c = (n− 1) log(d)− d . (3) (a) Write a function of the form fn=function(c,alpha,n) { ... } which Copyright© 2021 The University of Sydney 1 • computes the appropriate d so that c and d satisfy (2); • then computes and outputs the difference between the left-hand side and right-hand side in (3). (b) Use the R function uniroot() to find the root (in c) of the equation fn(c,0.04,5)=0. In your code you will need a command along the lines of uniroot(fn,lower=0,upper=...,alpha=0.04,n=5) Consult the week 7 exercise for some hints as to how to choose the upper=.... When you have worked out the right commands, wrap it all in a function of the form norm.var.umpu=function(alpha,n) { ... } which returns a list containing elements $c and $d. (c) Recreate your plot from part (b) of the previous question and add to it the power function of the UMPU test. 3. The GLRT test of (1) above uses the statistic Ln = `(X¯, Y/n;X)− `(X¯, 1;X) = −n 2 log ( Y n ) − n 2 + Y 2 which is an increasing function of Y − n log Y (as opposed to the UMPU which rejects for large Y − (n− 1) log Y ). Adapt your code for the previous question to compute the power of the exact GLRT, recreate your earlier plot and add a power curve to it so it shows all 3 power curves on the 1 graph. Add an informative heading, legend, etc.. Comment on the main differences between the 3 tests. 4. As a final step, recreate your last plot but use an extended range for the parameter: sig.sq=(1:400)/100. 2