Question

Suppose X1, .., Xn is a random sample from a N(0, a2) population, where variance o are known. Consider testing Ho : 0 = O0 vs. Hi 0 700 Q1(4pt): Using Likelihood Ratio Test (LRT) to obtain a level a test that rejects Ho if Уп(X — во) VП(х — во) <21-a/2 21-a/2 or о Q2(1pt): Is the two-sided test derived in 1) an uniformly most powerful test? If not, briefly state your reasons Q3 (1pt): Note that the test statistic derived in 1) has the form T(X) =Vn(X - 00)/a. Using the definition of p-value in notes, we know that the p-value has the form p(x) = Preo [T(X) T(x) in this case since the null hypothesis is a simple hypothesis, where X represents the random sample and represents the observed sample. Now simplify the formula of the p-value p(x) and find an expression of this p-value in terms of CDF functions of standard normal random variable. Q4(2pt): Monte Carlo Simulation) Suppose that sample size n 20, 00 0, a2 = 1, and a 0.05. Use the following algorithm to simulate 5000 repeated sampling of sample size 20 from N(0, 1) under null hypothesis Но- 1.1 Simulate a random sample X1, X2, ..., X20 from the null distribution N(0, 1) 1.2 Based on the simulated sample, calculate its corresponding p-value named p1 (x) 1.3 Repeat steps 1-2 5000 times, thus generate 5000 sample values of p-values, e.g., p'(2), p2(x),,p5000 () Plot the histogram of p'(x), p2 (x),. .., p5000 (x) in R and attach your R code. Q5 (1pt): Based on the Monte Carlo Simulation result in 4), what is the distribution of p-value p(X) under the null hypothesis in this example? (Interested students should know that the formal proof of this result is a direct application of "Probability Integral Transformation" of continuous random variables, a result we mentioned this Wednsday.) Q6(1pt) Based on the above results, is the p-value defined in 3) a valid p-value? Briefly state your reasons (Note that if a p-value is valid, we can use it to construct level a statistical test.)

Answer 1

(I; - 0) 202 Q1: The likelihood function for = ) = 1 f(1,0) = 11 + V20 I “ (ti - 0)2 exp - (27) on 1202 We know MLE of 6=ī if i #fo and if ī= f, then MLE of 6 = 60. (Ii-02 Likelihood ratio = A(0) = supeee. LO supeee, L(0) = exp (G; - 72 202 202 +) = 60 (+225) VnT -001 We reject H, if A(0) <c or, - >k, where k is such that p(Vīlz=001 > /10) – a or, k = 31-0/2.

Hence we reject H, at level a if Vnī - 00) Vnī - 00) -> 21-a/2 or, v 0 5-21-a/2 Q2. It not uniformly most power ful test. Since uniformly most power ful test - for testing Ho : 0 = do us. H1 : 0 > Ao at level a is Vn(3 – %). > 21-a and uniformly most power ful test. Since uniformly most powerful test (nt - 0) <- for testing Ho : 0 = fous. H1 : 0 > Ho at level a is Q3: p-value = P(T(X) >T(2) Ho) = P(VTCN – ) > T(2)|HQ) + P (VM(x – Pol 5 –T():10) =1- (T(2)) + (-11)) = 2(1 - (TL))).

R code:

p=1:5000*0
for(i in 1:5000)
{
x=rnorm(20,0,1)
t=abs(sqrt(20)*mean(x))
p[i]=2*(1-pnorm(t))
}
hist(p)

Histogram of p 500 400 300 Frequency 200 100 0 0.0 0.2 0.4 0.6 0.8 1.0

Q5: Uniform(0,1).

Q6. Yes.

When T(2) = 21-a/2 then p– value = P(T(X) > 21-a/2 H.) (Vn(X – 60) p7n(x – 0%). > 21-a/2H0 +P -S-21-a/2|H, = a