Suppose that X_1, ..., X_n form a random sample from the normal distribution wit

Question

Suppose that X_1, ..., X_n form a random sample from the normal distribution with mean mu and variance sigma^2. (a) When sigma^2 is known to be 1, a 95% confidence interval for mu is X plusminus 1.96/Squareroot n. Let p denote the probability that an additional observation, X_n + 1, will fall in this interval. Is p greater than, less than or equal to 0.95? First find p = Pr (X - 1.96/Squarerootn lessthanorequalto X_n + 1 lessthanorequalto X + 1.96/Squareroot n) and answer the question. (b) Suppose that mu is known to be 0. Show that the most powerful test for H_0: sigma = 1 versus H_1: sigma = 2 rejects H_0 when Sigma^n _i = 1 X^2_i greaterthanorequalto c, and find the value of c for n = 10 and alpha_0 = 0.05.

Explanation / Answer

a. Y = Xn+1 - Xbar

p = Pr(Xbar - 1.96/sqrt(n)<=Xn+1<=1.96/sqrt(n)) = Pr(-1.96/sqrt(n)<=Y<=1.96/sqrt(n))

Xn+1 follows N(mu,1). So, Y=Xn+1-Xbar follows N(0,sqrt(1+1/n))

p = Pr(-1.96/sqrt(n+1)<=z<=1.96/sqrt(n+1))

sqrt(n+1)>sqrt(n). SO, interval range is decreased and probability value is decreased. SO, p <005

b. The c value here is the CHi square value at n=10, alpha = 0.05

We get chi square(n, alpha) = 3.94 = c

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