We are going to test the equality of the means of two Normal populations (withou

Question

We are going to test the equality of the means of two Normal populations (without knowing the values of such means). Let X_1, ..., X_m, ~ N(mu_X, sigma^2_X), and Y_1, ..., Y_n ~ N(mu_Y, sigma^2_Y): all of the random variables are all independent. Note that the two samples have different sizes, m and n. The parameters and sigma^2_X and ^2_Y are known, whereas mu X and mu Y are unknown. Consider the two hypotheses: H_0: mu_X = mu_Y (equivalently: mu_X - mu_Y = 0) H_1: mu_X notequalto mu_Y (equivalently: mu_X - mu_Y notequalto 0) We will design a test based on the statistic W = X - Y (the difference between the two sample means). (a) What is the probability density function of W? (Be careful with the signs!) (b) Consider the hypothesis test: IF |W|

Explanation / Answer

Part (a)

X ~ N(µX, X2) => Xbar (based on m observations) ~ N(µX, X2/m)………… (1)

Similarly, Ybar (based on n observations) ~ N(µY, Y2/n)…………………….(2)

Since X and Y are independent, W = (Xbar - Ybar) ~ N[(µX - µY), {(X2/m) + (Y2/n)}]

So,

W is distributed as Normal with mean (µX - µY) and variance {(X2/m) + (Y2/n)}

=> pdf of W is: {1/(2)}e^-[(1/2){(x - µ)/}2] ANSWER, where

= sqrt{(X2/m) + (Y2/n)} and µ = (µX - µY)

Part (b)

Given the rejection criterion as | W | k, alternative being two-sided (i.e., type) and level of significance = , P(| W | k/under H0 ) = or P(W k/under H0 ) = /2 [because W is Normal which is symmetric about its mean]

Now, P(W k/under H0 ) = /2 => P[Z {(k - µ)/)}] = /2, where µ and are as defined in Part (a).

=> {(k - µ)/)} = Z/2 or k = µ + . Z/2) = (µX - µY) + [sqrt{(X2/m) + (Y2/n)} x Z/2]

But, under H0, (µX - µY) = 0.

So, k = Z/2 x sqrt{(X2/m) + (Y2/n)} ANSWER

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.