Suppose that we have a random sample of sizenfrom a normal population with known

Question

Suppose that we have a random sample of sizenfrom a normal population with known variance 2. Show that we can test the null hypothesis H0:=0 against the alternative hypothesis Ha:6=0using a 2-distribution. In particular, show that the test using Z= (Y0)/(/n) and rejection region|Z|> z/2 is equivalent to a test using a 2-statistic X (with 1 degree of freedom) and rejection region X > 2,1; here 2,1denotes the constant such thatP(X > 2,1) =. Hint: consider the test statistic obtained by squaring Z.

Problem 2 Suppose that we have a random sample of size n from a normal population with known variance Show that we can test the null hypothesis Ho : = 0 against the alternative hypothesis Ha : 140 using a 2-distribution. In particular, show that the test using Z = ( )/(o/vn) and rejection region Z > za/2 is equivalent to a test using a 2-statistic X (with 1 degree of freedom) and rejection region X X2.1. here X2,1 denotes the constant such that P(X > Hint: consider the test statistic obtained by squaring Z. X:1) = a.

Explanation / Answer

To test Null H0: µ = µ0 Vs Alternative HA: µ µ0, when is known

Test statistic is

Z = (n)(Xbar - µ0)/, and

Reject H0, if | Zcal | > Zcrit where Zcrit = upper (/2)% point of N(0, 1).

So, we have P[|(n)(Xbar - µ0)/| > Z/2 ] = .

Equivalent to saying:

P[{(n)(Xbar - µ0)/}2 > Z2/2 ] = …………………………………………… (1)

But, under H0, X = {(n)(Xbar - µ0)/}2 ~ 21…………………………………..(2)

[because under H0, {(n)(Xbar - µ0)/} ~ N(0, 1) and square of a Standard Normal Variate has a Chi-square distribution with 1 degree of freedom]

(1) and (2) => Z2/2 = 21,

Thus, H0: µ = µ0 Vs Alternative HA: µ µ0, when is known, can be tested using a Chi-square statistic with 1 degree of freedom.

DONE

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