Testing Hypothesis What is the difference between these two estimators: E[y_i] a

Question

Testing Hypothesis

What is the difference between these two estimators: E[y_i] and cap y_i where y_i is the response variable of a linear model? Given the samples, S_1 = {6 2 1 3 3} and S_2 = {5 3 2 4 2}, test the hypothesis: H_0: mu_1 = mu_2 vs H_1: mu_1 notequalto mu_2 at the .05 level of significance. Using the same two samples of problem 3, find the 95% CI of mu_2 - mu_1. Let t_0 represent a particular value of Student's t distribution. Find the value of t_0 which will make the following statement true: P(t lessthanorequalto -t_0 or t greaterthanorequalto t_0) = .10 with df = 12.

Explanation / Answer

Solution

Back-up Theory

Let X ~ N(µ1, 12) and Y ~ N(µ1, 12).

Test Statistic to test H0: µ1 = µ2 Vs H1: µ1 µ2 is:

t = |Xbar – Ybar|/[s{(1/n1) + (1/n2)}] …………………………………………(2)

where Xbar = sample average of X, Ybar = sample average of Y,

s = sq.rt[{(n1 - 1)s12 + (n2 - 1)s22}/(n1 + n2 - 2)], s12 , s22 are the respective sample variances and n1, n2 are the respective sample sizes.

Under H0, t ~ tn1 + n2 - 2 i.e., t-distribution with (n1 + n2 - 2) degrees of freedom. …(3)

H0 is rejected at % level of significance if tcal > tn1 + n2 – 2,/2, where tcal and tn1 + n2 – 2,/2 are respectively the calculated value of t and upper /2 percent point of t-distribution with (n1 + n2 - 2) degrees of freedom. (4)

100(1 – ) % confidence interval for (1 - 2) is:

|Xbar – Ybar| ± (t(2n – 2)/2){s(2/n)}, where n = common sample size…....................................…………..(5)

Now, to work out the solution,

Part (2)

yicap is the estimate of an individual value of Y, whereas {E(yi)}cap is the estimate of the mean or average of y values. ANSWER

Part (3)

Given n1 = n2 = 5, Xbar = 3, Ybar = 3.2, s12 = 3.5, s22 = 1.7, s2 = 2.6 or s = 1.6124,

[vide (2) under Back-up Theory],

t = (3.2 – 3.0)/{1.6124x (0.2 + 0.2)} = 0.2/1.0197 = 0.1961.

[vide (3) under Back-up Theory], t8,0.025 = 2.306 [using Excel Function]

[vide (4) under Back-up Theory], H0 is accepted at 5% level of significance ANSWER

Part (4)

[vide (5) under Back-up Theory], 95% CI for (µ2 - µ1) is:

(3,2 – 3.0) ± {(2.306)(1.6124x0.6324)} = 0.2 ± 2.3514 ANSWER

Part (d)

We want t0 such that P(t12 - t0 or t12 t0) = 0.1 or

2P(t12 t0) = 0.1 [because t-distribution is symmetric]

Or P(t12 t0) = 0.05

[using Excel Function], t0 = 1.782 ANSWER

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