Question

Reserve Problems Chapter 10 Section 1 Problem 1 Consider the hypothesis test Ho: 41 - My = 0 against H : H1Hy samples below: I 35 40 31 32 33 30 32 30 39 38 30 38 36 30 39 30 36 40 II 32 29 33 33 31 29 30 36 31 34 30 30 29 33 33 34 Variances: o = 4.0, 62 = 2.0. Use a = 0.05. (a) Test the hypothesis and find the P-value. Find the test statistic. Round your answers to four decimal places (e.g. 98.7654). 20 2.5313 P-value = (b) Explain how this test could be conducted with a confidence interval. Round your answers to three decimal places (e.g. 98.765). We are 95% confident that value 1 exceeds value 2 by between (c) What is the power of the test for the true difference in means of 2? Round your answer to three decimal places (e.g. 98.765). The power of the test is 1-B = SH1-42 s

Answer 1

Sample 1 :

∑x = 619

n1 = 18

Mean , x̅1 = Ʃx/n = 619/18 = 34.3889

σ1 = 4

Sample 2 :

∑x = 507

n2 = 16

Mean , x̅2 = Ʃx/n = 507/16 = 31.6875

σ1 = 2

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a) Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 ≠ µ2

Test statistic:

z = (x̅1 - x̅2)/√(σ1²/n1 + σ2²/n2 ) = (34.3889 - 31.6875)/√(4²/18 + 2²/16) = 2.5313

p-value = 2*(1-NORM.S.DIST(ABS(2.5313, 1) = 0.0114

b)

95% Confidence interval :

At α = 0.05, two tailed critical value, z_c = ABS(NORM.S.INV(0.05/2)) = 1.960

Lower Bound = (x̅1 - x̅2) - z_c*√(σ1²/n1 +σ2²/n2) = (34.3889 - 31.6875) - 1.96*√(4²/18 + 2²/16) = 0.610

Upper Bound = (x̅1 - x̅2) + z_c*√(σ1²/n1 +σ2²/n2) = (34.3889 - 31.6875) + 1.96*√(4²/18 + 2²/16) = 4.793

0.610 < µ1 - µ2 < 4.793

c)

Hypothesized mean, Δₒ = 0

True mean, Δ' = 2

Standard error = √(σ1²/n1 +σ2²/n2) = √(4²/18 + 2²/16) = 1.0672

Critical value, z crit = NORM.S.INV(0.05/2) = 1.960

Power = 1- P(z_α - (Δ'-Δₒ)/se)

= 1 - P(1.96 - (2/1.0672))

= 1 - P(0.0859)

= 1 - NORM.S.DIST(0.0859,1)

= 0.4658