Question

Reserve Problems Chapter 10 Section 1 Problem 1

Consider the hypothesis test  H0: μ1-μ2=0 against  H1: μ1-μ2≠0 samples below:

I	35	38	32	32	33	30	32	29	38	38	31	38	36	32	39	31	35	38
II	34	29	33	32	31	29	30	38	32	34	30	30	30	33	33	35

Variances: σ1=3.4, σ2=-0.5. Use α=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).

z0≈ Enter your answer; z0

P-value⁢= Enter your answer; 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 Enter your answer; lower bound ≤μ1-μ2≤ Enter your answer; upper bound

(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-β= Enter your answer in accordance to the item c) of the question statement

Reserve Problems Chapter 10 Section 1 Problem 1 Consider the hypothesis test HO : H1 – H2 = 0 against Hi : H1 – 42 7 0 samples below: I 35 38 32 32 33 30 32 29 38 38 31 38 36 32 39 31 35 38 II 34 29 33 32 31 29 30 38 32 34 30 30 30 33 33 35 Variances: 61 = 3.4, 62 = -0.5. 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). ZO 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). <H-123 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 =

Answer 1

Answer : Given ng =18 A2 = 16 Exii zey li ni + 35 +38 35+ 38+32+ 13 18 G17 18 = 34.278 se , X2 Zxzi ne 34 +29+33+-.... +33 +35 16 513 16 ser 32.063 -0.5 23.4, TI is Zo a Test statistic z, xz Zo 4 no ne 34.978- 32.063 (314) (2015) + 16 18 2.215 0.8111 Zo = 2. 7309 Zo 2.7309 .: Test statistic P-value ap(7.7 2.7309) a [- P(Z < 2.7309) -2 (1-0.9968) (from 2- Z-table) 2 * 0.0032 Pavalue = 0.0064

b z critical value is 2005 x old Zoos 2 ZO.025 Za = 1.96 (from 2-table) 95% confidence interval for Mi - Mz is CD - Je, - X - Za + n2 (314)? 60.5) = 34.278-32.063 + 1,96. + 18 16 2.215 11:96 * 0.8111 = 2.215 7 10 5898 = (2.215-1.5898, 2.215+1.5898) ci 2 0. 626, 3.805 i's 95%. C.I is 0.626 < M - My £ 3.805 C given co 2 co S Z Zorn to 24 2 (3.4) Fo (0.5) (314) (-os) 18 + 16 $(1.96–2:47) - 0(-196–2:47) $ 1-0.51) - 0(-4.43) O 3050 - 0 1 - o (From Z-table) ß 0 3050 power of test is 1-B = 1-0.3050 0.695 I-ß 0.695