Question

Reserve Problems Chapter 10 Section 1 Problem 1 Consider the hypothesis test Ho: M1 My = 0 against H : H1 – 70 samples below: I 36 39 32 32 33 30 32 29 39 38 31 38 36 30 39 31 35 40 II 34 29 34 32 31 29 30 38 32 34 30 29 31 33 33 34 Variances: 6 = = 4.0, 02 = 0.3. 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). 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 3? Round your answer to three decimal places (e.g. 98.765). The power of the test is 1 – B= ISH-M25

Answer 1

Minitab is used for the analysis.

The output is:

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
1     18 34.44 4.00 0.94
2 16 32.030 0.300 0.075

Difference = μ (1) - μ (2)
Estimate for difference: 2.410
95% CI for difference: (0.415, 4.405)
T-Test of difference = 0 (vs ≠): T-Value = 2.55 P-Value = 0.021 DF = 17

$z_{0}=\frac{34.44-32.03}{\sqrt{\frac{4^{2}}{18}+\frac{0.3^{2}}{16}}}=2.5481$

p-value = 2*P[t₁₇>2.55] = 0.021

95% C.I.=

$0.415\leq \mu_{1}-\mu_{2}\leq 4.405$

$\beta=P[t_{17}>\frac{(34.44-32.03)-3}{\sqrt{\frac{4^{2}}{18}+\frac{0.3^{2}}{16}}}]$

=0.728

power = 1-0.728 = 0.272

2) The output of the test is

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
1 8 1.130 0.110 0.039
2 8 1.0800 0.0900 0.032

Difference = μ (1) - μ (2)
Estimate for difference: 0.0500
95% CI for difference: (-0.0586, 0.1586)
T-Test of difference = 0 (vs ≠): T-Value = 1.00 P-Value = 0.338 DF = 13

The P value is 0.338

95% CI for difference: (-0.0586, 0.1586)