10-1. a) 1) The parameter of interest is the difference in means *μ*_{1} – *μ*_{2}. Note that Δ_{0 =} 0.

2) H_{0} : *μ*_{1} – *μ*_{2} = 0 or *μ*_{1} = *μ*_{2}

3) H_{1} : *μ*_{1} – *μ*_{2} ≠ 0 or *μ*_{1} ≠ *μ*_{2}

4) The test statistic is

5) Reject H_{0} if z_{0} < –z_{α/2} = –1.96 or z_{0} > z_{α/2} = 1.96 for α = 0.05.

6) _{1} = 4.7 _{2} = 7.8

*σ*_{1} = 10 *σ*_{2} = 5

n_{1} = 10 n_{2} = 15

7) Conclusion: Because –1.96 < –0.9 < 1.96, do not reject the null hypothesis. There is not sufficient evidence to conclude that the two means differ at α = 0.05.

*P*-value = 2(1 – Φ(0.9)) = 2(1 – 0.8 159 50) = 0.36 8

With 95% confidence, the true difference in the means is between –9.79 and 3.59. Because zero is contained in this interval, we conclude there is no significant difference between the means. We fail to reject the null hypothesis.

Power = 1 – 0.86 = 0.14

d) Assume the sample sizes are to be equal, use α = 0.05, ...

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