Question

求专业英语达人翻译

16. Trout experiment，continued
Suppose the trout experiment of Exercise 15 is to be repeated with the same v=4 treatments, and suppose that the same hypothesis, that the treatments have no effect on hemoglobin content, is to be tested.
(a) For calculating the number of observations needed on each treatment, what would
you use as a guess for σ2?
(b) Calculate the sample sizes needed for an analysis of variance test with α=0.05
to have power 0.95 if (i) Δ=1.5. (ii) Δ=1.0. (iii) Δ=2.0.
17. Soap experiment, continued
In Example 3.6.2, page 53, a sample size calculation was made for the number of observations needed to detect, with probability π (0.25)=0.90, a difference in weight loss of at least ) Δ=0.25 grams in v=3 difference types of soap, using an analysis of variance with a probability of α=0.05 of a Type I error. The calculation used an estimate of 0.007 grams2for σ2and showed that r=4 observations were needed on each type of soap. The experiment was run with r=4, and the least squares estimate for σ2 was 0.0772. If the true value for σ2 was, in fact, 0.08, what power did the test actually have for detecting a difference of )Δ=0.25 grams in the weight loss of the three soaps?

Answer 1

16. Trout experiment，continued

16。鳟鱼试验，继续

Suppose the trout experiment of Exercise 15 is to be repeated with the same v=4 treatments, and suppose that the same hypothesis, that the treatments have no effect on hemoglobin content, is to be tested.

假设鳟实验习题15是重复相同，4治疗，并假设相同的假设，这种治疗没有影响血红蛋白含量，是进行测试。

(a) For calculating the number of observations needed on each treatment, what would

（一）计算的若干意见需要在每个治疗，会有什么

you use as a guess for σ2?

你用一个猜想σ2？

(b) Calculate the sample sizes needed for an analysis of variance test with α=0.05

（二）计算样本大小需要一个方差分析测试α=0.05

to have power 0.95 if (i) Δ=1.5. (ii) Δ=1.0. (iii) Δ=2.0.

有功率0.95如果（我）Δ=1.5。（二）Δ=1。（三）Δ=2。

17. Soap experiment, continued

17。肥皂试验，继续

In Example 3.6.2, page 53, a sample size calculation was made for the number of observations needed to detect, with probability π (0.25)=0.90, a difference in weight loss of at least ) Δ=0.25 grams in v=3 difference types of soap, using an analysis of variance with a probability of α=0.05 of a Type I error. The calculation used an estimate of 0.007 grams2for σ2and showed that r=4 observations were needed on each type of soap. The experiment was run with r=4, and the least squares estimate for σ2 was 0.0772. If the true value for σ2 was, in fact, 0.08, what power did the test actually have for detecting a difference of )Δ=0.25 grams in the weight loss of the three soaps?

例如3.6.2，页53，样本大小的计算了观测的数目需要检测，概率π（0.25）=0.90，不同体重下降至少）Δ=0.25克，有3种不同类型的肥皂，使用方差分析与概率α=0.05型错误。使用的计算估计0.007grams2forσ2=4的观测表明，需要对每种类型的肥皂。实验运行=4，与最小二乘估计σ2是0.0772。如果真正的价值σ2，事实上，0.08，什么力量测试实际上有用于检测不同的）Δ=0.25克的重量损失的三块肥皂？