/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 If the sample size for each trea... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 21

If the sample size for each treatment is \(n_{t}\) and if \(s^{2}\) is based on \(12 d f\), find \(\omega\) in these cases: a. \(\alpha=.05, k=4, n_{t}=5\) b. \(\alpha=.01, k=6, n_{t}=8\)

Short Answer

Expert verified
Answer: In case A, 蠅 is approximately 0.55. In case B, 蠅 is approximately 1.08.

Step by step solution

01

Find critical value for F-distribution

Given \(\alpha = .05\), \(k = 4\), and \(n_{t} = 5\), we need to find the critical value for \(F_{\alpha, df_1, df_2}\) using the F distribution table. Here, \(df_1 = k - 1 = 3\), and \(df_2 = 12df - 4 + 1 = 12 * 3 - 4 + 1 = 33\). So, we need to find \(F_{.05, 3, 33}\). Using the F distribution table, we find that \(F_{.05, 3, 33} \approx 2.95\).
02

Calculate \(\omega\)

With the critical value from step 1, we can now calculate \(\omega\) using the equation \(\omega = \frac{s^2 \cdot SS_{treatments}}{n_t \cdot MS_{treatments}}\). The formula becomes: \(\omega = \frac{2.95 \cdot 5}{4 \cdot 4}\) \(\omega \approx 0.55\) #Case B#
03

Find critical value for F-distribution

Given \(\alpha = .01\), \(k = 6\), and \(n_{t} = 8\), we need to find the critical value for \(F_{\alpha, df_1, df_2}\) using the F distribution table. Here, \(df_1 = k - 1 = 5\), and \(df_2 = 12df - 6 + 1 = 12 * 5 - 6 + 1 = 54\). So, we need to find \(F_{.01, 5, 54}\). Using the F distribution table, we find that \(F_{.01, 5, 54} \approx 4.86\).
04

Calculate \(\omega\)

With the critical value from step 1, we can now calculate \(\omega\) using the equation \(\omega = \frac{s^2 \cdot SS_{treatments}}{n_t \cdot MS_{treatments}}\). The formula becomes: \(\omega = \frac{4.86 \cdot 8}{6 \cdot 6}\) \(\omega \approx 1.08\) In conclusion, the values of \(\omega\) for the given cases are approximately \(0.55\) for case A and \(1.08\) for case B.

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