In a completely randomized design, eight experimental units were used for each of the five levels of the factor. Consider the following ANOVA table.**ANOVA Table:**| Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | F | p-value ||---------------------|----------------|--------------------|-------------|-------|---------|| Treatments | 300 | 4 | 75 | 23.86 | 0.0000 || Error | 110 | 35 | 3.14 | | || Total | 410 | 39 | | | |---**(a) What hypotheses are implied in this problem?**- \( H_0 \): Not all the population means are equal. \( H_a \): \(\mu_1 = \mu_2 = \mu_3 = \mu_4 = \mu_5\)- \( H_0 \): \(\mu_1 = \mu_2 = \mu_3 = \mu_4 = \mu_5\) \( H_a \): Not all the population means are equal.- \( H_0 \): At least two of the population means are equal. \( H_a \): At least two of the population means are different.- \( H_0 \): \(\mu_1 \ne \mu_2 \ne \mu_3 \ne \mu_4 \ne \mu_5\) \( H_a \): \(\mu_1 = \mu_2 = \mu_3 = \mu_4 = \mu_5\)- \( H_0 \): \(\mu_1 = \mu_2 = \mu_3 = \mu_4 = \mu_5\) \( H_a \): \(\mu_1 \ne \mu_2 \ne \mu_3 \ne \mu_4 \ne \mu_5\)---**(b) At the \(\alpha = 0.05\) level of significance, can we reject the null hypothesis in part (a)? Explain.**- Because the p-value \(> \alpha = 0.05\), we cannot reject \( H_0 \).- Because the p-value \(\le \alpha = 0.05\

p-value is probability of measure of extremity of observed result.P value smaller than level of…

Answered: In a completely randomized design,…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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