In a completely randomized design, seven experimental units were used for each of the five levels of the factor. Complete the following ANOVA table. If your answer is zero enter "0". Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square (to 2 decimals) | F (to 2 decimals) | p-value (to 4 decimals) Treatments | 360 | | | | Error | | | | Total | 500 | | The p-value is - Select your answer - .
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Since there are seven experimental units for each of the five levels of the factor, there are 7*5 = 35 total observations. So, the df for the total is 35 - 1 = 34. Show more…
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In a completely randomized design, seven experimental units were used for each of the five levels of the factor. Complete the following ANOVA table.
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Completely randomized design: Eight experimental units were used for each of the five levels of the factor. Consider the following ANOVA table: Source | Sum of Squares | Degrees of Freedom | Mean Square | F | p-value ---|---|---|---|---|--- Treatments | 380 | 4 | 95 | 36.94 | 0.0000 Error | 90 | 35 | 2.57 | | Total | 470 | 39 | | | (a) What hypotheses are implied in this problem? H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: Not all the population means are equal. (b) At the α = 0.05 level of significance, can we reject the null hypothesis in part (a)? Explain. Because the p-value ≤ α = 0.05, we can reject H0.
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