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Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments 300 75 14.07 Error 160 30 5.333 Total 460 34 a. What hypotheses are implied in this problem? Họ: All five treatment means are equal H: Not all five treatment means are equal b. At the a = 0.05 level of significance, can we reject the null hypothesis in part (a)? Calculate the value of the test statistic (to 2 decimals). The p-value is between 0.01 and 0.025 What is your conclusion? Conclude not all five treatment means are equal v

Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments 300 75 14.07 Error 160 30 5.333 Total 460 34 a. What hypotheses are implied in this problem? Họ: All five treatment means are equal H: Not all five treatment means are equal b. At the a = 0.05 level of significance, can we reject the null hypothesis in part (a)? Calculate the value of the test statistic (to 2 decimals). The p-value is between 0.01 and 0.025 What is your conclusion? Conclude not all five treatment means are equal v

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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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 (to 2 decimals, if necessary). If your answer is zero enter "0". Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments 300 4 75 14.07 Error 160 30 5.333 Total 460 34 a. What hypotheses are implied in this problem? Ho: All five treatment means are equal H: Not all five treatment means are equal v b. At the a = 0.05 level of significance, can we reject the null hypothesis in part (a)? Calculate the value of the test statistic (to 2 decimals). The p-value is between 0.01 and 0.025 v What is your conclusion? Conclude not all five treatment means are equal v
Transcribed Image Text:In a completely randomized design, seven experimental units were used for each of the five levels of the factor. Complete the following ANOVA table (to 2 decimals, if necessary). If your answer is zero enter "0". Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments 300 4 75 14.07 Error 160 30 5.333 Total 460 34 a. What hypotheses are implied in this problem? Ho: All five treatment means are equal H: Not all five treatment means are equal v b. At the a = 0.05 level of significance, can we reject the null hypothesis in part (a)? Calculate the value of the test statistic (to 2 decimals). The p-value is between 0.01 and 0.025 v What is your conclusion? Conclude not all five treatment means are equal v
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