Question

Exercise 13.03 Question 1 of 14 Check My Work eBook 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". O Source of Variation Sum of Squares Degrees of Freedom Mean Square p-value Treatments 300 7 O Error O O Total 460 a. What hypotheses are implied in this problem Ho: - Select your answer - H: -Select your answer - b. At the am.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 - Select your answer - What is your conclusion? - Select your answer - Check My Work

Answer 1

= 35 nk= 75 Sumet Squares reatemet SSToeat 300 SStotal 460 SS Ensor SSTotal Sstecment 460-300 160. 4f of treatment k-1=5-4 =) dftatal-d freatmut 35-1-34 34-4- 30 SStreatmut Preatment Hean Sam squie df treatmant 300 75 160 SS EYOY Grer Mean Sum Squaru 30 = 5.33333

F-Stalistics MSS treatment 75 14.06.86 5.3333 Source of Varict.on SS P-Value P ns Treatments 14-06 4 O00 45 300 Grror 30 5-33 160 Total 34 460 ALL tive treatment means a cqual Ca) Ho Ha: Not all tivefreatmut meart ar equal b) at o.o5 level of rita test statutic = 14-06 lan tham O.01 PValue at not all tive- treatmut Cofleluds We can au egeral mears