Question

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). Round p-value to four decimal places. If your answer is zero enter "0". F p-value Source of Variation Sum of Squares Degrees of Freedom Mean Square Treatments 300 Error Total 460 a. What hypotheses are implied in this problem? Ho: Select H. Select b. At the - .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 What is your conclusion? Select

Answer 1

CRD Oaah oF the experoirmental units Seven Factors the OF 1eve Fve 35 52 N = tota d:F N I - 35-) total 34 beroeent taectrnent 5-1 dr toeatme 4 total df dfAearoert dform 34-4 30 Jraton Ounrces of SS taeatmert SS total - 800 tnat we SSt +SSE SStoeat 300 460 460 300 SSErsor 160 Meon squoroe MStorat SSet df rat 300 MS toeat

- SSEoSD6 M SE afeaoon - 160 MSES 5.83 MSteat Seroo 75 14. 07129 Fcal 14.03 O O000 p. value O O0O0 p. voJue AHONA Lable P.Salue f-Value Surmif DF rlean Soursce of Squore Squone 300 4 traearment 75 Pvalue o 16030 eroos toray 34

ipoiresis ae ohat this PODblem 9 toeatro en means aR Ho The equas two 0 t0e Hve toeatroet AT H! eas cliffer Sianiconti Means LOS level oF Signifcone) po1nesis in porst@ At the dO- 05 Con Pejet 4he ve Test Stakstic Fcal 14.07 P. Value = OO 5 P.value O.0000 Heae ce Reeat Signiconce Ho level of at Coclusio jeaist at Pejeut Ho, meoins differs 4eatrments OF the 5 Sigrificant