Pad 8:09 PM D finaiddocs.s D Chris Rock\'s open x D twenty one pilo X YouTube ts

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Pad 8:09 PM D finaiddocs.s D Chris Rock's open x D twenty one pilo X YouTube ts: x hom HW 5: Text 11.1 x C https xlitemprod.pearsoncmg.com /api/v1/print/math 5. The ANOVA summary table for an experiment with five groups, with seven values in each Degrees of Sum of Mean Square group, is shown to the right. Complete parts (a) through (d) below. Source Freedom Squares Variance) Among SSA 160 MSA 40 groups n c 30 SSW 600 MSWE 20 Within groups Total n 1 34 SST 760 Click here to view page 1of the F table.9 Click here to view page 2 of the F table,10 Click here to view page 3 of the F table 1 Click here to view page 4 of the F table a. At the 0.05 level of significance, state the decision rule for testing the null hypothesis that all five groups have equal population means. Determine the hypotheses. Choose the correct answer below. O A. Ho: B. Ho: H 7 H1: Not all the means are equal. H5 Ho: D. Ho: H H7 H1: Not all the means are equal. Find the test statistic. (Round to two decimal places as needed.) STAT Determine the critical value. (Round to two decimal places as needed.) b. What is your statistical decision? Ho. There is 2) evidence to conclude that there is a difference in the population means of the groups. c. At the 0.05 level of significance, what is the upper-ta critical value from the Studentized range distribution? (Round to two decimal places as needed.) 100% Print STAT 2.00

Explanation / Answer

a.There are five groups hence there are 5 means

and from the definition: we take second option

i.e B

b. F stat =2

c. F from the table df1=4 df2=30 F value from the table= 4.62

d. as F < F stat we accept null hypothesis. There is no significant evidence that means are differing each other

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