Respond to each of the following questions using this partially completed one-way ANOVA table. Source SS DF MS F Between 470 Within 40 Total 1264 44 (a). How many different populations are being compared? (b). Fill in the ANOVA table with the missing (c). State the appropriate null and the alternative (d). Based on the analysis of variance F-test, what conclusion should be reached regarding the null hypothesis? Test Using an α =0.05.
The ANOVA table is -
Source | SS | DF | MS | F |
Between | 470 | 4 | 117.5 | 5.92 |
Within | 794 | 40 | 19.85 | |
Total | 1264 | 44 |
SS(Within) = SS(Total) - SS(Between) = 1264 - 470 = 794
MS = SS/DF
The value of F statistic
P-value =
a) Number of different populations are being compared = DF(Between) + 1 = 4 + 1 = 5
b) The null hypothesis is H_{0} : mean of all 5 populations are same
The alternative hypothesis is H_{1} : at least one mean is different
c) Since P-value = 0.000772 < 0.05, so we reject the null hypothesis and we can conclude that the five group means are significantly different.
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