multiple c Consider the following ANOVA table: The number of treatments is 13 5
ID: 3253239 • Letter: M
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multiple c Consider the following ANOVA table: The number of treatments is 13 5 3 12 In one-way analysis of variance, within-treatments variation is measured by: sum of squares for error. sum of squares for treatments. total sum of squares. standard deviation. Consider the following ANOVA table: The total number of observations is: 25 29 30 32 In one-way analysis of variance, if all the sample means are equal, then the: total sum of squares is zero. sum of squares for error is zero. sum of squares for treatments is zero. sum of squares for error equals sum of squares for treatments. The F-test statistic in a one-way ANOVA is equal to: MST/MSE SST/SSE MSE/MST SSE/SST The numerator and denominator degrees of freedom for the F-test in a one-way A are, respectively, (n - k) and (k - 1) (k - 1) and (n - k) (k - n) and (n - 1) (n - 1)and (k - n) In a single-factor analysis of variance, MST is the mean square for treatments and MSE is the mean square for error. The null hypothesis of equal population means is rejected if: MST is much smaller than MSE. MST is much larger than MSE. MST is equal to MSE. None of these choices.Explanation / Answer
Q.19 Number of treatments are = 4+ 1 = 5 (option b)
Q.20 Within treatments avariation are measured by sum of squares for treatments (option b)
Q.21 The total number of observations are 25 + 4 + 1 = 30 (option c0
Q.22 all sample mean are equal than sum of square for treatments is zero. (otion c)
Q>23 F statisitc = MST/ MSE (option A)
Q.24 Numerator degree of freedom = k-1
denominator degree of freedom = n - k (option b)
Q.25 If MST is much larger than MSE than null hypothesis for equal means can be rejected. (option B)
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