Question

1 7 1 7 4 2. 8 5 3 5 8 6 3 3 9 6 5 9 6 LO

What is the calculated F-value for this dataset?

What is the critical F-value for this problem (assuming an α-level of 0.05?

Answer 1

Group 1 Group 2 Group 3 1 7 4 1 7 4 2 8 5 3 8 5 3 8 6 3 9 6 5 9 6 6 10 7 Sum= 24 66 43 3 8.25 5.375 Average = Σ Χ2 = 94 552 239 St. Dev. = 1.773 1.035 1.061 SS = 22 7.5 7.875 n = 8 8 8 The total sample size is N = 24. Therefore, the total degrees of freedom are dftotal = 24 -1 =23 Also, the between-groups degrees of freedom are dfbetween = 3-1=2, and the within-groups degrees of freedom are: dfwithin = dftotal - dfbetween = 23 - 2 = 21 First, we need to compute the total sum of values and the grand mean. The following is obtained x = 24+66 + 43 = 133 Also, the sum of squared values is = 94 + 552 +239 = 885 Based on the above calculations, the total sum of squares is computed as follows S Stotal = = 885 1332 = 147.958 24 The within sum of squares is computed as shown in the calculation below: SS within = S Swithin groups = 22 + 7.5+ 7.875 = 37.375 The between sum of squares is computed directly as shown in the calculation below:

Now that sum of squares are computed, we can proceed with computing the mean sum of squares: MSbetween = S Sbetween dfbetween 110.583 2 55.292 MS within SS within dfwithin 37.375 21 1.78 Now that sum of squares are computed, we can proceed with computing the mean sum of squares: M Sbetween = SSbetween dfbetween 110.583 2 = 55.292 37.375 MS within = S Swithin dfwithin = 1.78 21 Finally, with having already calculated the mean sum of squares, the F- statistic is computed as follows: F= M Sbetween M Swithin 55.292 1.78 31.067 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: 1 = 2 = 3 Ha: Not all means are equal The above hypotheses will be tested using an F-ratio for a One-Way ANOVA. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are dfi = 2 and df2 = 2, therefore, the rejection region for this F-test is R={F:F> F = 3.467) (3) Test Statistics F= M Sbetween MS within 55.292 1.78 = 31.067 (4) Decision about the null bypothesis Since it is observed that F = 31.067 > F. = 3.467, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p =0, and since p=0 < 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that not all 3 population means are equal, at the a = 0.05 significance level.

F = 31.067

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