ANOVA Analysis of Treatment Means: Comparing Three Groups, Exercises of Statistics

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ANOVA

Source of Variation

SS Df MS F P-value F critic

Between Groups 46.9583 2 23.4792 9.3031 0.0013 3. Within Groups 53 21 2.

Total 99.9583 23

(b) Reject Ho if F > 3.47 (= Frit)

(c) SST = 46.96, SSE = 53.00 and Total SS = 99.96 (per the output above)

(d) See the ANOVA table above.

(e)

Reject Ho since F-test statistic = 9.30 is GREATER than our critical value of 3.47. So, based on the sample data, there IS sufficient evidence to indicate that at least one of the population mean significantly differ from the others.

(f)

t = 2.080 <-- using the t-table, do down from .05/2 = .025 and across from DF Error = 21.

Sp = sqrt (MSE) = 2.

(xbar2-xbar3) +/- t * sp * sqrt (1/n2 + 1/n3) (6-4.25) +/- 2.080 * 1.5887 * sqrt (1/10 + 1/8) 1.75 +/- 1. 1.75 - 1.5675 = +. 1.75 + 2.4901 = +3.

95% CI is: (.18, 3.32)

We can conclude that the treatment means for 2 and 3 are different since 0 is NOT in our CI.

e. What is your conclusion regarding the null hypothesis?

We reject the null hypothesis if F > 4..

Here F = 8 > 4.26.

Since the calculated value is greater than critical value, we reject Ho and conclude that at least one treatment mean is different from the others.

4. (a)

Source Sum of Squares df Mean Square F Treatment 127.1 2 63.6 3. Error 669.9 32 20. Total 797.0 34

Ho is not rejected since 3.04<3.29. hence there is no difference in the mean fight times.