Temperature Participant 40 Degrees 60 Degrees 80 Degrees Participant Totals A 3.22 3.42 3.46 P = 10.10 n 5 B 3.27 3.31 3.35 P = 9.93 k = 3 C ດ 3.47 3.93 3.69 P = 11.09 N = 15 D 3.43 3.91 3.74 P = 11.08 G 53.53 E 3.35 4.16 3.82 P = 11.33 ΣΧ = 192.1709 T = 16.74 SS = 0.0441 T = 18.73 SS = 0.5286 T = 18.06 SS 0.1575 The three treatments in the experiment define three populations of interest. You use an ANOVA to test the hypothesis that the three population means are equal. Present the results of your analysis in the following ANOVA table by entering the missing df values and selecting the correct values for the missing SS, MS, and F entries. (Note: For best results, retain at least two additional decimal points throughout your calculations. Depending on the order in which you do these calculations and the number of digits you retain, you may find slight rounding differences in the last digit between your answers and the answer choices.) ANOVA Table Source SS df MS F Between treatments Within treatments Between subjects 0.0609 0.1372 Error Within treatments Between subjects Error Total F Distribution Numerator Degrees of Freedom = 26 Denominator Degrees of Freedom = 26 kkkk 14 .5000 .5000 0.0609 0.1372 0 1 1.00 2 3 4 5 6 7 00 8 Use the Distributions tool to find the critical value for a = 0.01. The critical value is F = At a significance level of a = You 9 F 0.01, evaluate the null hypothesis that the population means for all treatments are equal. The null hypothesis is conclude that temperature affects spatial reasoning abilities.

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