Question

Color affects us in many ways. For example, Exercise C.92 on page 498 describes an experiment showing that the color red appears to enhance men's attraction to women. Previous studies have also shown that athletes competing against an opponent wearing red perform worse, and students exposed to red before a test perform worse. \(^{3}\) Another study \(^{4}\) states that "red is hypothesized to impair performance on achievement tasks, because red is associated with the danger of failure." In the study, US college students were asked to solve 15 moderately difficult, five-letter, single-solution anagrams during a 5-minute period. Information about the study was given to participants in either red, green, or black ink just before they were given the anagrams. Participants were randomly assigned to a color group and did not know the purpose of the experiment, and all those coming in contact with the participants were blind to color group. The red group contained 19 participants and they correctly solved an average of 4.4 anagrams. The 27 participants in the green group correctly solved an average of 5.7 anagrams and the 25 participants in the black group correctly solved an average of 5.9 anagrams. Work through the details below to test if performance is different based on prior exposure to different colors. (a) State the hypotheses. (b) Use the fact that sum of squares for color groups is 27.7 and the total sum of squares is 84.7 to complete an ANOVA table and find the F-statistic. (c) Use the F-distribution to find the p-value. (d) Clearly state the conclusion of the test.

Short answer

The null hypothesis is rejected. There is significant statistical evidence to say that the performance is different based on exposure to different colors. The mean number of anagrams solved varies between different color groups.

Step-by-step solution

State the Hypotheses

Null hypothesis (H0): The means are all equal, there is no color effect, i.e., μ1 = μ2 = μ3. Alternative hypothesis (Ha): At least one mean is different, i.e., not all means are equal, demonstrating the effect of color.

Calculate the F-statistic

The formula for F-statistic in an ANOVA test is (Between group variability)/(Within group variability). The between group variability (SSB) is given as 27.7 and the total variability (SST) is 84.7. The within group variability (SSW) can be calculated by subtracting the SSB from SST, i.e., SSW = SST - SSB = 84.7 - 27.7 = 57. We know, degree of freedom for SSB (d.f_B) = 3 - 1 = 2 and for SSW (d.f_W) = total number of observations - number of groups = 71 - 3 = 68. Now we can calculate the mean square for between (MSB) = SSB / d.f_B = 27.7/2 = 13.85 and mean square for within (MSW) = SSW / d.f_W = 57/68 = 0.838. With these we can calculate, F = MSB / MSW = 13.85 / 0.838 = 16.52.

Calculate the P-value

Using an F-distribution table or statistical software to find the P-value associated with F(2,68), we get a P-value less than 0.01.

Conclusion

Since the P-value obtained in the previous step is less than any conventional level of significance (such as 0.05 or 0.01), the conclusion is to reject the null hypothesis. There is sufficient evidence to conclude that performance is different based on prior exposure to different colors.