[1.5%] ANOVA: Three groups of 5 students were tested in their ability to correctly answer a 10-question quiz under different formats. Group 1 had a True/False quiz, Group 2 had a Fill-In quiz, and Group 3 had a Multiple Choice quiz. Their scores were as follows:
True/False Fill-In Multiple Choice
6 3 6
8 5 10
9 6 6
7 7 8
10 5 7
Test whether there was a difference in performance on answering the same general questions under different formats. Use a .05 level of significance.
6a) P-value = ________________
6b) REJECT THE NULL or DO NOT REJECT THE NULL? (circle one)
6c) Can we conclude a difference in means? YES or NO (circle one)
6d) If the null is rejected, which specific differences are statistically significant?
T/F vs. Fill-in Fill-in vs. Multiple Choice T/F vs. Multiple Choice
(circle all that apply)
To test whether there was a difference in performance on answering the same general questions under different formats, we can use a one-way ANOVA test. The steps to conduct the test are as follows:
1. Calculate the mean score for each group.
Group 1 (True/False): (6 + 8 + 9 + 7 + 10) / 5 = 8
Group 2 (Fill-In): (3 + 5 + 6 + 7 + 5) / 5 = 5.2
Group 3 (Multiple Choice): (6 + 10 + 6 + 8 + 7) / 5 = 7.4
2. Calculate the overall mean score.
Overall mean = (8 + 5.2 + 7.4) / 3 = 6.533
3. Calculate the sum of squares between groups (SSB) and sum of squares within groups (SSW).
SSB = (5 * (8 - 6.533)^2) + (5 * (5.2 - 6.533)^2) + (5 * (7.4 - 6.533)^2) = 3.066
SSW = (6 - 8)^2 + (8 - 8)^2 + (9 - 8)^2 + (7 - 8)^2 + (10 - 8)^2 + (3 - 5.2)^2 + (5 - 5.2)^2 + (6 - 5.2)^2 + (7 - 5.2)^2 + (5 - 5.2)^2 + (6 - 7.4)^2 + (10 - 7.4)^2 + (6 - 7.4)^2 + (8 - 7.4)^2 + (7 - 7.4)^2 = 40.8
4. Calculate the degrees of freedom between groups (dfB) and degrees of freedom within groups (dfW).
dfB = k - 1 = 3 - 1 = 2 (k = number of groups)
dfW = N - k = 15 - 3 = 12 (N = total number of observations)
5. Calculate the mean squares between groups (MSB) and mean squares within groups (MSW).
MSB = SSB / dfB = 3.066 / 2 = 1.533
MSW = SSW / dfW = 40.8 / 12 = 3.4
6. Calculate the F-statistic.
F = MSB / MSW = 1.533 / 3.4 = 0.451
7. Look up the critical F-value from the F-distribution table with dfB = 2 and dfW = 12 at a significance level of 0.05. The critical F-value is 3.885.
8. Determine the p-value associated with the calculated F-statistic. You can use statistical software or an online calculator to find the p-value. For this example, the p-value is 0.651.
Now let's answer the given questions:
6a) P-value = 0.651
6b) DO NOT REJECT THE NULL
6c) Can we conclude a difference in means? NO
6d) If the null is rejected, which specific differences are statistically significant?
None of the differences are statistically significant since we do not reject the null hypothesis.
Note: Since the p-value (0.651) is greater than the significance level (0.05), we do not reject the null hypothesis. This means that we do not have enough evidence to conclude that there is a difference in performance on answering the same general questions under different formats.