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Ch 16 - ANOVA

# Explain no since the p value is only 00003 much less

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Unformatted text preview: of words recalled for each subject was recorded. The sample mean percentage of words recalled was 68.3% for the short list, 48% for the medium list, and 39.2%for the long list. A one‐ way ANOVA was used to assess whether the length of the list had a significant effect on the percentage of words recalled. SS df Mean Square F Sig. Between Groups 2668.8 2 1334.4 15.77 .0003 Within Groups 1184.1 14 84.6 Total 3852.9 16 a. Some values in the ANOVA table are missing. Complete the above table. b. State the null and alternative hypotheses that the above F statistic is testing. H0: 1 = 2 = 3 versus Ha: at least one i is different c. Suppose the necessary assumptions hold. Using a 5% significance level, does it appear that the average percentage of words recalled is the same for the three different lengths of lists? Explain. No, since the p-value is only 0.0003 (much less than 0.05) we would reject H0. d. Multiple comparisons were performed using Tukey’s method. y (I) GROUP short list Use the results and circle the pairs that are significantly different at a 1% level. short versus medium medium list long list Mean Difference (I-J) 20.33 29.17 -20.33 8.83 -29.17 -8.83 (J) GROUP medium list long list short list long list short list medium list short versus long 99% Confidence Interval Lower Bound Upper Bound 1.06 39.61 10.79 47.54 -39.61 -1.06 -10.44 28.11 -47.54 -10.79 -28.11 10.44 medium versus long e. Give a 99% confidence interval for the population mean percentage of words recalled for the long list group. Recall that the sample mean based on the 6 subjects in the long list group was 39.2 percent. 39.2 +/‐ (2.98)(9.198/sqrt(6)) 39.2 +/‐ 11.19 (28.01, 50.39) 180 What if some conditions do not hold? You probably won’t be surprised to learn that the necessary conditions for using an analysis of variance F‐test don’t hold for all data sets. In Section 16.3, two methods are discussed that can...
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