One-Way ANOVA: Inference for Comparing Means and Pooling Variances - Prof. Ellen Gundlach, Study notes of Data Analysis & Statistical Methods

Download One-Way ANOVA: Inference for Comparing Means and Pooling Variances - Prof. Ellen Gundlach and more Study notes Data Analysis & Statistical Methods in PDF only on Docsity! 1 Chapter 12 Inference for One-Way ANOVA and Comparing the Means Learning goals for this chapter: Know how one-way ANOVA and 2-sample comparison of means techniques are related. Test the standard deviations to see if it is appropriate to pool the variances. Understand why it is important to pool the variances in one-way ANOVA. Explain and check the assumptions for doing one-way ANOVA. Calculate 2R and the estimate for . Write the correct hypotheses (including the words “population mean”) for one- way ANOVA. Use the F test statistic and P-value from SPSS to perform the one-way ANOVA test. State the conclusion to a one-way ANOVA test in terms of the story. Know when to use a Bonferroni multiple comparisons test. Use SPSS to perform the Bonferroni multiple comparisons test and interpret the output (both P-values and confidence intervals). State the conclusions to a Bonferroni multiple comparisons test in terms of the story. Interpret side-by-side boxplots and means plots in terms of the story. Recognize the response variable, factors, number of levels for each factor, and the total number of observations for a story. Identify from reading a story whether the scenario is one-way ANOVA. Use One-Way ANOVA when you have one categorical and one quantitative variable and you want to compare the means. If the categorical variable has 2 groups (gender = male or female, for example), use Ch. 7 two-sample comparison of means t-test. If the categorical variable has more than 2 groups (eye color = blue, brown, black, green, hazel, other), then use Ch 12 one-way ANOVA. ANOVA: ANalysis Of Variance: the method for comparing several means One-way ANOVA: F test for H0: 1= 2=. . . = I (all the population means are equal) Ha: not all the population means are equal (at least one is different) 2 Is there at least one population mean that is statistically significantly different from the others? When you first approach a problem which involves comparing more than 2 groups, here is what you should do: 1. Find the size (n), sample mean, and sample standard deviation of each group. You can then plot the means on a graph. Do histograms of each group to look for outliers and overall shape. 2. Find the 5-number summary (Min, Q1, Median, Q3, Max) for each group, and do side-by-side box plots to see how much overlap there is between the groups. 3. Run ANOVA. Standard Deviations: The standard deviation, , is assumed to be the same for all of the groups even though the sample sizes, ni, may be different. If the largest s < 2 * smallest s, we can use the methods based on the assumption of equal standard deviations (above). If we assume all the standard deviations are equal, each s is an estimate of . We combine these into a Pooled Estimator of . 2 2 2 1 1 2 2 1 2 ( 1) ( 1) ... ( 1) ( 1) ( 1) ... ( 1) I I P I n s n s n s s n n n (In the SPSS ANOVA output, Ps MSE ) The ANOVA output (see pg. 764 for more information): Source Sum of Squares Degrees of Freedom Mean Square F Sig. Groups (Between Groups) SSG DFG = I - 1 MSG = SSG DFG MSG MSE P-value Error (Within Groups) SSE DFE = N - I MSE = SSE DFE = sP 2 Total SST DFT = N - 1 MST = SST DFT 5 Click “OK.” d) Run the analysis of variance. Report the F statistic and P-value. Write the hypotheses that go with this information. What do you conclude? Using SPSS: Analyze  Compare Means  One-Way ANOVA. Move “density” into the “Dependent List” box. Move “treatment” into the “Factor” box. (If you want a plot of the means, you can click “options” and “Means plot.”) Click “OK.” e) What is the pooled estimate for the standard deviation? f) What is R2? ANOVA Bone Dens ity 7433.867 2 3716.933 7.978 .002 12579.500 27 465.907 20013.367 29 Between Groups Within Groups Total Sum of Squares df Mean Square F Sig. 6 This means that _______% of the variation in bone density is explained by membership in the groups of high jump, low jump, and control. The other _____% of the variation is due to rat-to-rat variation within each of these groups. If H0 is rejected in One-Way ANOVA, that means that we have evidence that at least one of the means is different. Which one(s)? Multiple Comparisons Use multiple comparisons method ONLY AFTER you have rejected H0 with the F test. To perform a multiple comparisons procedure, compute t statistics for all pairs of means using the formula: 1 1 i j ij p i j x x t s n n If **ijt t , we declare that the population means and i j are different. Otherwise, we conclude that the data do not distinguish between them. The value for t ** depends on which multiple comparisons procedure we choose. We will use Bonferroni’s multiple comparisons. SPSS will do this for you and give you the P-value. Another multiple comparisons method is simultaneous confidence intervals for all the possible differences: Find ** 1 1 ( )i j p i j x x t s n n for all the different pairs. If an interval contains 0, then that pair of means will not be declared significantly different. Example: Going back to the rat jumping data, use the Bonferroni multiple comparisons procedure to determine which pairs of means differ significantly. Summarize your results in a short report. Using SPSS: Analyze  Compare Means  One-Way ANOVA. Move “density” into “Dependent List” box. Move “treatment” into “Factor” box. 7 Click “Post Hoc” box. Click “Bonferroni.” Click “Continue.” Click “OK.” (Remember Control=1, Lowjump=2, Highjump=3.) Multiple Comparisons Dependent Variable: Bone Density Bonferroni -11.400 9.653 .744 -36.04 13.24 -37.600* 9.653 .002 -62.24 -12.96 11.400 9.653 .744 -13.24 36.04 -26.200* 9.653 .034 -50.84 -1.56 37.600* 9.653 .002 12.96 62.24 26.200* 9.653 .034 1.56 50.84 (J) Treatment 2 3 1 3 1 2 (I) Treatment 1 2 3 Mean Difference (I-J) Std. Error Sig. Lower Bound Upper Bound 95% Confidence Interval The mean difference is significant at the .05 level.*.