In minitab x choose stat anova one way x enter score

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In Minitab x Choose Stat > ANOVA > One-Way . x Enter Score as the response variable and Disability as the Factor variable. x Click OK . Note: The “Response data are in a separate column for each factor level” choice would be used if each sample occurred in its own column. Scroll up to see the ANOVA table, which includes the Mean Square column, first for the treatment groups (the disability row) and then for the error term. In Comparing Groups (Quantitative) applet x Paste data into Sample data box, being sure to clarify which is the explanatory variable. Press Use Data . x Check the Show ANOVA Table box. (s) Verify that the F -statistic is then equal to the ratio of these two Mean Square values. Also verify the df values (degrees of freedom) for each row. Report the p-value. MST: MSE: F -statistic: df numerator: df denominator: p-value:
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Chance/Rossman, 2015 ISCAM III Investigation 5.4 344 Technical Conditions There are several technical conditions required for this randomization distribution to be well modeled by the F distribution: x The distribution for each group comes from a normal population. x The population standard deviation is the same for all the groups. x The observations are independent. When using this ANOVA F -test to compare several population means, we will check each condition as follows: x The normal probability plot (or dotplot or histogram) for each sample ’s responses is reasonably well-behaved. x The ratio of the largest standard deviation to the smallest standard deviation is at most 2. x The samples are independent random samples from each population. For simplicity, we will apply the same checks for a randomized experiment as well (with the last condition being met if the treatments are randomly assigned). If the first two conditions are not met, then suitable transformations may be useful. (t) Is there evidence of non-normality in these sample data? Calculate the ratio of the largest to smallest standard deviation; is this less than 2? (In the applet, check the box to Overlay F distribution to help assess the fit.) (u) Write a paragraph summarizing your conclusions for this study. Be sure to address issues of causation and generalizability as well as statistical significance. Study Conclusions The distributions of qualification scores in the five treatment groups look reasonably symmetric with similar standard deviations, so it is appropriate to apply the Analysis of Variance procedure. There is moderate evidence that the mean qualification ratings differ depending on the type of disability (p-value = 0.030). Descriptively, the candidates with crutches appear to have higher ratings on average, and the candidates with hearing impairments tend to have slightly lower ratings. (Other follow-up procedures could be used to determine which group means differ significantly from which others.) This was a randomized experiment, so we can attribute these differences in qualification scores to the disability shown. But we must be cautious about considering the students in this study to be representative of a larger population, particularly a population of employers who make actual hiring decisions.
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