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hw_02_sol - Solution Homework 2 Use the data from Homework...

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Solution - Homework 2 Use the data from Homework 1 to complete this assignment and regress Y on X and store the residuals. 1. Create boxplots for both X and Y. Are there any outliers? No outliers identified. See boxplot below. 8 6 4 2 0 8 6 4 2 0 X Y Boxplot of X, Y 2. Make a Scatterplot with Regression . Does there appear to be a linear relationship? What two points appear to be potential outliers? Yes there does appear to be a linear relationship with two points, row 18 (X=9, Y=6) and row 19 (X=5, Y=9), representing possible outliers. 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 X Y Scatterplot of Y vs X 1
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3. Check for outliers using the semi-studentized method. Are there any outliers? What are the absolute values of the semi-studentized residuals you identified in question 2? No as all semi-studentized residuals have an absolute value less than four. From the two points from question two the absolute semi-studentized values are 1.97819 (row 18) and 2.06186 (row 19). 4. Do a check of normality by using a probability plot of the residuals. Include: a) the null and alternative hypotheses, b) the p-value of the test, c) your decision based on a 0.05 level of significance, and d) Minitab copy of your plot. a) Ho: The residuals come from a normal distribution Ha: The residuals do not come from a normal distribution b) p-value is 0.942 c) Since p-value is greater than 0.05 we fail to reject Ho and will conclude the assumption of normality is plausible. d) 5.0 2.5 0.0 -2.5 -5.0 99 95 90 80 70 60 50 40 30 20 10 5 1 RESI1 Percent Mean -2.13163E-15 StDev 1.556 N 20 AD 0.158 P-Value 0.942 Probability Plot of RESI 1 Normal - 95% CI 5. Do a check of equal variances by performing a Modified Levene Test . Include: a) the null and alternative hypotheses, b) the p-value of the test, c) your decision based on a 0.05 level of significance, and d) Minitab copy of your plot. a) Ho: The variances are equal Ha: The variances are not equal b) The p-value is 0.533 NOTE: Remember that the Levene’s test is more robust against violations to normality than is the F-test making the Levene test a better overall test of equal variances. The only condition for the Levene test is that the variable being tested is continuous.
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