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Unformatted text preview: Stat 462, Spring 2008 Homework 5 Solution key Due: March 7, 2008 Prepared by: Esra Kurum 3.2 (5 points) (1) (2) 50 40 30 20 10 200 150 100 5050 Fitted Value Residual Residuals versus fits 2000 1500 1000 500 300 200 100100200300400500600 Fitted Value Residual Versus Fits (response is Y) 3.3(25 points, each part 5 points) a) The boxplot generated by Minitab: The ACT scores seem to be slightly left skewed with no apparent outliers. b) Dot plot of the residuals: According to the dot plot we can say, residuals are fairly symmetric and there may be two outliers. 51 Stat 462, Spring 2008 Homework 5 Solution key Due: March 7, 2008 Prepared by: Esra Kurum c) Plot generated by Minitab: Linearity, nonconstant error variance and sometimes outliers can be studied from this plot. In this case the residuals look fairly random and distributed in a horizontal band along the fitted values. There is no obvious pattern, so we can say they have a constant variance. d) The RyanJoiner statistic is 0.974 and its pvalue is < = 0.01, therefore the residuals are not normally distributed. Value from Table B6 for n = 100, = 0.05 is 0.987 or less The Normal Probability Plot generated by Minitab: RESI1 Percent 2 1123 99.9 99 95 90 80 70 60 50 40 30 20 10 5 1 0.1 Mean2.32777E15 StDev 0.6205 N 120 RJ 0.974 PValue <0.010 Probability Plot of RESI 1 Normal Also from the graph it can be seen that the residuals show a big departure from normality since they dont seem to form a straight line. e) H : errors have constant variance H a : errors have nonconstant variance 52 Stat 462, Spring 2008 Homework 5 Solution key Due: March 7, 2008 Prepared by: Esra Kurum Decision rule: Reject the null hypothesis if pvalue < = 0.01 Fail to reject null hypothesis if pvalue > = 0.01 Conclusion: Fail to reject the null hypothesis since pvalue (0.072)> Conclusion: Fail to reject the null hypothesis since pvalue (0....
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 Spring '08
 SENTURK,DAMLA
 Regression Analysis

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