STAT 462 - HW 5 Solution Key

# STAT 462 - HW 5 Solution Key - Stat 462 Spring 2008...

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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 50 0 -50 Fitted Value Residual Residuals versus fits 2000 1500 1000 500 0 300 200 100 0 -100 -200 -300 -400 -500 -600 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. 5-1

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Stat 462, Spring 2008 Homework 5 Solution key Due: March 7, 2008 Prepared by: Esra Kurum c) Plot generated by Minitab: Linearity, non-constant 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 Ryan-Joiner statistic is 0.974 and its p-value is < α = 0.01, therefore the residuals are not normally distributed. Value from Table B6 for n = 100, α = 0.05 is 0.987 or less The Normal Probability Plot generated by Minitab: RESI1 Percent 2 1 0 -1 -2 -3 99.9 99 95 90 80 70 60 50 40 30 20 10 5 1 0.1 Mean -2.32777E-15 StDev 0.6205 N 120 RJ 0.974 P-Value <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 don’t seem to form a straight line. e) H 0 : errors have constant variance H a : errors have non-constant variance 5-2
Stat 462, Spring 2008 Homework 5 Solution key Due: March 7, 2008 Prepared by: Esra Kurum Decision rule: Reject the null hypothesis if p-value < α = 0.01 Fail to reject null hypothesis if p-value > α = 0.01 Conclusion: Fail to reject the null hypothesis since p-value (0.072)>

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