Level and include your interpretation of these intervals. Find Bonferroni joint confidence intervals for Bo and B1 with a 90% family confidence student version 14 doesĪ) Estimated value is 1.07 and rounded lambda is 1.00ī) The rounded value implies one raise Y to power of 1.00 which means noĨ. This can only be done using Minitab Version 15 or higher – i.e. Rounded lambda values, b) the interpretation of this value, and c) the Box-Cox plot. Even though you may not have found any assumption violations perform a Box-CoxĪnalysis on Y to see if any transformation is suggested. Linear regression function is appropriate.ħ. Ha: The linear regression function is not appropriateī) F-statistic is 0.95 and p-value is 0.507Ĭ) Since p-value is greater than 0.05 we fail to reject Ho and conclude plausible that
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Your decision based on a 0.05 level of significance, and d) Minitab copy of your ANOVAĪ) Ho: The linear regression function is appropriate
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Include:Ī) the null and alternative hypotheses, b) the correct F-statistic and p-value of the test, c) Perform a Lack of Fit Test to check if linear regression function is appropriate. Te s t for Equa l V a r ia nce s for R ES I1Ħ. The only condition for the Levene test is that the variable being tested is continuous.Ĭ) Since the p-value is greater than 0.05 we conclude that the assumption of equalĩ 5 % B onfe r r oni C onfide nce Inte r v a ls for StD e v s Violations to normality than is the F-test making the Levene test a better overall test of equal Significance, and d) Minitab copy of your plot.ī) The p-value is 0.533 NOTE: Remember that the Levene’s test is more robust against Include: a) the null andĪlternative hypotheses, b) the p-value of the test, c) your decision based on a 0.05 level of
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Do a check of equal variances by performing a Modified Levene Test. Ha: The residuals do not come from a normal distributionĬ) Since p-value is greater than 0.05 we fail to reject Ho and will conclude the assumptionĥ. Level of significance, and d) Minitab copy of your plot.Ī) Ho: The residuals come from a normal distribution Include: a) the nullĪnd alternative hypotheses, b) the p-value of the test, c) your decision based on a 0.05 Do a check of normality by using a probability plot of the residuals. Points from question two the absolute semi-studentized values are 1.97819 (row 18) and 2.06186Ĥ. No as all semi-studentized residuals have an absolute value less than four. The absolute values of the semi-studentized residuals you identified in question 2? Check for outliers using the semi-studentized method. Row 19 (X=5, Y=9), representing possible outliers.ģ. Yes there does appear to be a linear relationship with two points, row 18 (X=9, Y=6) and What two points appear to be potential outliers? Does there appear to be a linear relationship?
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