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Chapter Ten The Analysis Of Variance

Chapter Ten The Analysis Of Variance. ANOVA Definitions > Factor The characteristic that differentiates the treatment or populations from one another. > Level (Treatments) The number of different treatments or populations.

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Chapter Ten The Analysis Of Variance

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  1. Chapter TenThe Analysis Of Variance

  2. ANOVA Definitions> Factor The characteristic that differentiates the treatment or populations from one another.> Level (Treatments) The number of different treatments or populations.

  3. Randomized Experiment Randomizing the order of sample observations will balance out any known or unknown nuisance variable that may influence the observed response.

  4. Mean Square for TreatmentsMSTr =J [(X1 – X)2 +…+ (XI – X)2]I – 1For I number of levelsFor J number of samples

  5. Mean Square for Error MSE = S21 + S22+…+ S2II

  6. Test Statistic for Single Factor ANOVAF = MSTr MSEWith 1 = I –1 & 2 = I(J-1)

  7. ANOVA on Single-Factor Experiment Several (I) Means All Normal (Same 2)Null Hypothesis: H0: u1= u2 =…= uITest Statistic:  = MSTr / MSEAlternative Hypothesis:Ha: (at least two means are not equal)Reject Region (upper tailed)F, I-1, I (J-1) Exp. (IJ – 1) DOF

  8. ANOVA on Single-Factor ExperimentExperiments were conducted to study whether commercial processing of various foods changes the concentration of essential elements for human consumption. One such experiment was to study the concentration of zinc in green beans. A batch of green beans was divided into 4 groups. The 4 groups were then randomly assigned to be measured (10) times each for zinc as follows: group 1 measured Raw; group 2 measured before Blanching; group 3 measured after Blanching; and group 4 measured after the final processing step. Ten independent measurements were taken from the 4 groups (treatments), yielding the following data: Zinc ConcentrationGroup 1 Group 2 Group 3 Group 4u1 = 2.01 u2 = 2.58 u3 = 2.10 u4 = 3.05S1 = 0.25 S2 = 0.50 S3 = 0.30 S4 = 1.00Test this hypothesis for significance at the 5% level.Measurements of this type are known to be Normal.

  9. ANOVA on Single-Factor Experiment ExampleThe coded values for the measure of elasticity (nt/m2) in plastic, prepared by two different processes A & B, for samples of (6) drawn randomly from each of the two processes are as follows:Group A Group B u1 = 7.28 u2 = 8.02S12 = 0.48 S22 = 0.71 Do the data present sufficient evidence to indicate a difference in mean elasticity for the two processes at a level of significance of α = .05? Measurements of this type are found to follow a Normal pdf.

  10. ANOVA on Single-Factor Experiment Several (I) Variances (Equal Samples J) All Normal Null Hypothesis: H0: 21 = 22 =…= 2ITest Statistic: 2 = (2.3026) Q / h“Bartlett’s Test”Alternative Hypothesis:Ha:(at least 2 variances are not equal)Reject Region (upper tailed test)22, I - 1Q = I(J–1)log(MSE) – (J-1)[log(S21)+…+log(S2I)]h = 1 + 1 I– 1 3(I-1) (J-1) I(J–1)

  11. ANOVA on Single-Factor Experiment Several (I) Variances (Equal Samples J) ExampleA study is designed to investigate the sulfur content of (5) major coal seams. Eight core samples are taken at randomly selected points within each seam. The measured response is the S% content. Before performing a Hypothesis Test on the data to detect any differences that might exist in the average sulfur content for these (5) seams, you are required to test the condition that the (5) seams all have the same population variance at a level of significance of .05. The summary statistics on the sulfur content of the (5) major coal seams follows:Seam 1 Seam 2 Seam 3 Seam 4 Seam 51= 1.66 2 = 1.17 3 = 1.46 4 = 0.88 5= 1.189S21=.175 S22=.144 S23=.115 S24=.123 S25=.074

  12. ANOVA on Single-Factor Experiment Several (I) Variances (Equal Samples J) ExampleUse Bartlett’s Hypothesis Test to determine whether it is reasonable to assume homogeneity of variances for the (4) treatment groups in the study whether commercial processing of various foods changes the concentration of essential elements for human consumption. Use  = .05.Rough rule of thumb: If the largest s is not much more than two times the smallest, it is reasonable to assume equal variances.

  13. ANOVA Multiple ComparisonsProcedures for identifying which ui’s significantly differ when H0 is rejected: > Tukey > Bonferroni > Duncan > Fisher LSD > Newman-Keuls

  14. Tukey’s T Method (Equal Samples)1. Select  & find Q, I, I(J-1)from Studentized Range Distribution Table A.10 on pg. 736. (m = I)2. Determine w = Q, I, I(J-1)*MSE/J3. List ui’s in increasing order & underline those pairs that differ by less than w. Any pair of ui’s not underscored by the same line corresponds to a pair of population or treatment means that are judged significantly different.

  15. Examples of Tukey’s MethodSummary Results: w = 5.37 x1 x5 x2 x3 x49.8 10.815.4 17.6 21.6Summary Results: w = 0.40 x5 x3 x2 x4 x16.1 6.3 6.8 7.3 7.5Summary Results: w = 0.40 x5 x3 x2 x4 x16.1 6.37.15 7.3 7.5

  16. ANOVA Multiple Comparison Tukey’s MethodA product development engineer is interested in maximizing the tensile strength of a new synthetic fiber. Previous experience indicates that the strength is affected by the % of cotton in the fiber. The engineer suspects that increasing the cotton content will increase the strength, at least initially. He decides to test (5) specimens at (5) levels of cotton content. Summary data follows:Cotton %: 15 20 25 30 35 Mean: 9.8 15.4 17.6 21.6 10.8 (psi) s : 3.35 3.13 2.07 2.61 2.86The Null Hypothesis H0 is rejected because the F statistic falls in the Reject Region. The % of cotton in the fiber significantly affects the mean tensile strength. Now use Tukey’s T method to find significant differences among the means. Use  = .05.

  17. ANOVA Multiple Comparison Tukey’s MethodAn experiment is developed to measure the effect that teaching methods have on a students’ performance. The following table lists the numerical grades on a standard arithmetic test given to 45 students divided randomly into (5) equal-sized groups. Groups 1 & 2 were taught by the current method. Groups 3, 4, & 5 were taught together for a number of days; on each day group 3 students were praised publicly for their previous work while group 4 students were criticized publicly. Group 5 students while hearing the praise and criticism of groups 3 & 4, were ignored. Group: 1 2 3 4 5Mean: 19.67 18.33 27.44 23.44 16.11 s2: 17.72 12.75 6.05 9.55 13.104 Test the null hypothesis that there is no difference in the mean grades produced by these teaching methods using  at .05. Then use Tukey’s T method to compare & illustrate the difference in the teaching methods.

  18. Least Significant Difference Method (Equal Samples1. Select  & find t/2, I(J-1) 2. Determine w = t/2, I(J-1) *2MSE/J3. Compare the observed difference between each pair of averages to the corresponding LSD. If | ui – uJ | > LSD, we conclude that the population mean ui and uJ differ.

  19. Example: Least Significant Difference MethodA manufacturer of paper used for making grocery bags is interested in improving the tensile strength of the product. Product engineering thinks that the tensile strength is a function of the hardwood concentration in the pulp and that the range of hardwood concentrations of practical interest is between 5 and 20%. You decide to investigate (4) levels of hardwood concentration. Six specimens at each of the (4) concentration levels are prepared and tested on a tensile tester in random order. The summary data from this experiment are shown in the following table:Hardwood (psi)Concentration > 5% 10% 15% 20% Mean: 10.00 15.67 17.00 21.17S2: 8.00 7.87 3.20 6.97Test the null hypothesis that there is no difference in the mean tensile strength produced by these (4) concentration levels using  at .01. Then use the LSD method at  = .05 to compare & illustrate the difference at each level of concentration.

  20. Example: Least Significant Difference MethodThe effective life of insulating fluids at an accelerated load of 50 m/sec2 is being studied. Test data have been obtained for (4) types of fluids. The summary results for (7) trials on each fluid are as follows: Life (in hours) at 50 m/sec2Fluid Type >> 1 2 3 4 Mean : 18.65 17.95 20.95 18.82 S2 : 3.81 3.44 3.53 2.42Is there any indication that the fluids differ at a significance level of .05?Which fluid or fluids would you select if the objective is long life? Use the Least Significance Difference method with an alpha of .05 to support you conclusion.

  21. -Error for Single Factor ANOVA F-TestNon-centrality parameter: = J  (I - )2 2For Non-central F distribution.With Degrees of Freedom:1 = I-12 = I(J-1)

  22. -Error for Single Factor ANOVA1) Find the value of 2 (Experience)2) Find the values of (i - ) 3) Compute 2 using: (Replaces ’) 2 = J  (i - )2I 24) Use Power Curves (pg. 422) to look-up power value:  = 1 – Power > Use appropriate set curves for 1 >  (with ) is on the horizontal axis > Move up to the curve associated with 2 > Find value of power value on vertical axis

  23. -Error ANOVA ExampleA product development engineer is interested in maximizing the tensile strength of a new synthetic fiber. Previous experience indicates that the strength is affected by the % of cotton in the fiber. The engineer suspects that increasing the cotton content will increase the strength, at least initially. He decides to test (5) specimens at (5) levels of cotton content. Summary data follows:Cotton %: 15 20 25 30 35 Mean: 9.8 15.4 17.6 21.6 10.8 (psi) s2 : 11.22 9.80 4.28 6.81 8.18What is the -error if the engineer is interested in rejecting the null hypothesis if the five treatment means are as follows: 15 = 11 20 = 12 25 = 15 30 = 18 35 = 19 Historically, the standard deviation of tensile strength is usually equal to 3 psi. Assume  = .01 for this test.

  24. -Error ANOVA ExampleSuppose that (5) means are being compared in a completely randomized experiment with  = .01. The design engineer would like to know how many samples to take if it is important to reject the Null Hypothesis with probability at least 0.90 if  (i - )2 = 25 & the population variance is known to be 5.0.

  25. -Error ANOVA ExampleSuppose that (4) Normal populations have common variance 2 = 25 and means 1 = 50, 2 = 60, 3 = 50, and 4 = 60. How many observations should be taken on each population so that the probability of rejecting the hypothesis of equality of means is at least 0.90? Use  = 0.05.

  26. Single-Factor ANOVA (Unequal Sample Sizes Ji)F = MSTr MSEWith 1 = I –1 & 2 = N-IWhere: MSTr = SSTrI – 1And MSE = SSE N -I

  27. ANOVA DefinitionsSum of Squares Treatment: SSTr =  Ji (i - )2iSum of Square Error: SSE =  (xij- i)2i jSum of Square Total: SST = SSTr + SSE

  28. Example of Unbalanced DesignTwenty-seven coins discovered in Cyprus were grouped into (4) classes, corresponding to (4) different coinages during the reign of King Manuel I Comnenus (1143-1180). Archaeologists are interested in whether there were significant differences in the Ag content of coins minted early and late in King Manuel’s reign. Test the H0 at  = .01. Summary data for testing the Ag content of early coins (group 1) to later coins (group 4) follows:Group Ji Mean SSE SSTr 1 9 6.74% 11.02 37.75 2 7 8.24% 3 4 4.88% 4 7 5.61%

  29. Multiple Comparisons (Unequal Samples)Tukey’s method modified:1. Select  & find Q, I, N-Ifrom Studentized Range Distribution Table A.10 on pg. 682 (m = I)2. Determine wij= Q, I, N-I*MSE x ( 1 + 1 ) 2 JiJjUses averages of pairs 1/Ji’s instead of 1/J.3. List ui’s in increasing order & underline those pairs that differ by less than wij.

  30. Example of Multiple Comparison(Unequal Sample Sizes)Use Tukey’s modified T method at  = .01 to compare & illustrate the difference in the means of Ag percentage in coins found on Cyprus.Group Ji Mean SSE SSTr 1 9 6.74% 11.02 37.75 2 7 8.24% 3 4 4.88% 4 7 5.61%

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