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.

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

Mean Square for Treatments MST r = J [(X 1 – X) 2 +…+ (X I – X) 2 ] I – 1 For I number of levels For J number of samples

Mean Square for Error MSE = S S 2 2 +…+ S 2 I I

Test Statistic for Single Factor ANOVA F = MST r MSE With 1 = I –1 & 2 = I(J-1)

ANOVA on Single-Factor Experiment Several (I) Means All Normal (Same  2 ) Null Hypothesis: H 0 : u 1 = u 2 =…= u I Test Statistic:  = MST r / MSE Alternative Hypothesis: H a : (at least two means are not equal) Reject Region (upper tailed)   F , I-1, I (J-1) Exp. (IJ – 1) DOF

ANOVA on Single-Factor Experiment Experiments 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 Concentration Group 1 Group 2 Group 3 Group 4 u 1 = 2.01 u 2 = 2.58 u 3 = 2.10 u 4 = 3.05 S 1 = 0.25 S 2 = 0.50 S 3 = 0.30 S 4 = 1.00 Test this hypothesis for significance at the 5% level. Measurements of this type are known to be Normal. ANOVA on Single-Factor Experiment Experiments 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 Concentration Group 1 Group 2 Group 3 Group 4 u 1 = 2.01 u 2 = 2.58 u 3 = 2.10 u 4 = 3.05 S 1 = 0.25 S 2 = 0.50 S 3 = 0.30 S 4 = 1.00 Test this hypothesis for significance at the 5% level. Measurements of this type are known to be Normal.

ANOVA on Single-Factor Experiment Example The coded values for the measure of elasticity (nt/m 2 ) 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 u 1 = 7.28 u 2 = 8.02 S 1 2 = 0.48 S 2 2 = 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.

ANOVA on Single-Factor Experiment Several (I) Variances (Equal Samples J) All Normal Null Hypothesis: H 0 :  2 1 =  2 2 =…=  2 I Test Statistic:  2 = (2.3026) Q / h “Bartlett’s Test” Alternative Hypothesis: H a :(at least 2 variances are not equal) Reject Region (upper tailed test)  2   2 , I - 1 Q = I(J–1)log(MSE) – (J-1)[log(S 2 1 )+…+log(S 2 I )] h = I – 1 3(I-1) (J-1) I(J–1)

ANOVA on Single-Factor Experiment Several (I) Variances (Equal Samples J) Example A 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 2Seam 3 Seam 4Seam 5  1 = 1.66  2 = 1.17  3 = 1.46  4 = 0.88  5 = S 2 1 =.175S 2 2 =.144S 2 3 =.115S 2 4 =.123 S 2 5 =.074

ANOVA on Single-Factor Experiment Several (I) Variances (Equal Samples J) Example Use 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.

ANOVA Multiple Comparisons Procedures for identifying which u i ’s significantly differ when H 0 is rejected: > Tukey > Bonferroni > Duncan > Fisher LSD > Newman-Keuls

Tukey’s T Method (Equal Samples) 1. Select  & find Q , I, I(J-1) from Studentized Range Distribution Table A.10 on pg (m = I) 2. Determine w = Q , I, I(J-1) *  MSE/J 3. List u i ’s in increasing order & underline those pairs that differ by less than w. Any pair of u i ’s not underscored by the same line corresponds to a pair of population or treatment means that are judged significantly different.

Examples of Tukey’s Method Summary Results: w = 5.37 x 1 x 5 x 2 x 3 x Summary Results: w = 0.40 x 5 x 3 x 2 x 4 x Summary Results: w = 0.40 x 5 x 3 x 2 x 4 x

ANOVA Multiple Comparison Tukey’s Method A 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 %: Mean: (psi) s : The Null Hypothesis H 0 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.

ANOVA Multiple Comparison Tukey’s Method An 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: Mean: s 2 : 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.

Least Significant Difference Method (Equal Samples 1. Select  & find t  /2, I(J-1) 2. Determine w = t  /2, I(J-1) *  2MSE/J 3. Compare the observed difference between each pair of averages to the corresponding LSD. If | u i – u J | > LSD, we conclude that the population mean u i and u J differ.

Example: Least Significant Difference Method A 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: S 2 : Test 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.

Example: Least Significant Difference Method The effective life of insulating fluids at an accelerated load of 50 m/sec 2 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/sec 2 Fluid Type >> Mean : S 2 : Is 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.

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

 -Error for Single Factor ANOVA 1) Find the value of  2 (Experience) 2) Find the values of  (  i -  ) 3) Compute  2 using: (Replaces  ’)  2 = J  (  i -  ) 2 I  2 4) 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

 -Error ANOVA Example A 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 %: Mean: (psi) s 2 : What 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.

 -Error ANOVA Example Suppose 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.

 -Error ANOVA Example Suppose 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.

Single-Factor ANOVA (Unequal Sample Sizes J i ) F = MSTr MSE With 1 = I –1 & 2 = N-I Where:MSTr = SSTr I – 1 AndMSE = SSE N -I

ANOVA Definitions Sum of Squares Treatment: SSTr =  J i (  i -  ) 2 i Sum of Square Error: SSE =  (x ij -  i ) 2 i j Sum of Square Total: SST = SSTr + SSE

Example of Unbalanced Design Twenty-seven coins discovered in Cyprus were grouped into (4) classes, corresponding to (4) different coinages during the reign of King Manuel I Comnenus ( ). 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 H 0 at  =.01. Summary data for testing the Ag content of early coins (group 1) to later coins (group 4) follows: Group J i Mean SSESSTr % % % %

Multiple Comparisons (Unequal Samples) Tukey’s method modified: 1. Select  & find Q , I, N- I from Studentized Range Distribution Table A.10 on pg (m = I) 2. Determine w ij = Q , I, N- I *  MSE x ( ) 2 J i J j Uses averages of pairs 1/ J i ’s instead of 1/ J. 3. List u i ’s in increasing order & underline those pairs that differ by less than w ij.

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 J i Mean SSESSTr % % % %