Chapter 13 Answers

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-3 (b) The computer result is shown below. One-way ANOVA: DC, DC-MK, MC Source DF SS MS F P Factor 2 1952.6 976.3 93.58 0.000 Error 33 344.3 10.4 Total 35 2296.9 S = 3.230 R-Sq = 85.01% R-Sq(adj) = 84.10% Because the P-value < 0.01 we reject H 0 and conclude that the type of chocolate has an effect on cardiovascular health at  = 0.05 or  = 0.01. (c) The computer result is shown below. Fisher 95% Individual Confidence Intervals All Pairwise Comparisons Simultaneous confidence level = 88.02% DC subtracted from: Lower Center Upper -+---------+---------+---------+-------- DC+MK -18.041 -15.358 -12.675 (---*----) MC -18.558 -15.875 -13.192 (----*---) -+---------+---------+---------+-------- -18.0 -12.0 -6.0 0.0 DC+MK subtracted from: Lower Center Upper -+---------+---------+---------+-------- MC -3.200 -0.517 2.166 (---*----) -+---------+---------+---------+-------- -18.0 -12.0 -6.0 0.0 The top intervals show differences between the mean antioxidant capacity for DC+MK – DC and MC – DC. Because these intervals are entirely within the negative range (do not include zero) there are significant differences between DC+MK and DC, and MC and DC. This implies that dark chocolate increases the mean antioxidant capacity of the subjects’ blood plasma. (d) The normal probability plot and the residual plots show that the model assumptions are reasonable. Residual Percent 5 0 -5 -10 99 90 50 10 1 Fitted Value Residual 116 112 108 104 100 5 0 -5 -10 Residual Frequency 6 4 2 0 -2 -4 -6 -8 16 12 8 4 0 Normal Probability Plot of the Residuals Residuals Versus the Fitted Values Histogram of the Residuals Residual Plots for DC, DC-MK, MC

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-6 (b) Residuals are acceptable. 13-7 The compressive strength of concrete is being studied, and four different mixing techniques are being investigated. The following data have been collected. (a) Test the hypothesis that mixing techniques affect the strength of the concrete. Use α = 0.05. (b) Find the P -value for the F -statistic computed in part (a). (c) Analyze the residuals from this experiment. (a) Analysis of Variance for STRENGTH Source DF SS MS F P TECHNIQU 3 489740 163247 12.73 0.000 200 160 125 5 4 3 FLOW UNIFORMIT -1 0 1 -2 -1 0 1 2 Normal Score Residual Normal Probability Plot of the Residuals (response is obs) 200 190 180 170 160 150 140 130 120 1 0 -1 FLOW Residual Residuals Versus FLOW (response is UNIFORMI) 4.5 4.0 3.5 1 0 -1 Fitted Value Residual Residuals Versus the Fitted Values (response is UNIFORMI)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-9 (b) There is some indication of that the variability of the response may be increasing as the mean response increases. There appears to be an outlier on the normal probability plot. (c) 95% Confidence interval on the mean of coating type 1 29 . 149 71 . 140 4 2 . 16 131 . 2 00 . 145 4 2 . 16 131 . 2 00 . 145 1 1 15 , 015 . 0 1 15 , 025 . 0 1              n MS t y n MS t y E i E 99% confidence interval on the difference between the means of coating types 1 and 4. 14 . 24 36 . 7 4 ) 2 . 16 ( 2 947 . 2 ) 25 . 129 00 . 145 ( 4 ) 2 . 16 ( 2 947 . 2 ) 25 . 129 00 . 145 ( 2 2 4 1 4 1 15 , 005 . 0 4 1 4 1 15 , 005 . 0 4 1                        n MS t y y n MS t y y E E 13-10 An article in Environment International [1992, Vol. 18(4)] described an experiment in which the amount of radon released in showers was investigated. Radon-enriched water was used in the experiment, and six different orifice diameters were tested in shower heads. The data from the experiment are shown in the following table. 5 4 3 2 1 5 0 -5 -10 COATINGT Residual Residuals Versus COATINGT (response is CONDUCTI) 145 140 135 130 5 0 -5 -10 Fitted Value Residual Residuals Versus the Fitted Values (response is CONDUCTI) -10 -5 0 5 -2 -1 0 1 2 Normal Score Residual Normal Probability Plot of the Residuals (response is CONDUCTI)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-12 13-12 An article in the Materials Research Bulletin [1991, Vol. 26(11)] investigated four different methods of preparing the superconducting compound PbMo 6 S 8 . The authors contend that the presence of oxygen during the preparation process affects the material’s s uperconducting transition temperature T c . Preparation methods 1 and 2 use techniques that are designed to eliminate the presence of oxygen, and methods 3 and 4 allow oxygen to be present. Five observations on T c (in °K) were made for each method, and the results are as follows: (a) Is there evidence to support the claim that the presence of oxygen during preparation affects the mean transition temperature? Use α = 0.05. (b) What is the P -value for the F -test in part (a)? (c) Analyze the residuals from this experiment. (d) Find a 95% confidence interval on mean T c when method 1is used to prepare the material. (a) Analysis of Variance of PREPARATION METHOD Source DF SS MS F P PREPMETH 3 22.124 7.375 14.85 0.000 Error 16 7.948 0.497 Total 19 30.072 Reject H 0 (b) P -value  0 (c) There are some differences in the amount variability at the different preparation methods and there is some curvature in the normal probability plot. There are also some potential problems with the constant variance assumption apparent in the fitted value plot. 25 20 15 10 100 0 -100 RODDING Residual Residuals Versus RODDING (response is STRENGTH) 1600 1550 1500 100 0 -100 Fitted Value Residual Residuals Versus the Fitted Values (response is STRENGTH) 4 3 2 1 1 0 -1 -2 PREPMETH Residual Residuals Versus PREPMETH (response is TEMPERAT) 15 14 13 12 1 0 -1 -2 Fitted Value Residual Residuals Versus the Fitted Values (response is TEMPERAT)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-15 (a) Is there a difference in the cross-linker level? Draw comparative box plots and perform an analysis of variance. Use α = 0.05. (b) Find the P -value of the test. Estimate the variability due to random error. (c) Plot average domain spacing against cross-linker level and interpret the results. (d) Analyze the residuals from this experiment and comment on model adequacy. (a) ANOVA Source DF SS MS F P Factor 5 2.5858 0.5172 18.88 0.000 Error 30 0.8217 0.0274 Total 35 3.4075 Yes, the box plot and ANOVA show that there is a difference in the cross-linker level. (b) Anova table in part (a) showed the p- value = 0.000 < α = 0.05 . Therefore there is at least one level of cross-linker is different. The variability due to random error is SS E = 0.8217 (c) Domain spacing seems to increase up to the 0.5 cross-linker level and declines once cross-linker level reaches 1. (d) The normal probability plot and the residual plots show that the model assumptions are reasonable. Data 1 0.5 0 -0.5 -0.75 Crosslinker Level -1 9.2 9.0 8.8 8.6 8.4 8.2 8.0 Boxplot of Crosslinker Level -1, -0.75, -0.5, 0, 0.5, 1 Cross-linker level Mean domain spacing 1.0 0.5 0.0 -0.5 -1.0 8.9 8.8 8.7 8.6 8.5 8.4 8.3 8.2 8.1 8.0 Scatterplot of Mean domain spacing vs Cross-linker level

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-18 13-16 An article in Journal of Food Science [2001, Vol. 66(3), pp. 472 – 477] reported on a study of potato spoilage based on different conditions of acidified oxine (AO), which is a mixture of chlorite and chlorine dioxide. The data follow: (a) Do the AO solutions differ in the spoilage percentage? Use α = 0.05. (b) Find the P -value of the test. Estimate the variability due to random error. (c) Plot average spoilage against AO solution and interpret the results. Which AO solution would you recommend for use in practice? (d) Analyze the residuals from this experiment. (a) From the analysis of variance shown below, 07 . 4 8 , 3 , 05 . 0  F > F 0 = 3.43 there is no difference in the spoilage percentage when using different AO solutions. ANOVA Source DF SS MS F P AO solutions 3 3364 1121 3.43 0.073 Error 8 2617 327 Total 11 5981 (b) From the table above, the P-value = 0.073 and the variability due to random error is SS E = 2617. (c) A 400ppm AO solution should be used because it produces the lowest average spoilage percentage. Fitted Value Residual 3.900 3.875 3.850 3.825 3.800 3.775 3.750 0.5 0.0 -0.5 -1.0 Residuals Versus the Fitted Values (response is protein content) Diet RESI1 Lupins Barley+Lupins Barley 0.5 0.0 -0.5 -1.0 Scatterplot of RESI1 vs Diet AO solution Avg. Spoilage 400 300 200 100 0 70 60 50 40 30 20 Scatterplot of Avg. Spoilage vs AO solution

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-21 One-way ANOVA: ANN1, ANN2, ANN3 Source DF SS MS F P Factor 2 1.848 0.924 6.02 0.009 Error 21 3.225 0.154 Total 23 5.073 S = 0.3919 R-Sq = 36.43% R-Sq(adj) = 30.37% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ---------+---------+---------+---------+ ANN1 8 0.0094 0.0130 (--------*---------) ANN2 8 0.0035 0.0075 (--------*---------) ANN3 8 0.5951 0.6786 (---------*--------) ---------+---------+---------+---------+ 0.00 0.30 0.60 0.90 Pooled StDev = 0.3919 From the ANOVA table. the P-value = 0.009. (b) Plots of residual follow. There is substantially lower variability in the measurements when the mean is low. This is evident in the plot of residuals versus the fitted values. (c) A confidence interval for the mean for ANN2 is obtained as follows. t 0.05/2, 21 = 2.0796, and the pooled standard deviation is 0.3919 0.0035 – 2.0796 × 0.3919 √8 < 𝜇 2 < 0.0035 + 2.0796 × 0.3919 √8 −0.285 < 𝜇 2 < 0.292 However, because the variance differs between the ANNs, one might use only the data from ANN2 to construct a confidence interval. Then t 0.05/2, 7 = 2.3646 and the standard deviation is 0.0075 0.0035 – 2.3646 × 0.0075 √8 < 𝜇 2 < 0.0035 + 2.2.3646 × 0.0075 √8 −0.00277 < 𝜇 2 < 0.00977

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-22 13-19 An article in Fuel Processing Technology (“ Application of the Factorial Design of Experiments to Biodiesel Production from Lard,” 2009, Vol. 90, pp. 1447– 1451) described an experiment to investigate the effect of potassium hydroxide in synthesis of biodiesel. It is suspected that potassium hydroxide (PH) is related to fatty acid methyl esters (FAME) which are key elements in biodiesel. Three levels of PH concentration were used, and six replicates were run in a random order. Data are shown in the following table. (a) Construct box plots to compare the factor levels. (b) Construct the analysis of variance. Are there any differences in PH concentrations at α = 0.05? Calculate the P -value. (c) Analyze the residuals from the experiment. (d) Plot average FAME against PH concentration and interpret your results. (e) Compute a 95% confidence interval on mean FAME when the PH concentration is 1.2. 1.5 1.0 0.5 0.0 -0.5 -1.0 99 95 90 80 70 60 50 40 30 20 10 5 1 Residual Percent Normal Probability Plot (responses are ANN1, ANN2, ANN3) 0.6 0.5 0.4 0.3 0.2 0.1 0.0 1.25 1.00 0.75 0.50 0.25 0.00 -0.25 -0.50 Fitted Value Residual Versus Fits (responses are ANN1, ANN2, ANN3)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-23 (a) (b) Anova: Single Factor SUMMARY Groups Count Sum Average Variance 0.6 6 515.8 85.96667 1.498667 0.9 6 534.4 89.06667 0.138667 1.2 6 540.2 90.03333 0.654667 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 54.16444 2 27.08222 35.44793 2.07E-06 3.68232 Within Groups 11.46 15 0.764 Total 65.62444 17 There are significant differences between the groups. The P=value = 2.07E-6. (c) PH1.2 PH0.9 PH0.6 91 90 89 88 87 86 85 84 Data Boxplot of PH0.6, PH0.9, PH1.2

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-24 There is some concern with the normal probability plot, but it does not indicate a serious departure from assumptions. There is also some difference in variation between the groups. (d) The mean of FAME tends to increase with PH. ( e) t 0.05/2, 15 = 2.1315 and the pooled estimate of variance from MS(Error) is 0.764. The 95% confidence interval is 2 1 0 -1 -2 99 95 90 80 70 60 50 40 30 20 10 5 1 Residual Percent Normal Probability Plot (responses are PH0.6, PH0.9, PH1.2) 90 89 88 87 86 1.0 0.5 0.0 -0.5 -1.0 -1.5 -2.0 Fitted Value Residual Versus Fits (responses are PH0.6, PH0.9, PH1.2)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-25 90.03333 +/- 2.1315(0.764/6) 1/2 = (89.273, 90.794) 13-20 For each of the following exercises, use the previous data to complete these parts. (a) Apply Fisher’s LSD method with α = 0.05 and determine which levels of the factor differ. (b) Use the graphical method to compare means described in this section and compare your conclusions to those from Fisher’s LSD method. Chocolate type in Exercise 13-4. Use α = 0.05. (a) Source DF SS MS F P Chocolate 2 1952.6 976.3 93.58 0.000 Error 33 344.3 10.4 Total 35 2296.9 S = 3.230 R-Sq = 85.01% R-Sq(adj) = 84.10% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ---+---------+---------+---------+------ DC 12 116.06 3.53 (---*---) DC+MK 12 100.70 3.24 (--*---) MC 12 100.18 2.89 (--*---) ---+---------+---------+---------+------ 100.0 105.0 110.0 115.0 Pooled StDev = 3.23 Grouping Information Using Fisher Method Chocolate N Mean Grouping DC 12 116.058 A DC+MK 12 100.700 B MC 12 100.183 B Means that do not share a letter are significantly different. Fisher 95% Individual Confidence Intervals All Pairwise Comparisons among Levels of Chocolate Simultaneous confidence level = 88.02% Chocolate = DC subtracted from: Chocolate Lower Center Upper DC+MK -18.041 -15.358 -12.675 MC -18.558 -15.875 -13.192 Chocolate -+---------+---------+---------+-------- DC+MK (---*----) MC (----*---) -+---------+---------+---------+-------- -18.0 -12.0 -6.0 0.0 Chocolate = DC+MK subtracted from:

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-26 Chocolate Lower Center Upper -+---------+---------+---------+-------- MC -3.200 -0.517 2.166 (---*----) -+---------+---------+---------+-------- -18.0 -12.0 -6.0 0.0 (b) The standard error of a mean is 3.230/12 1/2 = 0.932. From the graphical method, group DC is significantly different from the others and this agrees with Fisher’s method. 13-21 Cotton percentage in Exercise 13-5. Use α = 0.05. Fisher's pairwise comparisons Family error rate = 0.264 Individual error rate = 0.0500 Critical value = 2.086 Intervals for (column level mean) - (row level mean) 15 20 25 30 20 -9.346 -1.854 25 -11.546 -5.946 -4.054 1.546 30 -15.546 -9.946 -7.746 -8.054 -2.454 -0.254 35 -4.746 0.854 3.054 7.054 2.746 8.346 10.546 14.546 Significant differences are detected between levels 15 and 20, 15 and 25, 15 and 30, 20 and 30, 20 and 35, 25 and 30, 25 and 35, and 30 and 35. 13-22 Flow rate in Exercise 13-6. Use α = 0.01. Fisher's pairwise comparisons Family error rate = 0.117 Individual error rate = 0.0500 Critical value = 2.131 Intervals for (column level mean) - (row level mean) 125 160 160 -1.9775 -0.2225 250 -1.4942 -0.3942 0.2608 1.3608 There are significant differences between levels 125 and 160. 13-23 Mixing technique in Exercise 13-7. Use α = 0.05. Fisher's pairwise comparisons Family error rate = 0.184 Individual error rate = 0.0500 Critical value = 2.179 Intervals for (column level mean) - (row level mean) 1 2 3 2 -360 -11 3 -137 48 212 397 4 130 316 93 479 664 442 Significance differences between levels 1 and 2, 1 and 4, 2 and 3, 2 and 4, and 3 and 4. 13-24 Circuit type in Exercise 13-8. Use α = 0.01.

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-27 Fisher's pairwise comparisons Family error rate = 0.0251 Individual error rate = 0.0100 Critical value = 3.055 Intervals for (column level mean) - (row level mean) 1 2 2 -18.426 3.626 3 -8.626 -1.226 13.426 20.826 No significant differences at α = 0.01. 13-25 Coating type in Exercise 13-9. Use α = 0.01. Fisher's pairwise comparisons Family error rate = 0.0649 Individual error rate = 0.0100 Critical value = 2.947 Intervals for (column level mean) - (row level mean) 1 2 3 4 2 -8.642 8.142 3 5.108 5.358 21.892 22.142 4 7.358 7.608 -6.142 24.142 24.392 10.642 5 -8.642 -8.392 -22.142 -24.392 8.142 8.392 -5.358 -7.608 Significant differences between 1 and 3, 1 and 4, 2 and 3, 2 and 4, 3 and 5, 4 and 5. 13-26 Preparation method in Exercise 13-12. Use α = 0.05. Fisher's pairwise comparisons Family error rate = 0.189 Individual error rate = 0.0500 Critical value = 2.120 Intervals for (column level mean) - (row level mean) 1 2 3 2 -0.9450 0.9450 3 1.5550 1.5550 3.4450 3.4450 4 0.4750 0.4750 -2.0250 2.3650 2.3650 -0.1350 There are significant differences between levels 1 and 3, 4; 2 and 3, 4; and 3 and 4. 13-27 Air voids in Exercise 13-13. Use α = 0.05. Fisher's pairwise comparisons Family error rate = 0.118 Individual error rate = 0.0500

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-28 Critical value = 2.080 Intervals for (column level mean) - (row level mean) 1 2 2 1.799 19.701 3 8.424 -2.326 26.326 15.576 Significant differences between levels 1 and 2; and 1 and 3. 13-28 Cross-linker Exercise 13-14. Use α = 0.05. (a) 1952 . 0 6 0274 . 0 2 042 . 2 2 25 , 025 . 0     b MS t LSD E Fisher 95% Individual Confidence Intervals All Pairwise Comparisons among Levels of Cross-linker level Simultaneous confidence level = 65.64% Cross-linker level = -0.5 subtracted from: Cross-linker level Lower Center Upper -0.75 -0.5451 -0.3500 -0.1549 -1 -0.7451 -0.5500 -0.3549 0 -0.1118 0.0833 0.2785 0.5 0.0215 0.2167 0.4118 1 -0.1451 0.0500 0.2451 Cross-linker level ---------+---------+---------+---------+ -0.75 (---*---) -1 (---*---) 0 (---*---) 0.5 (---*---) 1 (---*---) ---------+---------+---------+---------+ -0.50 0.00 0.50 1.00 Cross-linker level = -0.75 subtracted from: Cross-linker level Lower Center Upper -1 -0.3951 -0.2000 -0.0049 0 0.2382 0.4333 0.6285 0.5 0.3715 0.5667 0.7618 1 0.2049 0.4000 0.5951 Cross-linker level ---------+---------+---------+---------+ -1 (---*---) 0 (---*---) 0.5 (---*---) 1 (---*---) ---------+---------+---------+---------+ -0.50 0.00 0.50 1.00 Cross-linker level = -1 subtracted from:

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-29 Cross-linker level Lower Center Upper ---------+---------+---------+---------+ 0 0.4382 0.6333 0.8285 (---*---) 0.5 0.5715 0.7667 0.9618 (---*---) 1 0.4049 0.6000 0.7951 (---*---) ---------+---------+---------+---------+ -0.50 0.00 0.50 1.00 Cross-linker level = 0 subtracted from: Cross-linker level Lower Center Upper 0.5 -0.0618 0.1333 0.3285 1 -0.2285 -0.0333 0.1618 Cross-linker level ---------+---------+---------+---------+ 0.5 (---*---) 1 (---*---) ---------+---------+---------+---------+ -0.50 0.00 0.50 1.00 Cross-linker level = 0.5 subtracted from: Cross-linker level Lower Center Upper 1 -0.3618 -0.1667 0.0285 Cross-linker level ---------+---------+---------+---------+ 1 (---*---) ---------+---------+---------+---------+ -0.50 0.00 0.50 1.00 Cross-linker levels -0.5, 0, 0.5 and 1 are not detected to differ. Cross-linker levels -0.75 and -1 are not detected to differ from one other, but both are significantly different to the others. (b) The mean values are 8.0667, 8.2667, 8.6167, 8.7, 8.8333, 8.6667 0676 . 0 6 0274 . 0 ˆ    b MS E X  The width of a scaled normal distribution is 6(0.0676) = 0.405 0 0.2 0.4 0.6 0.8 1 8 8.2 8.4 8.6 8.8 9

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-30 With a scaled normal distribution over this plot, the conclusions are similar to those from the LSD method. 13-29 Diets in Exercise 13-15. Use α = 0.01. (a) There is no significant difference in protein content between the three diet types. Fisher 99% Individual Confidence Intervals All Pairwise Comparisons among Levels of C4 Simultaneous confidence level = 97.33% C4 = Barley subtracted from: C4 Lower Center Upper Barley+lupins -0.3207 -0.0249 0.2709 lupins -0.4218 -0.1260 0.1698 C4 -------+---------+---------+---------+-- Barley+lupins (-----------*-----------) lupins (-----------*-----------) -------+---------+---------+---------+-- -0.25 0.00 0.25 0.50 C4 = Barley+lupins subtracted from: C4 Lower Center Upper -------+---------+---------+---------+-- lupins -0.3911 -0.1011 0.1889 (-----------*-----------) -------+---------+---------+---------+-- -0.25 0.00 0.25 0.50 (b) The mean values are: 3.886, 3.8611, 3.76 (barley, b+l, lupins) From the ANOVA the estimate of  can be obtained Source DF SS MS F P C4 2 0.235 0.118 0.72 0.489 Error 76 12.364 0.163 Total 78 12.599 S = 0.4033 R-Sq = 1.87% R-Sq(adj) = 0.00% The minimum sample size could be used to calculate the standard error of a sample mean 081 . 0 25 163 . 0 ˆ    b MS E X  The graph would not show any differences between the diets. 13-30 Suppose that four normal populations have common variance σ 2 = 25 and means μ 1 = 50, μ 2 = 60, μ 3 = 50, and μ 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 α = 0.05. 55   ,  1 = -5,  2 = 5,  3 = -5,  4 = 5. ) 1 ( 4 ) 1 ( 3 1 , ) 25 ( 4 ) 100 ( 2         n n a a n n Various choices for n yield: n  2  a(n-1) Power=1-  4 4 2 12 0.80 5 5 2.24 16 0.90 Therefore, n = 5 is needed.

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-31 13-31 Suppose that five normal populations have common variance σ 2 = 100 and means μ 1 = 175, μ 2 = 190, μ 3 = 160, μ 4 = 200, and μ 5 = 215. How many observations per population must be taken so that the probability of rejecting the hypothesis of equality of means is at least 0.95? Use α = 0.01. 188   ,  1 = -13,  2 = 2,  3 = -28,  4 = 12,  5 = 27. ) 1 ( 5 ) 1 ( 4 1 , ) 100 ( 5 ) 1830 ( 2         n n a a n n Various choices for n yield: n  2  a(n – 1) Power =1-  2 7.32 2.7 5 0.55 3 10.98 3.13 10 0.95 Therefore, n = 3 is needed. 13-32 Suppose that four normal populations with common variance σ 2 are to be compared with a sample size of eight observations from each population. Determine the smallest value for ∑ 𝜏 ? 2 4 ?=1 / σ 2 that can be detected with power 90%. Use α = 0.05. From the top chart with v 1 = 3 (4 treatments – 1), v 2 = 28 (degrees of freedom for error = 4(7) = 28) , φ = 2. Then the smallest value of ∑ 𝜏 𝑖 2 4 𝑖=1 𝜎 2 is a φ 2 / n = φ 2 /2 = 4/2 = 2, where a = 4 and n = 8. 13-33 Suppose that five normal populations with common variance σ 2 are to be compared with a sample size of seven observations from each. Suppose that 𝜏 1 =…= 𝜏 4 = 0. What is the smallest value for 𝜏 5 2 / σ 2 that can be detected with power 90% and α = 0.01? From the bottom chart with v 1 = 4 (5 treatments – 1), v 2 = 30 (degrees of freedom for error = 5(6) = 30), obtain φ is approximately 2.2. Then the smallest value of ∑ 𝜏 𝑖 2 5 𝑖=1 𝜎 2 is a φ 2 / n = 5φ 2 /7 = 3.5, where a = 5 and n = 7. Section 13-3 13-34 An article in the Journal of the Electrochemical Society [1992, Vol. 139(2), pp. 524 – 532)] describes an experiment to investigate the low-pressure vapor deposition of polysilicon. The experiment was carried out in a large-capacity reactor at Sematech in Austin, Texas. The reactor has several wafer positions, and four of these positions were selected at random. The response variable is film thickness uniformity. Three replicates of the experiment were run, and the data are as follows: (a) Is there a difference in the wafer positions? Use α = 0.05. (b) Estimate the variability due to wafer positions. (c) Estimate the random error component. (d) Analyze the residuals from this experiment and comment on model adequacy. (a) Analysis of Variance for UNIFORMITY Source DF SS MS F P WAFERPOS 3 16.220 5.407 8.29 0.008 Error 8 5.217 0.652 Total 11 21.437 Reject H 0 , and conclude that there are significant differences among wafer positions.

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-32 (b) 585 . 1 3 652 . 0 407 . 5 ˆ 2      n MS MS E Treatments   (c) 652 . 0 ˆ 2   E MS  (d) Greater variability at wafer position 1. There is some slight curvature in the normal probability plot. 13-35 A textile mill has a large number of looms. Each loom is supposed to provide the same output of cloth per minute. To investigate this assumption, five looms are chosen at random, and their output is measured at different times. The following data are obtained: (a) Are the looms similar in output? Use α = 0.05. (b) Estimate the variability between looms. (c) Estimate the experimental error variance. (d) Analyze the residuals from this experiment and check for model adequacy. (a) Analysis of Variance for OUTPUT Source DF SS MS F P LOOM 4 0.3416 0.0854 5.77 0.003 Error 20 0.2960 0.0148 Total 24 0.6376 Reject H 0 , there are significant differences among the looms. 4 3 2 1 1 0 -1 WAFERPOS Residual Residuals Versus WAFERPOS (response is UNIFORMI) 4 3 2 1 1 0 -1 Fitted Value Residual Residuals Versus the Fitted Values (response is UNIFORMI) -1 0 1 -2 -1 0 1 2 Normal Score Residual Normal Probability Plot of the Residuals (response is uniformi)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-33 (b) 01412 . 0 5 0148 . 0 0854 . 0 ˆ 2      n MS MS E Treatments   (c) 0148 . 0 ˆ 2   E MS  (d) Residuals are acceptable. 13-36 In the book Bayesian Inference in Statistical Analysis (1973, John Wiley and Sons) by Box and Tiao, the total product yield for five samples was determined randomly selected from each of six randomly chosen batches of raw material. (a) Do the different batches of raw material significantly affect mean yield? Use α = 0.01. (b) Estimate the variability between batches. (c) Estimate the variability between samples within batches. (d) Analyze the residuals from this experiment and check for model adequacy. (a) Yes, the different batches of raw materi al significantly affect mean yield at α = 0.01 because the P-value is small. Source DF SS MS F P Batch 5 56358 11272 4.60 0.004 Error 24 58830 2451 -0.2 -0.1 0.0 0.1 0.2 -2 -1 0 1 2 Normal Score Residual Normal Probability Plot of the Residuals (response is OUTPUT) 5 4 3 2 1 0.2 0.1 0.0 -0.1 -0.2 LOOM Residual Residuals Versus LOOM (response is OUTPUT) 4.1 4.0 3.9 3.8 0.2 0.1 0.0 -0.1 -0.2 Fitted Value Residual Residuals Versus the Fitted Values (response is OUTPUT)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-34 Total 29 115188 (b) Variability between batches 2 . 1764 5 2451 11272 ˆ 2      n MS MS E Treatments   (c) Variability within batches 2451 ˆ 2   MSE  (d) The normal probability plot and the residual plots show that the model assumptions are reasonable. 13-37 An article in the Journal of Quality Technology [1981, Vol. 13(2), pp. 111 – 114)] described an experiment that investigated the effects of four bleaching chemicals on pulp brightness. These four chemicals were selected at random from a large population of potential bleaching agents. The data are as follows: (a) Is there a difference in the chemical types? Use α = 0.05. (b) Estimate the variability due to chemical types. (c) Estimate the variability due to random error. (d) Analyze the residuals from this experiment and comment on model adequacy. (a) Analysis of Variance for BRIGHTNENESS Source DF SS MS F P CHEMICAL 3 54.0 18.0 0.75 0.538 Error 16 384.0 24.0 Total 19 438.0 Fail to reject H 0 , there is no significant difference among the chemical types. Residual Percent 100 50 0 -50 -100 99 90 50 10 1 Fitted Value Residual 1600 1550 1500 100 50 0 -50 -100 Residual Frequency 100 75 50 25 0 -25 -50 -75 8 6 4 2 0 Observation Order Residual 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 100 50 0 -50 -100 Normal Probability Plot of the Residuals Residuals Versus the Fitted Values Histogram of the Residuals Residuals Versus the Order of the Data Residual Plots for Yield

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-35 (b) 2 . 1 5 0 . 24 0 . 18 ˆ 2       set equal to 0 (c) 0 . 24 ˆ 2   (d) Variability is smaller in chemical 4. There is some curvature in the normal probability plot. 13-38 Consider the vapor-deposition experiment described in Exercise 13-34. (a) Estimate the total variability in the uniformity response. (b) How much of the total variability in the uniformity response is due to the difference between positions in the reactor? (c) To what level could the variability in the uniformity response be reduced if the position-to-position variability in the Reactor could be eliminated? Do you believe this is a substantial reduction? (a) 237 . 2 ˆ ˆ ˆ 2 2 2       position total (b) 709 . 0 ˆ ˆ 2 2  total position   (c) It could be reduced to 0.6522. This is a reduction of approximately 71%. 13-39 Consider the cloth experiment described in Exercise 13-35. (a) Estimate the total variability in the output response. (b) How much of the total variability in the output response is due to the difference between looms? 4 3 2 1 10 5 0 -5 CHEMICAL Residual Residuals Versus CHEMICAL (response is BRIGHTNE) 82 81 80 79 78 10 5 0 -5 Fitted Value Residual Residuals Versus the Fitted Values (response is BRIGHTNE) 10 5 0 -5 2 1 0 -1 -2 Normal Score Residual Normal Probability Plot of the Residuals (response is BRIGHTNE)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-36 (c) To what level could the variability in the output response be reduced if the loom-to-loom variability could be eliminated? Do you believe this is a significant reduction? From 13-35 we have 01412 . 0 5 0148 . 0 0854 . 0 ˆ 2      n MS MS E Treatments   0148 . 0 ˆ 2   E MS  (a) The total variability is 0.0141 + 0.0148 = 0.0289. (b)The proportion due to the difference in looms is 0.01412/0.0289 = 48.8% (c) Variability could be reduced to 0.0148 if loom differences are eliminated. This is a substantial reduction of approximately one half. 13-40 Reconsider Exercise 13-8 in which the effect of different circuits on the response time was investigated. Suppose that the three circuits were selected at random from a large number of circuits. (a) How does this change the interpretation of the experiment? (b) What is an appropriate statistical model for this experiment? (c) Estimate the parameters of this model. (a) Now this is a random effects experiment. (b) ij i ij y       where i  and ij  are random variables. (c) We obtain the following ANOVA table. Source DF SS MS F P Factor 2 260.9 130.5 4.01 0.046 Error 12 390.8 32.6 Total 14 651.7 S = 5.707 R-Sq = 40.04% R-Sq(adj) = 30.04% 58 . 19 5 6 . 32 5 . 130 ˆ 2      n MS MS E Treatments   6 . 32 ˆ 2   E MS  13-41 Reconsider Exercise 13-15 in which the effect of different diets on the protein content of cow’s milk was investigated. Suppose that the three diets reported were selected at random from a large number of diets. To simplify, delete the last two observations in the diets with n = 27 (to make equal sample sizes). (a) How does this change the interpretation of the experiment? (b) What is an appropriate statistical model for this experiment? (c) Estimate the parameters of this model. (a) Instead of testing the hypothesis that the individual treatment effects are zero, we are testing whether there is variability in protein content between all diets. 0 : 0 : 2 1 2 0       H H (b) The statistical model is         n j a i y ij i ,..., 2 , 1 ,..., 2 , 1    ) , 0 ( ~ 2   N i and ) , 0 ( ~ 2    N i (c) The last TWO observations were omitted from two diets to generate equal sample sizes with n = 25. ANOVA: Protein versus DietType Analysis of Variance for Protein

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-37 Source DF SS MS F P DietType 2 0.2689 0.1345 0.82 0.445 Error 72 11.8169 0.1641 Total 74 12.0858 S = 0.405122 R-Sq = 2.23% R-Sq(adj) = 0.00% 1641 . 0 2   E MS  001184 . 0 25 1641 . 0 1345 . 0 2       n MS MS E tr   Section 13-4 13-42 Consider the following computer output from a RCBD. (a) How many levels of the factor were used in this experiment? (b) How many blocks were used in this experiment? (c) Fill in the missing information. Use bounds for the P -value. (d) What conclusions would you draw if α = 0.05? What would you conclude if α = 0.01? (a) MS factor = factor factor DF SS , DF factor = 3 6 . 64 8 . 193   factor factor MS SS The levels of the factor = DF for the factor + 1 = 3 + 1 = 4. Therefore, 4 levels of the factor are used in this experiment. This can also be obtained from the result that the error degrees of freedom equal the product of the degrees of freedom for factor and block. Let dfF and dfB denote the degrees of freedom for factors and blocks, respectively. Therefore, the total degrees of freedom = 15 = dfT + dfB + (dfT)(dfB) = dfT + 3 + 3(dfT). Therefore dfT = 3. (b) Because the number of blocks = DF of block +1 = 3 + 1 =4. There are 4 blocks used in this experiment. (c) From part a), DF factor = 3. F = 4713 . 14 464 . 4 6 . 64   error factor MS MS P-value < 0.0 DF error = DF Total – DF Factor - DF Block = 15 – 3 – 3 = 9. MS block = block block DF SS , SS block = MS block DF block = 4.464(9) = 40.176 (d) Because the P-value < 0.01, we reject H 0 . There are significance differences in the factor level means at  = 0.05 or  = 0.01.

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-38 13-43 Consider the following computer output from a RCBD. There are four levels of the factor and five blocks. (a) Fill in the missing information. Use bounds for the P -value. (b) What conclusions would you draw if α = 0.05? What would you conclude if α = 0.01? (a) Because there are 4 levels and 5 blocks, there are 20 trials. Therefore, df(Total)= 19 and df(Factor) = 3 and df(Block) = 4. Therefore, df(Error) = df(Total) – df(Factor) – df(Block) = 19 – 3 – 4 = 12. Also, SS(Factor) = MS(Factor)(df(Factor)) = 115.2067(3) = 345.620 Also, SS(Block) = MS(Block)(df(Block)) = 71.9775(4) = 345.620 Because F = 3.49809 = MS(Factor)/MS(Error), MS(Error) = 115.2067/3.49809 = 32.934 Therefore, SS(Error) = MS(Error) (df(Error)) = 32.934(12) = 395.210 Because F = 3.49809 with 3 numerator and 12 denominator degrees of freedom, the P -value = 0.0497. (b) Because the 0.01 < P -value < 0.05, reject H 0 at  = 0.05, but fail to reject H  at  = 0.01. 13-44 Exercise 13-4 introduced you to an experiment to investigate the potential effect of consuming chocolate on cardiovascular health. The experiment was conducted as a completely randomized design, and the exercise asked you to use the ANOVA to analyze the data and draw conclusions. Now assume that the experiment had been conducted as an RCBD with the subjects considered as blocks. Analyze the data using this assumption. What conclusions would you draw (using α = 0.05) about the effect of the different types of chocolate on cardiovascular health? Would your conclusions change if α = 0.01? The output from computer software follows. Source DF SS MS F P Factor 2 1952.64 976.322 147.35 0.000 Block 11 198.54 18.049 2.72 0.022 Error 22 145.77 6.626 Total 35 2296.95 S = 2.574 R-Sq = 93.65% R-Sq(adj) = 89.90% Because the P -value for the factor is near zero, there are significant differences in the factor level means at  = 0.05 or  = 0.01. 13-45 Reconsider the experiment of Exercise 13-5. Suppose that the experiment was conducted as a RCBD with blocks formed by days (denoted as columns in the data table). In the experiment, the primary interest is still in the effect of cotton percentage and day is considered a nuisance factor. (a) Consider day as a block, and re-estimate the ANOVA. (b) Does cotton percentage still affect strength at α = 0.05? (c) Compare the conclusions here with those obtained from the analysis without blocks. (a) Source DF SS MS F P Cotton 2 161.733 80.8667 11.78 0.004 Block 4 46.267 11.5667 1.68 0.246 Error 8 54.933 6.8667 Total 14 262.933

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-39 S = 2.620 R-Sq = 79.11% R-Sq(adj) = 63.44% (b) Because the cotton P -value = 0.004 < 0.05, the cotton percentage is a significant factor for strength of the fabric. (c) The ANOVA for the analysis without blocks follows. The cotton P-value is still much smaller than 0.05 so the cotton percentage is a significant factor for strength of the fabric. The MS error is slightly lower than for the blocked analysis. Therefore, blocking is not necessary in an experiment such as this. Source DF SS MS F P Cotton 2 161.73 80.87 9.59 0.003 Error 12 101.20 8.43 Total 14 262.93 S = 2.904 R-Sq = 61.51% R-Sq(adj) = 55.10% (d) The residual plots follow. There are no serious departures from the assumptions. 5.0 2.5 0.0 -2.5 -5.0 99 95 90 80 70 60 50 40 30 20 10 5 1 Residual Percent Normal Probability Plot (response is Strength)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-40 13-46 An article in Quality Engineering [“ Designed Experiment to Stabilize Blood Glucose Levels” (1999– 2000, Vol. 12, pp. 83 – 87)] described an experiment to minimize variations in blood glucose levels. The treatment was the exercise time on a Nordic Track cross-country skier (10 or 20 min). The experiment was blocked for time of day. The data were as follows: (a) Is there an effect of exercise time on the average blood glucose? Use α = 0.05. (b) Find the P -value for the test in part (a). (c) Analyze the residuals from this experiment. (a) Analysis of variance for Glucose Source DF SS MS F P Time 1 36.13 36.125 0.06 0.819 Min 1 128.00 128.000 0.21 0.669 Error 5 3108.75 621.750 Total 7 3272.88 No, there is no effect of exercise time on the average blood glucose. (b) P -value = 0.819 (c) The normal probability plot and the residual plots show that the model assumptions are reasonable. 20.0 17.5 15.0 12.5 10.0 7.5 5.0 4 3 2 1 0 -1 -2 -3 -4 -5 Fitted Value Residual Versus Fits (response is Strength)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-41 13-47 In “The Effect of Nozzle Design on the Stability and Performance of Turbulent Water Jets” ( Fire Safety Journal , August 1981, Vol. 4), C. The obald described an experiment in which a shape measurement was determined for several different nozzle types at different levels of jet efflux velocity. Interest in this experiment focuses primarily on nozzle type, and velocity is a nuisance factor. The data are as follows: (a) Does nozzle type affect shape measurement? Compare the nozzles with box plots and the analysis of variance. (b) Use Fisher’s LSD method to determine specific differences among the nozzles. Does a graph of the average (or standard deviation) of the shape measurements versus nozzle type assist with the conclusions? (c) Analyze the residuals from this experiment. (a) Analysis of Variance for SHAPE Source DF SS MS F P NOZZLE 4 0.102180 0.025545 8.92 0.000 VELOCITY 5 0.062867 0.012573 4.39 0.007 Error 20 0.057300 0.002865 Total 29 0.222347 Reject H 0 , nozzle type affects shape measurement. Residual Percent 50 25 0 -25 -50 99 90 50 10 1 Fitted Value Residual 110 105 100 20 0 -20 -40 Residual Frequency 20 10 0 -10 -20 -30 -40 3 2 1 0 Observation Order Residual 8 7 6 5 4 3 2 1 20 0 -20 -40 Normal Probability Plot of the Residuals Residuals Versus the Fitted Values Histogram of the Residuals Residuals Versus the Order of the Data Residual Plots for Glucose

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-42 (b) Fisher's pairwise comparisons Family error rate = 0.268 Individual error rate = 0.0500 Critical value = 2.060 Intervals for (column level mean) - (row level mean) 1 2 3 4 2 -0.15412 0.01079 3 -0.20246 -0.13079 -0.03754 0.03412 4 -0.24412 -0.17246 -0.12412 -0.07921 -0.00754 0.04079 5 -0.11412 -0.04246 0.00588 0.04754 0.05079 0.12246 0.17079 0.21246 There are significant differences between levels 1 and 3, 4; 2 and 4; 3 and 5; and 4 and 5. (c) The residual analysis shows that there is some inequality of variance. The normal probability plot is acceptable. 28.74 23.46 20.43 16.59 14.37 11.73 1.15 1.05 0.95 0.85 0.75 VELOCITY SHAPE 5 4 3 2 1 1.15 1.05 0.95 0.85 0.75 NOZZLE SHAPE 30 20 10 0.1 0.0 -0.1 VELOCITY Residual Residuals Versus VELOCITY (response is SHAPE) 1.0 0.9 0.8 0.7 0.1 0.0 -0.1 Fitted Value Residual Residuals Versus the Fitted Values (response is SHAPE) 5 4 3 2 1 0 . 1 0 . 0 - 0 . 1 N O Z Z L E R e s i d u a l R e s i d u a l s V e r s u s N O Z Z L E ( r e s p o n s e i s S H A P E )

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-43 13-48 In Design and Analysis of Experiments , 8th edition (John Wiley & Sons, 2012), D. C. Montgomery described an experiment that determined the effect of four different types of tips in a hardness tester on the observed hardness of a metal alloy. Four specimens of the alloy were obtained, and each tip was tested once on each specimen, producing the following data: (a) Is there any difference in hardness measurements between the tips? (b) Use Fisher’s LSD method to investigate spe cific differences between the tips. (c) Analyze the residuals from this experiment. (a) Analysis of Variance of HARDNESS Source DF SS MS F P TIPTYPE 3 0.38500 0.12833 14.44 0.001 SPECIMEN 3 0.82500 0.27500 30.94 0.000 Error 9 0.08000 0.00889 Total 15 1.29000 Reject H 0 , and conclude that there are significant differences in hardness measurements between the tips. (b) Fisher's pairwise comparisons Family error rate = 0.184 Individual error rate = 0.0500 Critical value = 2.179 Intervals for (column level mean) - (row level mean) 1 2 3 2 -0.4481 0.3981 3 -0.2981 -0.2731 0.5481 0.5731 4 -0.7231 -0.6981 -0.8481 0.1231 0.1481 -0.0019 Significant difference between tip types 3 and 4 (c) Residuals are acceptable. 0.1 0.0 -0.1 2 1 0 -1 -2 Normal Score Residual Normal Probability Plot of the Residuals (response is SHAPE)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-44 13-49 An article in the American Industrial Hygiene Association Journal (1976, Vol. 37, pp. 418 – 422) described a field test for detecting the presence of arsenic in urine samples. The test has been proposed for use among forestry workers because of the increasing use of organic arsenics in that industry. The experiment compared the test as performed by both a trainee and an experienced trainer to an analysis at a remote laboratory. Four subjects were selected for testing and are considered as blocks. The response variable is arsenic content (in ppm) in the subject’ s urine. The data are as follows: (a) Is there any difference in the arsenic test procedure? (b) Analyze the residuals from this experiment. (a) Analysis of Variance for ARSENIC Source DF SS MS F P TEST 2 0.0014000 0.0007000 3.00 0.125 SUBJECT 3 0.0212250 0.0070750 30.32 0.001 Error 6 0.0014000 0.0002333 Total 11 0.0240250 Fail to reject H 0 , there is no evidence of differences between the tests. 4 3 2 1 0.15 0.10 0.05 0.00 -0.05 -0.10 SPECIMEN Residual Residuals Versus SPECIMEN (response is HARDNESS) 4 3 2 1 0.15 0.10 0.05 0.00 -0.05 -0.10 TIPTYPE Residual Residuals Versus TIPTYPE (response is HARDNESS) 10.2 10.1 10.0 9.9 9.8 9.7 9.6 9.5 9.4 9.3 9.2 0.15 0.10 0.05 0.00 -0.05 -0.10 Fitted Value Residual Residuals Versus the Fitted Values (response is HARDNESS) -0.10 -0.05 0.00 0.05 0.10 0.15 -2 -1 0 1 2 Normal Score Residual Normal Probability Plot of the Residuals (response is hardness)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-45 (b) Some indication of variability increasing with the magnitude of the response. 13-50 An article in the Food Technology Journal (1956, Vol. 10, pp. 39 – 42) described a study on the protopectin content of tomatoes during storage. Four storage times were selected, and samples from nine lots of tomatoes were analyzed. The protopectin content (expressed as hydrochloric acid soluble fraction mg/kg) is in Table 13E-1. (a) The researchers in this study hypothesized that mean protopectin content would be different at different storage times. Can you confirm this hypothesis with a statistical test using α = 0.05? (b) Find the P -value for the test in part (a). (c) Which specific storage times are different? Would you agree with the statement that protopectin content decreases as storage time increases? (d) Analyze the residuals from this experiment. 4 3 2 1 0.02 0.01 0.00 -0.01 -0.02 -0.03 SUBJECT Residual Residuals Versus SUBJECT (response is ARSENIC) 3 2 1 0.02 0.01 0.00 -0.01 -0.02 -0.03 TEST Residual Residuals Versus TEST (response is ARSENIC)) 0.15 0.10 0.05 0.02 0.01 0.00 -0.01 -0.02 -0.03 Fitted Value Residual Residuals Versus the Fitted Values (response is ARSENIC) -0.03 -0.02 -0.01 0.00 0.01 0.02 -2 -1 0 1 2 Normal Score Residual Normal Probability Plot of the Residuals (response is ARSENIC)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-46 (a) Analysis of Variance of PROPECTIN Source DF SS MS F P STORAGE 3 1972652 657551 4.33 0.014 LOT 8 1980499 247562 1.63 0.169 Error 24 3647150 151965 Total 35 7600300 Reject H 0 , and conclude that the storage times affect the mean level of propectin. (b) P-value = 0.014 (c) Fisher's pairwise comparisons Family error rate = 0.196 Individual error rate = 0.0500 Critical value = 2.037 Intervals for (column level mean) - (row level mean) 0 7 14 7 -171 634 14 -214 -445 592 360 21 239 8 50 1045 813 856 There are differences between 0 and 21 days; 7 and 21 days; and 14 and 21 days. The propectin levels are significantly different at 21 days from the other storage times so there is evidence that the mean level of propectin decreases with storage time. However, differences such as between 0 and 7 days and 7 and 14 days were not significant so that the level is not simply a linear function of storage days. (d) Observations from lot 3 at 14 days appear unusual. Otherwise, the residuals are acceptable. 1000 500 0 -500 2 1 0 -1 -2 Normal Score Residual Normal Probability Plot of the Residuals (response is propecti)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-47 13-51 An experiment was conducted to investigate leaking current in a SOS MOSFETS device. The purpose of the experiment was to investigate how leakage current varies as the channel length changes. Four channel lengths were selected. For each channel length, five different widths were also used, and width is to be considered a nuisance factor. The data are as follows: (a) Test the hypothesis that mean leakage voltage does not depend on the channel length using α = 0.05. (b) Analyze the residuals from this experiment. Comment on the residual plots. (c) The observed leakage voltage for channel length 4 and width 5 was erroneously recorded. The correct observation is 4.0. Analyze the corrected data from this experiment. Is there evidence to conclude that mean leakage voltage increases with channel length? A version of the electronic data file has the reading for length 4 and width 5 as 2. It should be 20. (a) Analysis of Variance for LEAKAGE Source DF SS MS F P LENGTH 3 72.66 24.22 1.61 0.240 WIDTH 4 90.52 22.63 1.50 0.263 Error 12 180.83 15.07 Total 19 344.01 Fail to reject H 0 , mean leakage voltage does not depend on the channel length. (b) One unusual observation in width 5, length 4. There are some problems with the normal probability plot, including the unusual observation. 9 8 7 6 5 4 3 2 1 0 1000 500 0 -500 LOT Residual Residuals Versus LOT (response is PROTOPE) 20 10 0 1000 500 0 -500 STORAGE Residual Residuals Versus STORAGE (response is PROTOPEC) 1500 1000 500 0 1000 500 0 -500 Fitted Value Residual Residuals Versus the Fitted Values (response is PROTOPEC)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-48 (c) Analysis of Variance for LEAKAGE VOLTAGE Source DF SS MS F P LENGTH 3 8.1775 2.7258 6.16 0.009 WIDTH 4 6.8380 1.7095 3.86 0.031 Error 12 5.3100 0.4425 Total 19 20.3255 Reject H 0 . And conclude that the mean leakage voltage does depend on channel length. By removing the data point that was erroneous, the analysis results in a conclusion. The erroneous data point that was an obvious outlier had a strong effect the results of the experiment. Supplemental Exercises 13-52 Consider the following computer output. 4 3 2 1 10 5 0 LENGTH Residual Residuals Versus LENGTH (response is LEAKAGE) 5 4 3 2 1 10 5 0 WIDTH Residual Residuals Versus WIDTH (response is LEAKAGE) 10 5 0 10 5 0 Fitted Value Residual Residuals Versus the Fitted Values (response is LEAKAGE) 0 5 10 -2 -1 0 1 2 Normal Score Residual Normal Probability Plot of the Residuals (response is LEAKAGE)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-49 (a) How many levels of the factor were used in this experiment? (b) How many replicates were used? (c) Fill in the missing information. Use bounds for the P -value. (d) What conclusions would you draw if α = 0.05? What if α = 0.01? (a) Note that df(Factor) = df(Total) – df(Error) = 19 – 15 = 4. Because the number of levels for a factor = df(Factor) + 1, 5 levels were used in the experiment. (b) Total number of observations = df(Total) + 1 = 19 + 1 = 20. Because there are 5 levels used in this experiment, the number of replicates = 20/5 = 4. (c) From part (a), the df(Factor) = 4. SS(Factor) = SS(Total) – SS(Error) = 326.2 – 167.5 = 158.7. MS(Factor) = SS(Factor)/DF(Factor) = 158.7/4 = 39.675. MS(Error) = SS(Error)/DF(Error) = 167.5/15 = 11.167. F = MS(Factor)/MS(Error) = 39.675/11.167 = 3.553 0.025 < P-value < 0.05. (d) Because the P-value <  = 0.05 we reject H 0 for  = 0.05. There are significance differences in the factor level means at  = 0.05. Because the P-value >  = 0.01 we fail to reject H 0 for  = 0.01. There are not significance differences in the factor level means at  = 0.01. 13-53 Consider the following computer output. (a) How many levels of the factor were used in this experiment? (b) How many blocks were used? (c) Fill in the missing information. Use bounds for the P -value. (d) What conclusions would you draw if α = 0.05? What if α = 0.01? (a) Because MS = SS/df(Factor), df(Factor) = SS/MS = 126.880/63.4401 = 2. The number of levels = df(Factor) + 1 = 2 + 1 = 3. Therefore, 3 levels of the factor were used. (b) Because df(Total) = df(Factor) + df(Block) + df(Error) 11 = 2 + df(Block) + 6. Therefore, df(Block) = 3. Therefore, 4 blocks were used in the experiment. (c) From parts (a) and (b), df(Factor) = 3 and df(Block) = 2 SS(Error) = df(Error)MS(Error) = (6)2.7403 = 16.4418 F = MS(Factor)/MS(Error) = 63.4401/2.7403 = 23.15 From Appendix Table VI, P-value < 0.01 (d) Because the P-value < 0.01 we reject H 0 . There are significant differences in the factor level means at  = 0.05 or  = 0.01. 13-54 An article in Lubrication Engineering (December 1990) described the results of an experiment designed to investigate the effects of carbon material properties on the progression of blisters on carbon face seals. The carbon face seals are used extensively in equipment such as air turbine starters. Five different carbon materials were tested, and the surface roughness was measured. The data are as follows:

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-50 (a) Does carbon material type have an effect on mean surface roughness? Use α = 0.05. (b) Find the residuals for this experiment. Does a normal probability plot of the residuals indicate any problem with the normality assumption? (c) Plot the residuals versus 𝑦 ̂ ?? . Comment on the plot. (d) Find a 95% confidence interval on the difference between mean surface roughness for the EC10 and the EC1 carbon grades. (e) Apply the Fisher LSD method to this experiment. Summarize your conclusions regarding the effect of material type on surface roughness. (a) Analysis of Variance for SURFACE ROUGNESS Analysis of Variance for y Source DF SS MS F P Material 3 0.2402 0.0801 4.96 0.020 Error 11 0.1775 0.0161 Total 14 0.4177 Reject H 0 (b) One observation is an outlier.

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-51 (c) There appears to be a problem with constant variance. This may be due to the outlier in the data. (d) 95% confidence interval on the difference in the means of EC10 and EC1 602 . 0 118 . 0 2 ) 0161 . 0 ( 4 ) 0161 . 0 ( 201 . 2 ) 130 . 0 490 . 0 ( 2 ) 0161 . 0 ( 4 ) 0161 . 0 ( 201 . 2 ) 130 . 0 490 . 0 ( 4 1 4 1 2 1 11 , , 025 . 0 4 1 3 1 2 1 11 , , 025 . 0 4 1                            n MS n MS t y y n MS n MS t y y E E E E 13-55 An article in the IEEE Transactions on Components, Hybrids, and Manufacturing Technology [(1992, Vol. 15(2), pp. 146 – 153)] described an experiment in which the contact resistance of a brake-only relay was studied for three different materials (all were silver-based alloys). The data are as follows. -0.2 -0.1 0.0 0.1 0.2 0.3 -2 -1 0 1 2 Normal Score Residual Normal Probability Plot of the Residuals (response is Surf Rou)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-52 (a) Does the type of alloy affect mean contact resistance? Use α = 0.01. (b) Use Fisher’s LSD method to determine which means differ. (c) Find a 99% confidence interval on the mean contact resistance for alloy 3. (d) Analyze the residuals for this experiment. (a) Analysis of Variance for RESISTANCE Source DF SS MS F P ALLOY 2 10941.8 5470.9 76.09 0.000 Error 27 1941.4 71.9 Total 29 12883.2 Reject H 0 , the type of alloy has a significant effect on mean contact resistance. (b) Fisher's pairwise comparisons Family error rate = 0.119 Individual error rate = 0.0500 Critical value = 2.052 Intervals for (column level mean) - (row level mean) 1 2 2 -13.58 1.98 3 -50.88 -45.08 -35.32 -29.52 There are differences in the mean resistance for alloy types 1 and 3; and types 2 and 3. (c) 99% confidence interval on the mean contact resistance for alloy 3 83 . 147 97 . 132 10 9 . 71 771 . 2 4 . 140 10 9 . 71 771 . 2 4 . 140 1 3 27 , 005 . 0 3 27 , 005 . 0 3              n MS t y n MS t y E i E (d) Variability of the residuals increases with the response. The normal probability plot has some curvature in the tails, indicating a problem with the normality assumption. A transformation of the response should be conducted. 3 2 1 30 20 10 0 -10 -20 ALLOY Residual Residuals Versus ALLOY (response is RESISTAN)) 140 130 120 110 100 30 20 10 0 -10 -20 Fitted Value Residual Residuals Versus the Fitted Values (response is RESISTAN)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-53 13-56 An article in the Journal of Quality Technology [(1982, Vol. 14(2), pp. 80 – 89)] described an experiment in which three different methods of preparing fish were evaluated on the basis of sensory criteria, and a quality score was assigned. Assume that these methods have been randomly selected from a large population of preparation methods. The data are in the following table: (a) Is there any difference in preparation methods? Use α = 0.05. (b) Calculate the P -value for the F -statistic in part (a). (c) Analyze the residuals from this experiment and comment on model adequacy. (d) Estimate the components of variance. (a) Analysis of Variance for SCORE Source DF SS MS F P METHOD 2 13.55 6.78 1.68 0.211 Error 21 84.77 4.04 Total 23 98.32 Fail to reject H 0 (b) P -value = 0.211 (c) There is some curvature in the normal probability plot. There appears to be some differences in the variability for the different methods. The variability for method one is larger than the variability for method 3. -20 -10 0 10 20 30 -2 -1 0 1 2 Normal Score Residual Normal Probability Plot of the Residuals (response is RESISTAN) 3 2 1 0 -1 -2 -3 -4 -5 2 1 0 -1 -2 Normal Score Residual Normal Probability Plot of the Residuals (response is score)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-54 (d) 342 . 0 8 04 . 4 78 . 6 ˆ 2      n MS MS E Treatments   04 . 4 ˆ 2   E MS  13-57 An article in the Journal of Agricultural Engineering Research (1992, Vol. 52, pp. 53 – 76) described an experiment to investigate the effect of drying temperature of wheat grain on baking quality bread. Three temperature levels were used, and the response variable measured was the volume of the loaf of bread produced. The data are as follows: (a) Does drying temperature affect mean bread volume? Use α = 0.01. (b) Find the P -value for this test. (c) Use the Fisher LSD method to determine which means are different. (d) Analyze the residuals from this experiment and comment on model adequacy. (a) Analysis of Variance for VOLUME Source DF SS MS F P TEMPERATURE 2 16480 8240 7.84 0.007 Error 12 12610 1051 Total 14 29090 Reject H 0 . (b) P-value = 0.007 (c) Fisher's pairwise comparisons Family error rate = 0.116 Individual error rate = 0.0500 Critical value = 2.179 Intervals for (column level mean) - (row level mean) 70 75 75 -16.7 72.7 80 35.3 7.3 124.7 96.7 There are significant differences in the mean volume for temperature levels 70 and 80; and 75 and 80. The highest temperature results in the smallest mean volume. 3 2 1 3 2 1 0 -1 -2 -3 -4 -5 METHOD Residual Residuals Versus METHOD (response is SCORE)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-55 (d) There are some relatively small differences in the variability at the different levels of temperature. The variability decreases with the fitted values. There is an unusual observation on the normal probability plot. 13-58 An article in Agricultural Engineering (December 1964, pp. 672 – 673) described an experiment in which the daily weight gain of swine is evaluated at different levels of housing temperature. The mean weight of each group of swine at the start of the experiment is considered to be a nuisance factor. The data from this experiment are as follows: (a) Does housing air temperature affect mean weight gain? Use α = 0.05. (b) Use Fisher’s LSD method to determine which temperature levels are different. (c) Analyze the residuals from this experiment and comment on model adequacy. (a) Analysis of Variance of Weight Gain Source DF SS MS F P MEANWEIG 2 0.2227 0.1113 1.48 0.273 AIRTEMP 5 10.1852 2.0370 27.13 0.000 Error 10 0.7509 0.0751 Total 17 11.1588 Reject H 0 and conclude that the air temperature has an effect on the mean weight gain. (b) Fisher's pairwise comparisons Family error rate = 0.314 Individual error rate = 0.0500 Critical value = 2.179 Intervals for (column level mean) - (row level mean) 50 60 70 80 90 60 -0.9101 -50 0 50 -2 -1 0 1 2 Normal Score Residual Normal Probability Plot of the Residuals (response is vol) 80 75 70 50 0 -50 TEMPERAT Residual Residuals Versus TEMPERAT (response is VOLUME) 1250 1240 1230 1220 1210 1200 1190 1180 1170 50 0 -50 Fitted Value Residual Residuals Versus the Fitted Values (response is VOLUME)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-56 0.1034 70 -1.2901 -0.8868 -0.2766 0.1268 80 -0.9834 -0.5801 -0.2001 0.0301 0.4334 0.8134 90 -0.3034 0.0999 0.4799 0.1732 0.7101 1.1134 1.4934 1.1868 100 1.0266 1.4299 1.8099 1.5032 0.8232 2.0401 2.4434 2.8234 2.5168 1.8368 There are significant differences in the mean air temperature levels 50 and 70, 100; 60 and 90, 100; 70 and 90, 100; 80 and 90, 100; and 90 and 100. The mean of temperature level 100 is different from all the other temperatures. (c) There appears to be some problems with the assumption of constant variance. 13-59 An article in Communications of the ACM [(1987, Vol. 30(5), pp. 53 – 76] reported on a study of different algorithms for estimating software development costs. Six algorithms were applied to eight software development projects and the percent error in estimating the development cost was observed. The data are in Table 13E-2. (a) Do the algorithms differ in mean cost estimation accuracy? Use α = 0.05. (b) Analyze the residuals from this experiment. 200 150 100 0.5 0.0 -0.5 MEANWEIG Residual Residuals Versus MEANWEIG (response is WEIGHTGA) 100 90 80 70 60 50 0.5 0.0 -0.5 AIRTEMP Residual Residuals Versus AIRTEMP (response is WEIGHTGA) 2 1 0 0.5 0.0 -0.5 Fitted Value Residual Residuals Versus the Fitted Values (response is WEIGHTGA) 0.5 0.0 -0.5 2 1 0 -1 -2 Normal Score Residual Normal Probability Plot of the Residuals (response is wt gain)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-57 (c) Which algorithm would you recommend for use in practice? (a) Analysis of Variance for PCTERROR Source DF SS MS F P ALGORITH 5 2825746 565149 6.23 0.000 PROJECT 7 2710323 387189 4.27 0.002 Error 35 3175290 90723 Total 47 8711358 Reject H 0 , the algorithms are significantly different. (b) The residuals look acceptable, except there is one unusual point. 8 7 6 5 4 3 2 1 1000 500 0 -500 PROJECT Residual Residuals Versus PROJECT (response is PCTERROR) 6 5 4 3 2 1 1000 500 0 -500 ALGORITH Residual Residuals Versus ALGORITH (response is PCTERROR) 1000 500 0 1000 500 0 -500 Fitted Value Residual Residuals Versus the Fitted Values (response is PCTERROR)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-58 (c) The best choice is algorithm 5 because it has the smallest mean and a low variability. 13-60 An article in Nature Genetics [(2003, Vol. 34(1), pp. 85 –90)], “Treatment -Specific Changes in Gene Expression Discriminate in vivo Drug Response in Human Leukemia Cells,” reported the results of a study of gene expression as a function of different treatments for leukemia. Three treatment groups are mercaptopurine (MP) only, low-dose methotrexate (LDMTX) and MP, and high-dose methotrexate (HDMTX) and MP. Each group contained ten subjects. The responses from a specific gene are shown in Table 13E-3. (a) Check the normality of the data. Can you assume that these samples are from normal populations? (b) Take the logarithm of the raw data and check the normality of the transformed data. Is there evidence to support the claim that the treatment means differ for the transformed data? Use α = 0.1. (c) Analyze the residuals from the transformed data and comment on model adequacy. (a) The normal probability plot shows that the normality assumption is not reasonable. (b) The normal probability plot shows that the normality assumption is reasonable. 1000 500 0 -500 2 1 0 -1 -2 Normal Score Residual Normal Probability Plot of the Residuals (response is PCTERROR) Obs Percent 1500 1000 500 0 -500 99 95 90 80 70 60 50 40 30 20 10 5 1 Mean <0.005 293.8 StDev 309.4 N 30 AD 2.380 P-Value Probability Plot of Obs Normal - 95% CI

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-59 Source DF SS MS F P Treatments 2 1.188 0.594 2.57 0.095 Error 27 6.237 0.231 Total 29 7.425 There is evidence to support the claim that the treatment means differ at α = 0.1 for the transformed data since the P - value = 0.095. (c) The normal probability plot and the residual plots show that the model assumptions are reasonable. 13-61 Consider an ANOVA situation with a = 5 treatments. Let σ 2 = 9 and α = 0.05, and suppose that n = 4. (a) Find the power of the ANOVA F -test when μ 1 = μ 2 = μ 3 = 1, μ 4 = 3, and μ 5 = 2. (b) What sample size is required if you want the power of the F -test in this situation to be at least 0.90? log(obs) Percent 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 99 95 90 80 70 60 50 40 30 20 10 5 1 Mean 0.066 2.209 StDev 0.5060 N 30 AD 0.687 P-Value Probability Plot of log(obs) Normal - 95% CI Residual Percent 1.0 0.5 0.0 -0.5 -1.0 99 90 50 10 1 Fitted Value Residual 2.5 2.4 2.3 2.2 2.1 1.0 0.5 0.0 -0.5 -1.0 Residual Frequency 0.8 0.4 0.0 -0.4 -0.8 8 6 4 2 0 Observation Order Residual 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 1.0 0.5 0.0 -0.5 -1.0 Normal Probability Plot of the Residuals Residuals Versus the Fitted Values Histogram of the Residuals Residuals Versus the Order of the Data Residual Plots for log(obs)

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-60 (a)  =1.6, 2  =0.284,  =0.5333 Numerator degrees of freedom = 1 4 1     a Denominator degrees of freedom = 2 15 ) 1 (     n a From Chart Figure 13-6,   0.8 and the power = 1 -  = 0.2 (b) n  2  a(n-1)  Power = 1-  50 3.56 1.89 245 0.05 0.95 The sample size should be approximately n = 50. 13-62 Consider an ANOVA situation with a = 4 means μ 1 = 1, μ 2 = 5, μ 3 = 8, and μ 5 = 4. Suppose that σ 2 = 4, n = 4, and α = 0.05. (a) Find the power of the ANOVA F -test. (b) How large would the sample size have to be if you want the power of the F -test for detecting this difference in means to be at least 0.90? (a)  = (1+5+8+4)/4 = 4.5 and 5 . 2 25 . 6 ) 4 ( 4 ] ) 5 . 4 4 ( ) 5 . 4 8 ( ) 5 . 4 5 ( ) 5 . 4 1 [( 4 2 2 2 2 2             Numerator degrees of freedom = 1 3 1     a Denominator degrees of freedom = 2 12 ) 1 (     n a From Figure 13-6,  = 0.05 and the power = 1 -  = 0.95 (b) n  a(n-1)  Power = 1-  4 6.25 2.5 12 0.05 0.95 3 4.6875 2.165 8 0.25 0.75 The sample size should be approximately n = 4. 13-63 An article in Marine Biology [“ Allozymes an Morphometric Characters of Three Species of Mytilus in the Northern and Southern Hemispheres” (1991, Vol. 111, pp. 323– 333)] discussed the ratio of the anterior adductor muscle scar length to shell length for shells from five different geographic locations. The following table is part of a much larger data set from their research. (a) Are there differences in the mean ratios due to different locations at α = 0.05? Calculate the P -value. (b) Analyze the residuals from the experiment. In particular, comment on the normality assumption.  2

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-61 (a) Source DF SS MS F P Factor 4 0.003562 0.000891 4.66 0.006 Error 25 0.004782 0.000191 Total 29 0.008345 S = 0.01383 R-Sq = 42.69% R-Sq(adj) = 33.52% Level N Mean StDev Tillamook 6 0.08083 0.01312 Newport 6 0.07333 0.00905 Petersburg 6 0.10267 0.01792 Magadan 6 0.07950 0.01460 Tvarminne 6 0.09533 0.01297 Individual 95% CIs For Mean Based on Pooled StDev Level ---------+---------+---------+---------+ Tillamook (-------*-------) Newport (-------*-------) Petersburg (------*-------) Magadan (-------*-------) Tvarminne (-------*------) ---------+---------+---------+---------+ 0.075 0.090 0.105 0.120 Pooled StDev = 0.01383 Because the location P-value = 0.006 < 0.05, the effect of location is significant. (b) Box plots and residual plots follow. The plots of residuals do not indicate serious departures from assumptions. Tvarminne Magadan Petersburg Newport Tillamook 0.14 0.13 0.12 0.11 0.10 0.09 0.08 0.07 0.06 0.05 Data Boxplot of Tillamook, Newport, Petersburg, Magadan, Tvarminne

Applied Statistics and Probability for Engineers, 6 th edition December 31, 2013 13-62 13-64 An article in Bioresource Technology [“ Preliminary Tests on Nisin and Pediocin Production Using Waste Protein Sources: Factorial and Kinetic Studies” (2006, Vol. 97(4), pp. 605 – 613)] described an experiment in which pediocin was produced from waste protein. Nisin and pediocin are bacteriocins (compounds produced by bacteria that inhibit related strains) used for food preservation. Three levels of protein (g/L) from trout viscera extracts were compared. (a) Construct box plots of the data. What visual impression do you have from these plots? (b) Does the level of protein have an effect on mean pediocin production? Use α = 0.05. (c) Would you draw a different conclusion if α = 0.01 had been used? (d) Plot the residuals from the experiment and comment. In particular, comment on the normality of the residuals. (e) Find a 95% confidence interval on mean pediocin production when the level of protein is 2.50 g/L. (a) The following box plot suggests that the protein at level 1.67 produces a difference response than the other levels. 0.04 0.03 0.02 0.01 0.00 -0.01 -0.02 -0.03 99 95 90 80 70 60 50 40 30 20 10 5 1 Residual Percent Normal Probability Plot (responses are Tillamook, Newport, Petersburg, Magadan, Tvarminne) 0.105 0.100 0.095 0.090 0.085 0.080 0.075 0.070 0.04 0.03 0.02 0.01 0.00 -0.01 -0.02 -0.03 Fitted Value Residual Versus Fits (responses are Tillamook, Newport, Petersburg, Magadan, Tvarminne)