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Chapter 14: Problem 5

An article in Technometrics ["Exact Analysis of Means with Unequal Variances" (2002, Vol. 44, pp. \(152-160\) ) ] described the technique of the analysis of means (ANOM) and presented the results of an experiment on insulation. Four insulation types were tested at three different temperatures. The data are as follows: (a) Write a model for this experiment. (b) Test the appropriate hypotheses and draw conclusions using the analysis of variance with \(\alpha=0.05\) (c) Graphically analyze the interaction. (d) Analyze the residuals from the experiment. (e) Use Fisher's LSD method to investigate the differences between mean effects of insulation type. Use \(\alpha=0.05 .\) $$ \begin{array}{lllllll} \hline && {\text { Temperature ( }{ }^{\circ} \mathbf{F} \text { ) }} \\ \text { Insulation } &&{\mathbf{1}} && {\mathbf{2}} && {3} \\ & 6.6 & 4 & 4.5 & 2.2 & 2.3 & 0.9 \\ & 2.7 & 6.2 & 5.5 & 2.7 & 5.6 & 4.9 \\ 1 & 6 & 5 & 4.8 & 5.8 & 2.2 & 3.4 \\ & 3 & 3.2 & 3 & 1.5 & 1.3 & 3.3 \\ & 2.1 & 4.1 & 2.5 & 2.6 & 0.5 & 1.1 \\ 2 & 5.9 & 2.5 & 0.4 & 3.5 & 1.7 & 0.1 \\ & 5.7 & 4.4 & 8.9 & 7.7 & 2.6 & 9.9 \\ & 3.2 & 3.2 & 7 & 7.3 & 11.5 & 10.5 \\ 3 & 5.3 & 9.7 & 8 & 2.2 & 3.4 & 6.7 \\ & 7 & 8.9 & 12 & 9.7 & 8.3 & 8 \\ & 7.3 & 9 & 8.5 & 10.8 & 10.4 & 9.7 \\ 4 & 8.6 & 11.3 & 7.9 & 7.3 & 10.6 & 7.4 \end{array} $$

Short Answer

Expert verified
(a) Specify a two-way ANOVA model. (b) Perform ANOVA; check significance at \( \alpha = 0.05 \). (c) Use an interaction plot. (d) Analyze residuals for randomness. (e) Use Fisher's LSD for pairwise comparisons.

Step by step solution

01

Model Specification

The model for this experiment can be written as \( Y_{ijk} = \mu + \alpha_i + \beta_j + (\alpha \beta)_{ij} + \epsilon_{ijk} \), where:- \( Y_{ijk} \) is the response variable (insulation effectiveness) for the \( i \)-th insulation type, the \( j \)-th temperature, and the \( k \)-th observation,- \( \mu \) is the overall mean,- \( \alpha_i \) is the effect of the \( i \)-th insulation type,- \( \beta_j \) is the effect of the \( j \)-th temperature,- \( (\alpha \beta)_{ij} \) is the interaction effect between the \( i \)-th insulation type and the \( j \)-th temperature,- \( \epsilon_{ijk} \) is the random error term.
02

ANOVA Test Setup

To analyze the data, we'll perform an ANOVA test to see if there are any significant differences in the insulation effectiveness between different types and temperatures. Define the null hypothesis \( H_0 \) as no effect of insulation type and temperature, and no interaction effect. The alternative hypothesis \( H_a \) is that at least one effect is significant. Set the significance level to \( \alpha = 0.05 \).
03

ANOVA Calculation

Perform the ANOVA calculation using statistical software or tools. Calculate the F-statistic and p-values for the main effects and the interaction. For instance, obtain sums of squares due to Insulation Type, Temperature, Interaction, and Error.
04

Hypothesis Testing and Conclusion

Compare the p-values from the ANOVA table with \( \alpha = 0.05 \). If a p-value is less than 0.05, reject the null hypothesis for that factor. Draw conclusions about which effects (insulation type, temperature, interaction) are statistically significant.
05

Interaction Plot

Graphically analyze the interaction effect between insulation type and temperature using an interaction plot. This plot will help visualize any interactions, where lines that converge or cross indicate a possible interaction effect.
06

Residual Analysis

Calculate residuals by taking the difference between observed and predicted values. Plot residuals to check for patterns. Look for randomness in the residual plots; non-randomness suggests model inadequacy.
07

Fisher's LSD Method

Conduct Fisher's Least Significant Difference (LSD) test to determine which insulation types differ from each other. The LSD method involves pairwise comparisons of means and checks which differences are statistically significant at \( \alpha = 0.05 \). Calculate the confidence interval for each pairwise comparison to assess significance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Statistical Models
When conducting an experiment, a statistical model provides a mathematical representation of the relationships between different factors or variables. In our insulation experiment, the statistical model helps us understand how insulation type and temperature affect insulation effectiveness. The model is expressed as:
\[ Y_{ijk} = \mu + \alpha_i + \beta_j + (\alpha \beta)_{ij} + \epsilon_{ijk} \]
  • \( Y_{ijk} \): The response variable, indicating insulation effectiveness for the specific conditions of insulation type \(i\), temperature \(j\), and observation \(k\).
  • \( \mu \): The overall mean of insulation effectiveness across all conditions.
  • \( \alpha_i \): The effect of the \(i\)-th insulation type on insulation effectiveness.
  • \( \beta_j \): The effect of the \(j\)-th temperature on insulation effectiveness.
  • \( (\alpha \beta)_{ij} \): The interaction effect between the \(i\)-th insulation type and the \(j\)-th temperature.
  • \( \epsilon_{ijk} \): The random error term, capturing variability not explained by the other components.
This model allows us to see how each factor contributes to the overall effectiveness of the insulation and whether these factors interact with each other.
Hypothesis Testing
Hypothesis testing is a crucial statistical method that allows us to make informed conclusions about the effects in an experiment. In this context, we perform an ANOVA (Analysis of Variance) test.
We define our hypotheses as follows:
  • **Null Hypothesis \( H_0 \):** There is no significant effect of insulation type, temperature, or their interaction on insulation effectiveness.
  • **Alternative Hypothesis \( H_a \):** At least one of these factors has a significant effect.
The ANOVA helps determine if variations between groups (insulation types or temperatures) are more than just random error. We set a significance level, \( \alpha = 0.05 \). If a p-value for any factor or interaction is less than \( 0.05 \), we reject the null hypothesis, suggesting that the factor significantly influences insulation effectiveness.
Interaction Plot
An interaction plot is a helpful graphical representation that shows how different factors in an experiment interact with each other. For the insulation study, an interaction plot would display the insulation effectiveness across different temperatures and insulation types.
In the plot:
  • Each line represents a different insulation type or temperature.
  • The x-axis typically displays one factor, while the y-axis shows the response variable.
  • Crossover or non-parallel lines suggest an interaction effect; the insulation type might perform differently under varying temperatures.
Using an interaction plot, we can readily identify any interactions that may exist and interpret how they impact the results, guiding us to consider interactions when drawing conclusions from the data.
Residual Analysis
Residual analysis involves examining the differences between observed data and the values predicted by our statistical model. This step checks the accuracy and appropriateness of the model used.
Key steps in residual analysis include:
  • Calculating residuals by subtracting predicted values from actual observations.
  • Plotting residuals to identify any patterns. Ideally, residuals should scatter randomly around zero.
  • Random residuals suggest the model fits well, while patterns may indicate issues such as omitted variables or incorrect model assumptions.
By performing residual analysis, we assess model adequacy and identify areas for improvement, ensuring our conclusions about the insulation effects are robust and reliable.
Fisher's Least Significant Difference
Fisher's Least Significant Difference (LSD) method provides a straightforward approach to compare group means after finding significant effects in ANOVA. It's particularly useful for determining which insulation types significantly differ in effectiveness.
The LSD procedure involves:
  • Performing pairwise comparisons between group means.
  • Calculating confidence intervals for these differences.
  • Checking if intervals include zero. If not, there is a statistically significant difference at the set \( \alpha = 0.05 \).
This method offers clarity on the specific pairs of insulation types with different mean effects, facilitating more informed decisions about which insulation types are truly superior under the experimental conditions.

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Most popular questions from this chapter

An article in Quality Engineering ["Mean and Variance Modeling with Qualitative Responses: A Case Study" (1998-1999, Vol. 11, pp. 141-148)] studied how three active ingredients of a particular food affect the overall taste of the product. The measure of the overall taste is the overall mean liking score (MLS). The three ingredients are identified by the variables \(x_{1}, x_{2}\), and \(x_{3}\). The data are shown in the following table. $$ \begin{array}{crrrr} \hline \text { Run } & x_{1} & x_{2} & x_{3} & \text { MLS } \\ \hline 1 & 1 & 1 & -1 & 6.3261 \\ 2 & 1 & 1 & 1 & 6.2444 \\ 3 & 0 & 0 & 0 & 6.5909 \\ 4 & 0 & -1 & 0 & 6.3409 \\ 5 & 1 & -1 & 1 & 5.907 \\ 6 & 1 & -1 & -1 & 6.488 \\ 7 & 0 & 0 & -1 & 5.9773 \\ 8 & 0 & 1 & 0 & 6.8605 \\ 9 & -1 & -1 & 1 & 6.0455 \\ 10 & 0 & 0 & 1 & 6.3478 \\ 11 & 1 & 0 & 0 & 6.7609 \\ 12 & -1 & -1 & -1 & 5.7727 \\ 13 & -1 & 1 & -1 & 6.1805 \\ 14 & -1 & 1 & 1 & 6.4894 \\ 15 & -1 & 0 & 0 & 6.8182 \end{array} $$ (a) Fit a second-order response surface model to the data. (b) Construct contour plots and response surface plots for MLS. What are your conclusions? (c) Analyze the residuals from this experiment. Does your analysis indicate any potential problems? (d) This design has only a single center point. Is this a good design in your opinion?

An article in Solid State Technology (1984, Vol. 29, pp. \(281-284\) ) described the use of factorial experiments in photolithography, an important step in the process of manufacturing integrated circuits. The variables in this experiment (all at two levels) are prebake temperature \((A),\) prebake time \((B),\) and exposure energy \((C),\) and the response variable is delta line width, the difference between the line on the mask and the printed line on the device. The data are as follows: \((1)=-2.30, a=-9.87, b=-18.20\), \(a b=-30.20, c=-23.80, a c=-4.30, b c=-3.80,\) and \(a b c=-14.70\) (a) Estimate the factor effects. (b) Use a normal probability plot of the effect estimates to identity factors that may be important. (c) What model would you recommend for predicting the delta line width response based on the results of this experiment? (d) Analyze the residuals from this experiment, and comment on model adequacy.

An experiment to study the effect of machining factors on ceramic strength was described at www.itl.nist.gov/div898/ handbook/. Five factors were considered at two levels each: \(A=\) Table Speed, \(B=\) Down Feed Rate, \(C=\) Wheel Grit, \(D=\) Direction, \(E=\) Batch. The response is the average of the ceramic strength over 15 repetitions. The following data are from a single replicate of a \(2^{5}\) factorial design. \(\begin{array}{rrrrrr}\text { A } & \text { B } & \text { C } & \text { D } & \text { E } & \text { Strength } \\ -1 & -1 & -1 & -1 & -1 & 680.45 \\ 1 & -1 & -1 & -1 & -1 & 722.48 \\ -1 & 1 & -1 & -1 & -1 & 702.14 \\ 1 & 1 & -1 & -1 & -1 & 666.93 \\ -1 & -1 & 1 & -1 & -1 & 703.67 \\ 1 & -1 & 1 & -1 & -1 & 642.14 \\ -1 & 1 & 1 & -1 & -1 & 692.98 \\ 1 & 1 & 1 & -1 & -1 & 669.26 \\\ -1 & -1 & -1 & 1 & -1 & 491.58 \\ 1 & -1 & -1 & 1 & -1 & 475.52 \\ -1 & 1 & -1 & 1 & -1 & 478.76 \\ 1 & 1 & -1 & 1 & -1 & 568.23 \\ -1 & -1 & 1 & 1 & -1 & 444.72\end{array}\) \(\begin{array}{rrrrrr}1 & -1 & 1 & 1 & -1 & 410.37 \\ -1 & 1 & 1 & 1 & -1 & 428.51 \\ 1 & 1 & 1 & 1 & -1 & 491.47 \\ -1 & -1 & -1 & -1 & 1 & 607.34 \\\ 1 & -1 & -1 & -1 & 1 & 620.8 \\ -1 & 1 & -1 & -1 & 1 & 610.55 \\ 1 & 1 & -1 & -1 & 1 & 638.04 \\ -1 & -1 & 1 & -1 & 1 & 585.19 \\ 1 & -1 & 1 & -1 & 1 & 586.17 \\ -1 & 1 & 1 & -1 & 1 & 601.67 \\ 1 & 1 & 1 & -1 & 1 & 608.31 \\ -1 & -1 & -1 & 1 & 1 & 442.9 \\ 1 & -1 & -1 & 1 & 1 & 434.41 \\ -1 & 1 & -1 & 1 & 1 & 417.66 \\ 1 & 1 & -1 & 1 & 1 & 510.84 \\ -1 & -1 & 1 & 1 & 1 & 392.11 \\\ 1 & -1 & 1 & 1 & 1 & 343.22 \\ -1 & 1 & 1 & 1 & 1 & 385.52 \\ 1 & 1 & 1 & 1 & 1 & 446.73\end{array}\) (a) Estimate the factor effects and use a normal probability plot of the effects. Identify which effects appear to be large. (b) Fit an appropriate model using the factors identified in part (a). (c) Prepare a normal probability plot of the residuals. Also, plot the residuals versus the predicted ceramic strength. Comment on the adequacy of these plots. (d) Identify and interpret any significant interactions. (e) What are your recommendations regarding process operating conditions?

Johnson and Leone (Statistics and Experimental Design in Engineering and the Physical Sciences, John Wiley, 1977\()\) described an experiment conducted to investigate warping of copper plates. The two factors studied were temperature and the copper content of the plates. The response variable is the amount of warping. The data are as follows: $$ \begin{array}{clccc} \hline {\begin{array}{c} \text { Temperature } \\ \left({ }^{\circ} \mathbf{C}\right) \end{array}} && {\text { Copper } \text { Content }(\%)} \\ \ \mathbf{4 0} & \mathbf{6 0} & \mathbf{8 0} & \mathbf{1 0 0} \\ 50 & 17,20 & 16,21 & 24,22 & 28,27 \\ 75 & 12,9 & 18,13 & 17,12 & 27,31 \\ 100 & 16,12 & 18,21 & 25,23 & 30,23 \\ 125 & 21,17 & 23,21 & 23,22 & 29,31 \end{array} $$ (a) Is there any indication that either factor affects the amount of warping? Is there any interaction between the factors? Use \(\alpha=0.05\) (b) Analyze the residuals from this experiment. (c) Plot the average warping at each level of copper content and compare the levels using Fisher's LSD method. Describe the differences in the effects of the different levels of copper content on warping. If low warping is desirable, what level of copper content would you specify? (d) Suppose that temperature cannot be easily controlled in the environment in which the copper plates are to be used. Does this change your answer for part (c)?

Heat-treating metal parts is a widely used manufacturing process. An article in the Journal of Metals (1989, Vol. 41 ) described an experiment to investigate flatness distortion from heat-treating for three types of gears and two heat-treating times. The data follow: $$ \begin{array}{lcc} \hline & \ {\text { Time (minutes) }} \\ \ \text { Gear Type } & \mathbf{9 0} & \mathbf{1 2 0} \\ \hline \text { 20-tooth } & 0.0265 & 0.0560 \\ & 0.0340 & 0.0650 \\ \text { 24-tooth } & 0.0430 & 0.0720 \\ & 0.0510 & 0.0880 \\ \text { 28-tooth } & 0.0405 & 0.0620 \\ & 0.0575 & 0.0825 \end{array} $$ (a) Is there any evidence that flatness distortion is different for the different gear types? Is there any indication that heat-treating time affects the flatness distortion? Do these factors interact? Use \(\alpha=0.05 .\) (b) Construct graphs of the factor effects that aid in drawing conclusions from this experiment. (c) Analyze the residuals from this experiment. Comment on the validity of the underlying assumptions.
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