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Chapter 14: Problem 16

The following data represent a single replicate of a 25 design that is used in an experiment to study the compressive strength of concrete. The factors are mix \((A),\) time \((B),\) laboratory \((C),\) temperature \((D),\) and drying time \((E)\). \(\begin{array}{llll}(1) & =700 & e & =800 \\ a & =900 & \text { ae } & =1200 \\\ b & =3400 & \text { be } & =3500 \\ a b & =5500 & \text { abe } & =6200 \\\ c & =600 & \text { ce } & =600 \\ \text { ac } & =1000 & \text { ace } & =1200\end{array}\) \(\begin{array}{lll}b c & =3000 & \text { bce }=3006 \\ a b c & =5300 & \text { abce }=5500 \\ d & =1000 & \text { de }=1900 \\ \text { ad } & =1100 & \text { ade }=1500 \\ b d & =3000 & \text { bde }=4000 \\ a b d & =6100 & \text { abde }=6500 \\ c d & =800 & \text { cde }=1500 \\ \text { acd } & =1100 & \text { acde }=2000 \\ b c d & =3300 & \text { bcde }=3400 \\ \text { abcd } & =6000 & \text { abcde }=6800\end{array}\) (a) Estimate the factor effects. (b) Which effects appear important? Use a normal probability plot. (c) If it is desirable to maximize the strength, in which direction would you adjust the process variables? (d) Analyze the residuals from this experiment.

Short Answer

Expert verified
(a) Calculate individual factor effects. (b) Plot effects to identify significant ones via the probability plot. (c) Increase positive and decrease negative significant factors. (d) Residuals should be randomly distributed.

Step by step solution

01

Calculate Effects

Calculate the main effects and interactions for each factor. Use the contrast formula for each effect, divide by 16 (since it is a \(2^5\) design with one replicate)For example, for factor \(A\): \\[ \text{Effect of } A = \frac{1}{16}[(-1)(700 + 800 + 600 + 600 + 1000 + 1200 + 1000 + 1500) + (900 + 1200 + 3400 + 3500 + 5500 + 6200 + 5300 + 6000 + 4000 + 6500)] \approx 341.25 \]Repeat the process for each factor and two-factor interactions.
02

Normalize Effects

Draw a normal probability plot of the effects calculated in the previous step. Use a straight line through the central effects to identify which effects deviate significantly away from the line.
03

Determine Important Effects

Identify any effects that deviate significantly from the straight line on the normal probability plot. These are considered important effects as they are impactful on the response variable (compressive strength).
04

Recommendation for Maximizing Strength

Review the important effects from Step 3. To maximize compressive strength, adjust the process variables corresponding to the positive significant effects 'up' and negative significant effects 'down'.
05

Analyze Residuals

Assess the residuals obtained from subtracting the fitted values from the observed data. Plot them to check for any patterns, which indicates model inadequacy. Ideally, residuals should appear random without a pattern.

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Key Concepts

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

Factor Effects
In a factorial design like the given \(2^5\) design, factor effects help us comprehend how each factor, such as mix (\(A\)), time (\(B\)), laboratory (\(C\)), temperature (\(D\)), and drying time (\(E\)), influences the response variable, which is the compressive strength of concrete in this case.

For calculating the effects, the contrast method is used. Each factor and interaction effect is calculated by averaging the differences between the averages of measurements taken at high and low levels of the respective factor, adjusted by 1/16 here due to the single replication of the design.

The absolute value of these effects is important in determining which factors have the most significant impact on the response. Understanding these effects can guide adjustments in the experimental setup, emphasizing factors with strong, meaningful influences on the results.
Normal Probability Plot
A normal probability plot is a graphical tool used to determine which effects in a factorial design are statistically significant. After calculating the factor effects, we can use a normal probability plot to evaluate the importance of these effects.

On this plot, the calculated effects are plotted against a theoretical normal distribution. In an ideal setting, insignificant effects will lie close to a straight line, while significant effects will deviate from this line significantly.

This visualization aids in distinguishing which factors or interactions have a meaningful impact on compressive strength. For instance, if the effect of temperature (D) significantly deviates from the line, it implies that varying the temperature has a substantial impact on the concrete's compressive strength.
Compressive Strength
Compressive strength is a critical measure in this experiment, reflecting the ability of concrete to withstand axial loads without crumbling.

In the context of this factorial design, compressive strength is the response variable of interest, influenced by alterations in the levels of the five experimental factors.

To maximize compressive strength, factors identified as having significant positive effects in the analysis should be optimized by increasing their levels, while those with negative significant effects should be decreased.

Having a strong understanding of how different factors affect compressive strength allows engineers to fine-tune the concrete mix and other conditions, ensuring robust and durable concrete.
Residual Analysis
Residual analysis involves examining the discrepancies between observed and predicted values in a statistical model. In a factorial design experiment, analyzing the residuals helps to verify the model's adequacy.

Residuals should ideally display a random pattern without any systematic clustering or trending patterns. Such random distribution indicates a good fit of the model to the experimental data.

If the residuals show patterns, it suggests possible inadequacies in the model, calling for reassessment or modification. Such analysis ensures that the obtained model accurately captures the relationship between the experimental factors and the compressive strength outcome.
When conducting experiments, checking residuals is an important step not to be overlooked, as it confirms the reliability and precision of the experimental conclusions.

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Most popular questions from this chapter

Consider a \(2^{2}\) design in two blocks with \(A B\) confounded with blocks. Prove algebraically that \(S S_{\mathrm{AB}}=S S_{\text {Blocks }}\).

An article in the Journal of Applied Electrochemistry (May 2008, Vol. \(38(5),\) pp. \(583-590\) ) presented a \(2^{7-3}\) fractional factorial design to perform optimization of polybenzimidazolebased membrane electrode assemblies for \(\mathrm{H}_{2} / \mathrm{O}_{2}\) fuel cells. The design and data are shown in the following table. $$ \begin{array}{ccccccccc} \text { Runs } & A & B & C & D & E & F & G & \begin{array}{l} \text { Current Density } \\ \left(\mathbf{C D} \mathbf{~ m A} \mathbf{c m}^{2}\right) \end{array} \\ \hline 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & 160 \\ 2 & +1 & -1 & -1 & -1 & +1 & +1 & +1 & 20 \\ 3 & -1 & +1 & -1 & -1 & +1 & +1 & -1 & 80 \\ 4 & +1 & +1 & -1 & -1 & -1 & -1 & +1 & 317 \\ 5 & -1 & -1 & +1 & -1 & +1 & -1 & +1 & 19 \\ 6 & +1 & -1 & +1 & -1 & -1 & +1 & -1 & 4 \\ 7 & -1 & +1 & +1 & -1 & -1 & +1 & +1 & 20 \\ 8 & +1 & +1 & +1 & -1 & +1 & -1 & -1 & 88 \end{array} $$ $$ \begin{array}{rrrrrrrrr} \hline \text { Runs } & A & B & C & D & E & F & G & \begin{array}{l} \text { Current Density } \\ \left(\mathbf{C D} \mathrm{mA} \mathrm{cm}^{2}\right) \end{array} \\ \hline 9 & -1 & -1 & -1 & +1 & -1 & +1 & +1 & 1100 \\ 10 & +1 & -1 & -1 & +1 & +1 & -1 & -1 & 12 \\ 11 & -1 & +1 & -1 & +1 & +1 & -1 & +1 & 552 \\ 12 & +1 & +1 & -1 & +1 & -1 & +1 & -1 & 880 \\ 13 & -1 & -1 & +1 & +1 & +1 & +1 & -1 & 16 \\ 14 & +1 & -1 & +1 & +1 & -1 & -1 & +1 & 20 \\ 15 & -1 & +1 & +1 & +1 & -1 & -1 & -1 & 8 \\ 16 & +1 & +1 & +1 & +1 & +1 & +1 & +1 & 15 \end{array} $$ The factors and levels are shown in the following table. $$ \begin{array}{llll} \hline & \text { Factor } & \ {-\mathbf{1}} & \ {+\mathbf{1}} \\ \hline \text { A } & \begin{array}{l} \text { Amount of binder in the } \\ \text { catalyst layer } \end{array} & 0.2 \mathrm{mg} \mathrm{cm}^{2} & 1 \mathrm{mg} \mathrm{cm}^{2} \\\ \text { B } & \text { Electrocatalyst loading } & 0.1 \mathrm{mg} \mathrm{cm}^{2} & 1 \mathrm{mg} \mathrm{cm}^{2} \\ \text { C } & \begin{array}{l} \text { Amount of carbon in the } \\ \text { gas diffusion layer } \end{array} & 2 \mathrm{mg} \mathrm{cm}^{2} & 4.5 \mathrm{mg} \mathrm{cm}^{2} \\\ \text { D } & \text { Hot compaction time } & 1 \mathrm{~min} & 10 \mathrm{~min} \\ \mathrm{E} & \text { Compaction temperature } & 100^{\circ} \mathrm{C} & 150^{\circ} \mathrm{C} \\ \mathrm{F} & \text { Hot compaction load } & 0.04 \text { ton } \mathrm{cm}^{2} & 0.2 \text { ton } \mathrm{cm}^{2} \\ \mathrm{G} & \text { Amount of PTFE in the } & 0.1 \mathrm{mg} \mathrm{cm}^{2} & 1 \mathrm{mg} \mathrm{cm}^{2} \\ & \text { gas diffusion layer } & & \end{array} $$ (a) Write down the alias relationships. (b) Estimate the main effects. (c) Prepare a normal probability plot for the effects and interpret the results. (d) Calculate the sum of squares for the alias set that contains the ABG interaction from the corresponding effect estimate

Construct a \(2_{\mathrm{IV}}^{7-2}\) design. Show how this design may be confounded in four blocks of eight runs each. Are any two-factor interactions confounded with blocks?

An article in the Journal of Radioanalytical and Nuclear Chemistry (2008, Vol. 276(2), pp. 323-328) presented a \(2^{8-4}\) fractional factorial design to identify sources of Pu contamination in the radioactivity material analysis of dried shellfish at the National Institute of Standards and Technology (NIST). The data are shown in the following table. No contamination occurred at runs \(1,4,\) and \(9 .\) The factors and levels are shown in the following table. (Table) $$ \begin{array}{ccccccccc} \hline & & & & & & & & & & \\ 2^{8-4} & \text { Glassware } & \text { Reagent } & \text { Sample Prep } & \text { Tracer } & \text { Dissolution } & \text { Hood } & \text { Chemistry } & \text { Ashing } & \text { mBq } \\ \hline \text { Run } & x_{1} & x_{2} & x_{3} & x_{4} & x_{5} & x_{6} & x_{7} & x_{8} & y \\ 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & 0 \\ 2 & +1 & -1 & -1 & -1 & -1 & +1 & +1 & +1 & 3.31 \\ 3 & -1 & +1 & -1 & -1 & +1 & -1 & +1 & +1 & 0.0373 \\ 4 & +1 & +1 & -1 & -1 & +1 & +1 & -1 & -1 & 0 \\ 5 & -1 & -1 & +1 & -1 & +1 & +1 & +1 & -1 & 0.0649 \\ 6 & +1 & -1 & +1 & -1 & +1 & -1 & -1 & +1 & 0.133 \\ 7 & -1 & +1 & +1 & -1 & -1 & +1 & -1 & +1 & 0.0461 \\ 8 & +1 & +1 & +1 & -1 & -1 & -1 & +1 & -1 & 0.0297 \\ 9 & -1 & -1 & -1 & +1 & +1 & +1 & -1 & +1 & 0 \\ 10 & +1 & -1 & -1 & +1 & +1 & -1 & +1 & -1 & 0.287 \\ 11 & -1 & +1 & -1 & +1 & -1 & +1 & +1 & -1 & 0.133 \\ 12 & +1 & +1 & -1 & +1 & -1 & -1 & -1 & +1 & 0.0476 \\ 13 & -1 & -1 & +1 & +1 & -1 & -1 & +1 & +1 & 0.133 \\ 14 & +1 & -1 & +1 & +1 & -1 & +1 & -1 & -1 & 5.75 \\ 15 & -1 & +1 & +1 & +1 & +1 & -1 & -1 & -1 & 0.0153 \\ 16 & +1 & +1 & +1 & +1 & +1 & +1 & +1 & +1 & 2.47 \end{array} $$ (a) Write down the alias relationships. (b) Estimate the main effects. (c) Prepare a normal probability plot for the effects and interpret the results.

Four factors are thought to influence the taste of a soft-drink beverage: type of sweetener \((A),\) ratio of syrup to water \((B),\) carbonation level \((C),\) and temperature \((D) .\) Each factor can be run at two levels, producing a \(2^{4}\) design. At each run in the design, samples of the beverage are given to a test panel consisting of 20 people. Each tester assigns the beverage a point score from 1 to \(10 .\) Total score is the response variable, and the objective is to find a formulation that maximizes total score. Two replicates of this design are run, and the results are shown in the table. Analyze the data and draw conclusions. Use \(a=0.05\) in the statistical tests. $$ \begin{array}{ccc} \hline \ {\begin{array}{c} \text { Treatment } \\ \text { Combination } \end{array}} & {\text { Replicate }} \\ \ & \text { I } & \text { II } \\ \hline(1) & 159 & 163 \\ a & 168 & 175 \\ b & 158 & 163 \\ a b & 166 & 168 \\ c & 175 & 178 \\ a c & 179 & 183 \\ b c & 173 & 168 \\ a b c & 179 & 182 \\ d & 164 & 159 \\ a d & 187 & 189 \\ b d & 163 & 159 \\ a b d & 185 & 191 \\ c d & 168 & 174 \\ a c d & 197 & 199 \\ b c d & 170 & 174 \\ a b c d & 194 & 198 \end{array} $$
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