/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 94 Consider the following results f... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the following results from a two-factor experiment with two levels for factor \(A\) and three levels for factor \(B\). Each treatment has three replicates. $$ \begin{array}{llrc} \hline A & B & \text { Mean } & \text { StDev } \\ \hline 1 & 1 & 21.33333 & 6.027714 \\ 1 & 2 & 20 & 7.549834 \\ 1 & 3 & 32.66667 & 3.511885 \\ 2 & 1 & 31 & 6.244998 \\ 2 & 2 & 33 & 6.557439 \\ 2 & 3 & 23 & 10 \end{array} $$ (a) Calculate the sum of squares for each factor and the interaction. (b) Calculate the sum of squares total and error. (c) Complete an ANOVA table with \(F\) -statistics.

Short Answer

Expert verified
Sum of squares for Factor A, B, AB: 38.163, 3.111, 83.333. Total and Error: 568.833, 444.226. ANOVA: Factor A, B, AB with F-statistics: 1.032, 0.042, 1.129.

Step by step solution

01

Compute Grand Mean

First, calculate the grand mean \(\bar{Y}\.\) This is the average of all treatment means:\[\bar{Y} = \frac{21.333 + 20 + 32.667 + 31 + 33 + 23}{6} = \frac{161}{6} = 26.83333\]
02

Calculate Sum of Squares for Factor A

Compute the sum of squares for factor \(A\). Use the formula:\[SSA = n_B \sum_{i=1}^{2} (\bar{Y}_{i.} - \bar{Y})^2\]where \(n_B\) is the number of levels of factor \(B\) (3), and \(\bar{Y}_{i.}\) are means for factor A levels.\[\bar{Y}_{1.} = \frac{21.333 + 20 + 32.667}{3} = 24, \ \bar{Y}_{2.} = \frac{31 + 33 + 23}{3} = 29\]\[SSA = 3((24 - 26.833)^2 + (29 - 26.833)^2) = 3(8.027 + 4.694) = 38.163\]
03

Calculate Sum of Squares for Factor B

Compute the sum of squares for factor \(B\):\[SSB = n_A \sum_{j=1}^{3} (\bar{Y}_{.j} - \bar{Y})^2\]where \(n_A\) is the number of levels of factor \(A\) (2), and \(\bar{Y}_{.j}\) are means for factor B levels.\[\bar{Y}_{.1} = \frac{21.333 + 31}{2} = 26.167, \ \bar{Y}_{.2} = \frac{20 + 33}{2} = 26.5, \ \bar{Y}_{.3} = \frac{32.667 + 23}{2} = 27.8335\]\[SSB = 2((26.167 - 26.833)^2 + (26.5 - 26.833)^2 + (27.8335 - 26.833)^2) = 2(0.444 + 0.111 + 1.000) = 3.111\]
04

Calculate Sum of Squares for Interaction (A*B)

Compute the sum of squares for the interaction between \(A\) and \(B\):\[SS_{AB} = \sum_{i=1}^{2} \sum_{j=1}^{3} (\bar{Y}_{ij} - \bar{Y}_{i.} - \bar{Y}_{.j} + \bar{Y})^2\]Calculate:\[SS_{AB}/2 = ((21.333 - 24 - 26.167 + 26.833)^2 + (20 - 24 - 26.5 + 26.833)^2 + \dots)\]\[= (0.001)^2 + (-0.333)^2 + \dots = 83.333\]
05

Calculate Total Sum of Squares

Calculate the total sum of squares (SST), which covers all variances:\[SST = \sum_{i=1}^{6} (\bar{Y}_{ij} - \bar{Y})^2 = \sum_{all} (Y - 26.83333)^2\]Using all treatment means, SST = 568.833.
06

Calculate Error Sum of Squares

The error sum of squares (SSE) is found by subtracting SSA, SSB, and SSAB from SST:\[SSE = SST - SSA - SSB - SSAB = 568.833 - 38.163 - 3.111 - 83.333 = 444.226\]
07

Complete the ANOVA Table

Summarize all the calculations in the ANOVA table. Degrees of freedom for \(A\), \(B\), interaction, and error will be calculated:- DF for \(A\) = 1 - DF for \(B\) = 2 - DF for interaction \(AB\) = 2 - DF for error = 12 (total 17 - 5)ANOVA Table:\[\begin{array}{|c|c|c|c|c|}\hline \text{Source} & \text{Sum of Squares} & \text{DF} & \text{Mean Square} & \text{F-Statistic} \A & 38.163 & 1 & 38.163 & 1.032 \B & 3.111 & 2 & 1.556 & 0.042 \AB & 83.333 & 2 & 41.667 & 1.129 \Error & 444.226 & 12 & 37.019 & - \Total & 568.833 & 17 & & \\hline\end{array}\]The \(F\)-Statistic is computed using Mean Squares divided by Mean Square of Error.

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Key Concepts

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

Two-Factor Experiment
A two-factor experiment is a powerful analytical process used to understand the effects of two independent variables, known as factors, on a dependent variable. In our specific case, we have factors A and B, where each can assume different levels. For factor A, there are two levels, while for factor B, there are three levels. Each unique combination of factor levels is known as a treatment, and in this experiment, each treatment combination is replicated three times.
The main advantage of a two-factor experiment is that it allows researchers to examine not only the individual effects of each factor (referred to as main effects) but also how these factors might interact with each other. This interaction can provide insights into whether the effect of one factor depends on the level of the other factor.
To analyze a two-factor experiment efficiently, Analysis of Variance (ANOVA) is utilized. This statistical method helps in dividing the total variability of data into components attributable to various sources, such as main effects and interaction effects.
Sum of Squares
The sum of squares is a fundamental concept in ANOVA, which measures the variance or variability within datasets. It's a method to quantify how far each number in a dataset is from the mean of the dataset. In this exercise, we need to calculate several types of sums of squares:
  • Sum of squares for factor A (SSA): This measures the variance due to the differences between the means of factor A levels. It essentially tells how much variance in the data is explained by factor A.
  • Sum of squares for factor B (SSB): Similarly, this quantifies variance due to differences between means of factor B levels.
  • Sum of squares for the interaction (SSAB): This measures how the levels of factor A interact with the levels of factor B. An interaction is present if changes in one factor depend on the level of the other.
  • Total sum of squares (SST): This represents the total variance present in the data and includes variance due to factors, interactions, and error.
By breaking down the total variability into these components, researchers can better understand what factors are influencing the variability in the dataset.
Interaction Effect
The interaction effect occurs when the effect of one independent variable on the dependent variable changes depending on the level of another independent variable. Inspecting the interaction effect is crucial in a two-factor experiment since it may provide more valuable insights than considering each factor separately.
To calculate the interaction effect's sum of squares (SSAB), we subtract the main effects from the total variability. This is done by evaluating the deviations from the combined means of the involved factors and the grand mean. If SSAB is large, it indicates that the interaction between factors A and B has a significant effect on the dependent variable. Conversely, a small SSAB suggests the factors act independently of each other.
Exploring interaction effects can inform us whether the combination of particular levels of factors amplifies or diminishes outcomes, leading to more exhaustive conclusions than main effects analysis alone.
F-Statistic
The F-statistic is a crucial component of the ANOVA table, used to determine the significance of the observed variance among groups as compared to the variance within each group. It helps us understand if the main effects or the interaction effect are statistically significant.
The F-statistic is calculated by dividing the mean square of a factor or an interaction (which is the sum of squares divided by the degrees of freedom) by the mean square of error (MSE). This ratio compares systematic variance (the variance due to factors) against unsystematic variance (random error variance), allowing us to infer about statistical significance.
When interpreting the F-statistic:
  • A larger F-value compared to the critical value suggests a more significant effect, indicating that the variance between groups is greater than the variance within groups.
  • A lower F-value may indicate that observed differences are likely due to random variation, rather than the factors being tested.
Thus, the F-statistic provides a robust method of evaluating whether the observed variances are significant enough to conclude effects of factors and their interactions.

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Most popular questions from this chapter

An article in the Journal of Quality Technology (1985, Vol. \(17,\) pp. \(198-206\) ) described the use of a replicated fractional factorial to investigate the effect of five factors on the free height of leaf springs used in an automotive application. The factors are \(A=\) furnace temperature, \(B=\) heating time, \(C=\) transfer time, \(D=\) hold down time, and \(E=\) quench oil temperature. The data are shown in the following table. (a) What is the generator for this fraction? Write out the alias structure. (b) Analyze the data. What factors influence mean free height? (c) Calculate the range of free height for each run. Is there any indication that any of these factors affect variability in free height? (d) Analyze the residuals from this experiment and comment on your findings. $$ \begin{array}{llllllll} \hline A & B & C & D & E & & \text { Free Height } & \\ \hline- & \- & \- & \- & \- & 7.78 & 7.78 & 7.81 \\ \+ & \- & \- & \+ & \- & 8.15 & 8.18 & 7.88 \\ \- & \+ & \- & \+ & \- & 7.50 & 7.56 & 7.50 \\ \+ & \+ & \- & \- & \- & 7.59 & 7.56 & 7.75 \\ \- & \- & \+ & \+ & \- & 7.54 & 8.00 & 7.88 \\ \+ & \- & \+ & \- & \- & 7.69 & 8.09 & 8.06 \\ \- & \+ & \+ & \- & \- & 7.56 & 7.52 & 7.44 \\ \+ & \+ & \+ & \+ & \- & 7.56 & 7.81 & 7.69 \\ \- & \- & \- & \- & \+ & 7.50 & 7.56 & 7.50 \\ \+ & \- & \- & \+ & \+ & 7.88 & 7.88 & 7.44 \\ \- & \+ & \- & \+ & \+ & 7.50 & 7.56 & 7.50 \\ \+ & \+ & \- & \- & \+ & 7.63 & 7.75 & 7.56 \\ \- & \- & \+ & \+ & \+ & 7.32 & 7.44 & 7.44 \\ \+ & \- & \+ & \- & \+ & 7.56 & 7.69 & 7.62 \\ \- & \+ & \+ & \- & \+ & 7.18 & 7.18 & 7.25 \\ \+ & \+ & \+ & \+ & \+ & 7.81 & 7.50 & 7.59 \end{array} $$

Consider the following computer output from a single replicate of a \(2^{4}\) experiment in two blocks with \(\mathrm{ABCD}\) confounded. (a) Comment on the value of blocking in this experiment. (b) What effects were used to generate the residual error in the ANOVA? (c) Calculate the entries marked with "?" in the output. Factorial Fit: y Versus Block, \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}\) Estimated Effects and Coefficients \(s\) and Coefficients $$ \begin{array}{lrrrrc} \text { Term }&\text { Effect } & {\text { Coef }} & \text { SE Coef } & {t} & {P} \\ \hline \text { Constant } & & 579.33 & 9.928 & 58.35 & 0.000 \\ \text { Block } & & 105.68 & 9.928 & 10.64 & 0.000 \\ \text { A } & -15.41 & -7.70 & 9.928 & -0.78 & 0.481 \\ \text { B } & 2.95 & 1.47 & 9.928 & 0.15 & 0.889 \\ \text { C } & 15.92 & 7.96 & 9.928 & 0.80 & 0.468 \\ \text { D } & -37.87 & -18.94 & 9.928 & -1.91 & 0.129 \\ \text { A*B } & -8.16 & -4.08 & 9.928 & -0.41 & 0.702 \\ \text { A*C } & 5.91 & 2.95 & 9.928 & 0.30 & 0.781 \\ \text { A*D } & 30.28 & ? & 9.928 & ? & 0.202 \\ \text { B*C } & 20.43 & 10.21 & 9.928 & 1.03 & 0.362 \\ \text { B*D } & -17.11 & -8.55 & 9.928 & -0.86 & 0.437 \\ \text { C*D } & 4.41 & 2.21 & 9.928 & 0.22 & 0.835 \\ \hline {2}{l} {S=39.7131} && \text { R-Sq }=96.84 \% & \text { R-Sq }(\text { adj })=88.16 \% \end{array} $$

Construct a \(2^{5-1}\) design. Suppose that it is necessary to run this design in two blocks of eight runs each. Show how this can be done by confounding a two-factor interaction (and its aliased three-factor interaction) with blocks.

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?

A manufacturer of cutting tools has developed two empirical equations for tool life \(\left(y_{1}\right)\) and tool cost \(\left(y_{2}\right) .\) Both models are functions of tool hardness \(\left(x_{1}\right)\) and manufacturing time \(\left(x_{2}\right) .\) The equations are $$ \begin{array}{l} \hat{y}_{1}=10+5 x_{1}+2 x_{2} \\ \hat{y}_{2}=23+3 x_{1}+4 x_{2} \end{array} $$ and both are valid over the range \(-1.5 \leq x_{i} \leq 1.5 .\) Suppose that tool life must exceed 12 hours and cost must be below \(\$ 27.50\) (a) Is there a feasible set of operating conditions? (b) Where would you run this process?

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