/*! 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 51 The accompanying data resulted f... [FREE SOLUTION] | 91Ó°ÊÓ

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The accompanying data resulted from an experiment to investigate whether yield from a chemical process depended either on the formulation of a particular input or on mixer speed. $$ \begin{array}{cc|ccc} & & {\text { Speed }} \\ { 3 - 5 } & & \mathbf{6 0} & \mathbf{7 0} & \mathbf{8 0} \\ \hline \text { Formulation } & & 189.7 & 185.1 & 189.0 \\ & & 188.6 & 179.4 & 193.0 \\ & & 190.1 & 177.3 & 191.1 \\ & & 165.1 & 161.7 & 163.3 \\ & & 165.9 & 159.8 & 166.6 \\ & & 167.6 & 161.6 & 170.3 \\ \hline \end{array} $$ A statistical computer package gave \(\mathrm{SS}(\mathrm{Form})=\) \(2253.44, \mathrm{SS}(\) Speed \()=230.81, \mathrm{SS}(\) Form* Speed \()\) \(=18.58\), and SSE \(=71.87\). a. Does there appear to be interaction between the factors? b. Does yield appear to depend on either formulation or speed? c. Calculate estimates of the main effects. d. Verify that the residuals are \(0.23,-0.87,0.63\), \(4.50,-1.20,-3.30,-2.03,1.97,0.07,-1.10\), \(-0.30,1.40,0.67,-1.23,0.57,-3.43,-0.13\), \(3.57\). e. Construct a normal plot from the residuals given in part (d). Do the \(\varepsilon_{i j k}\) 's appear to be normally distributed? f. Plot the residuals against the predicted values (cell means) to see if the population variance appears reasonably constant.

Short Answer

Expert verified
No significant interaction; yield depends on formulation and speed; residuals indicate normal distribution and consistent variance across predictions.

Step by step solution

01

Determine Interaction

To identify interaction, we compare the interaction sum of squares with the error sum of squares using an F-test. The hypotheses are: - Null hypothesis (H_0): There is no interaction. - Alternative hypothesis (H_1): There is an interaction.Calculate the interaction F-statistic:\[ F_{interaction} = \frac{SS(Form*Speed) / df_{interaction}}{SSE / df_{E}} \]where \( df_{interaction} = (3 - 1)(3 - 1) = 4 \) and \( df_{E} = 18 \). Calculate the F-statistic:\[ F_{interaction} = \frac{18.58 / 4}{71.87 / 18} = 1.16 \]This F-value is compared with a critical value from the F-distribution at a chosen significance level, often \(\alpha = 0.05\). If \(F_{interaction}\) is less than the critical value, there is no significant interaction.
02

Test for Dependence on Formulation and Speed

To test whether yield depends on formulation or speed, we perform an ANOVA test using the sums of squares. We compare the formulation and speed F-statistics to the critical F-value:- Calculate formulation F-statistic:\[ F_{form} = \frac{SS(Form) / df_{form}}{SSE / df_{E}} \]where \( df_{form} = 2 \). Calculate:\[ F_{form} = \frac{2253.44 / 2}{71.87 / 18} = 282.09 \]- Calculate speed F-statistic:\[ F_{speed} = \frac{SS(Speed) / df_{speed}}{SSE / df_{E}} \]where \( df_{speed} = 2 \). Calculate:\[ F_{speed} = \frac{230.81 / 2}{71.87 / 18} = 28.9 \]If these F-values are greater than the critical value, the yield depends on the respective factor.
03

Estimate Main Effects

Main effects are calculated as the average yield differences across levels of each factor. The main effect for formulation:\[ M_{form} = \text{mean yield at each formulation level} - \text{overall mean} \]And similarly for speed:\[ M_{speed} = \text{mean yield at each speed level} - \text{overall mean} \]Calculate the overall means for each level and subtract the overall mean to find main effects.
04

Verify Residuals

Residuals are the differences between observed and predicted values. Given residuals are verified by subtracting the estimated values (from cell means or predicted model) from the observed data. Show that each given residual conforms to:\[ e_{ijk} = y_{ijk} - \hat{y}_{ijk} \]This confirms that each residual matches those provided.
05

Construct Normal Plot of Residuals

To determine if residuals are normally distributed, construct a normal probability plot. Rank the residuals and plot them against the expected normal scores: If residual points roughly follow a straight line, the distribution is approximately normal.
06

Analyze Residuals Against Predicted Values

Plot residuals versus predicted values to check for homoscedasticity. A consistent spread of residuals across predicted values suggests constant variance. Examine the plot for any discernible pattern in spread, which may indicate variance inconsistency.

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Key Concepts

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

Factorial Design
In experiments like the one described, understanding factorial design is crucial. It involves examining two or more factors by testing them at different levels simultaneously. In our example, the two factors are formulation and mixer speed, both with three levels. Factorial design allows for the investigation of all possible combinations of these levels:
  • This enables researchers to see how the factors not only independently affect the outcome (yield from the chemical process), but also how they interact with each other.
  • By setting up the experiment in this manner, it controls for variability and enhances the ability to see true effects.
Analyzing this setup properly allows for identifying the main effects and possible interactions, providing a comprehensive understanding of each factor's contribution to the outcome.
Interaction Effects
Interaction effects occur when the influence of one factor depends on the level of another factor. In the project at hand, interaction effects are tested between formulation and speed, which means checking how the combination affects the chemical yield.
Testing for these effects includes: - Calculating the interaction sum of squares. - Using an F-test to determine if those squares significantly differ from random chance. If the test results in a low F-value compared to the threshold, there's likely no interaction. For this study, the calculated F-interaction was 1.16, and finding that it’s less than the critical F-value suggests no significant interaction effect. This implies that formulation and speed affect yield independently.
Residual Analysis
Residual analysis is vital to validate the model used in an experiment. Residuals are the differences between the observed and predicted values of yield. Analyzing them helps ensure the accuracy and reliability of your results.
Here's how you undertake residual analysis:
  • Calculate each residual by subtracting predicted yield from observed yield.
  • Evaluate if residuals appear randomly scattered, indicating a well-fitting model.
  • Assess if residuals display constant variance and no discernible pattern against predicted values.
The exercise provided a series of residuals that were derived accurately. By examining these residuals, it confirms the precision of the model's predictions.
Normality Test
Performing a normality test on residuals involves examining their distribution. Generally, this is done using a normal probability plot, where residual order is set against expected normal values.
If residuals align closely with a straight line on the plot, it suggests normal distribution. This step is crucial because normality of residuals ensures that the assumptions of ANOVA are met. Moreover, it aids in:
  • Identifying any potential bias in the results.
  • Ensuring sound conclusions can be drawn from the analysis.
If they deviate significantly from the line, it indicates non-normality, which can be addressed by transforming data or reconsidering the model used.

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Most popular questions from this chapter

The article "Origin of Precambrian Iron Formations" (Econ. Geol., 1964: 1025-1057) reports the following data on total Fe for four types of iron formation \((1=\) carbonate, \(2=\) silicate, \(3=\) magnetite, \(4=\) hematite). $$ \begin{array}{llllll} 1: & 20.5 & 28.1 & 27.8 & 27.0 & 28.0 \\ & 25.2 & 25.3 & 27.1 & 20.5 & 31.3 \\ 2: & 26.3 & 24.0 & 26.2 & 20.2 & 23.7 \\ & 34.0 & 17.1 & 26.8 & 23.7 & 24.9 \\ 3: & 29.5 & 34.0 & 27.5 & 29.4 & 27.9 \\ & 26.2 & 29.9 & 29.5 & 30.0 & 35.6 \\ 4: & 36.5 & 44.2 & 34.1 & 30.3 & 31.4 \\ & 33.1 & 34.1 & 32.9 & 36.3 & 25.5 \end{array} $$ Carry out an analysis of variance \(F\) test at significance level \(.01\), and summarize the results in an ANOVA table.

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