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Chapter 11: Problem 2

Four different coatings are being considered for corrosion protection of metal pipe. The pipe will be buried in three different types of soil. To investigate whether the amount of corrosion depends either on the coating or on the type of soil, 12 pieces of pipe are selected. Each piece is coated with one of the four coatings and buried in one of the three types of soil for a fixed time, after which the amount of corrosion (depth of maximum pits, in \(.0001\) in.) is determined. The data appears in the table. \begin{tabular}{|l|llll} & & \multicolumn{4}{|c}{ Soil Type \((\boldsymbol{B})\)} \\ \cline { 2 - 5 } & \(\mathbf{1}\) & \(\mathbf{1}\) & \(\mathbf{2}\) & \(\mathbf{3}\) \\\ \hline \multirow{3}{*}{ Coating (A) } & \(\mathbf{2}\) & 64 & 49 & 50 \\ & \(\mathbf{3}\) & 53 & 51 & 48 \\ & \(\mathbf{4}\) & 47 & 45 & 50 \\ & & 51 & 43 & 52 \\ \hline \end{tabular} a. Assuming the validity of the additive model, carry out the ANOVA analysis using an ANOVA table to see whether the amount of corrosion depends on either the type of coating used or the type of soil. Use \(\alpha=.05\). b. Compute \(\hat{\mu}, \hat{\alpha}_{1}, \hat{\alpha}_{2}, \hat{\alpha}_{3}, \hat{\alpha}_{4}, \hat{\beta}_{1}, \hat{\beta}_{2}\), and \(\hat{\beta}_{3}\).

Short Answer

Expert verified
Perform an ANOVA, calculate F-statistics, compare with critical values. Compute estimates of effects from sample means.

Step by step solution

01

Identify the Hypotheses for ANOVA

To determine if there's significant influence of coating or soil type on corrosion, we pose two hypotheses. Null Hypothesis(H0): There is no effect of coating or soil type on corrosion. Alternative Hypothesis(H1): There is an effect. We鈥檒l conduct two one-way ANOVAs, one for coatings and another for soil types, using 伪 = 0.05 for significance.
02

Compute the Sum of Squares (SS)

Calculate Total Sum of Squares (SST), Sum of Squares for Coatings (SSA), and Sum of Squares for Soil Type (SSB). For each, use the formula: SST = 危(y_{ij} - y虅)^2, SSA = b危(y虅i. - y虅)^2, SSB = a危(y虅.j - y虅)^2, where 'i' and 'j' are the levels of coating and soil types, respectively.
03

Calculate the Mean Squares (MS)

With degrees of freedom for coatings (dof_A = a - 1), soil types (dof_B = b - 1), and total (dof_total = ab - 1), determine the mean squares: MSA = SSA/dof_A, MSB = SSB/dof_B, and MSE = SSE/dof_E, where SSE = SST - SSA - SSB and dof_E is error degrees of freedom.
04

Perform ANOVA Test and Check F-statistic

Compute F-statistics: F_A = MSA/MSE and F_B = MSB/MSE. Compare calculated F values with critical F values from F-distribution tables. If calculated F is greater, reject the null hypothesis for that factor, showing the factor significantly affects corrosion.
05

Calculate Estimates of Effects

Estimate coefficients where \( \hat{\mu} = y虅 \), \( \hat{\alpha}_i = y虅_i. - \hat{\mu} \), and \( \hat{\beta}_j = y虅_.j - \hat{\mu} \). This shows how each coating and soil type deviates from the overall mean.

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Key Concepts

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

Sum of Squares in ANOVA
In ANOVA analysis, understanding the Sum of Squares (SS) is crucial as it helps partition the total variability in the data into components attributable to different sources. Broadly, in this exercise, we focus on three main sums of squares:
  • Total Sum of Squares (SST): This measures the overall variability in the corrosion data across all coatings and soils. It is calculated by summing the squared differences between each data point and the overall mean.
  • Sum of Squares for Coatings (SSA): This quantifies the variability due to different coatings. Here, the mean corrosion caused by each coating is compared to the overall mean to see how coatings influence corrosion.
  • Sum of Squares for Soil Type (SSB): This examines the differences in corrosion based on soil types by comparing the mean corrosion in each soil type to the overall mean.
Each of these components helps in understanding how much of the total variation in corrosion can be explained by the type of coating or soil.
Hypothesis Testing in Statistics
Hypothesis testing in statistics is a method used to make decisions based on data analysis. For ANOVA, we utilize hypothesis testing to determine if our independent variables (coating and soil) significantly affect the dependent variable (corrosion). The process typically involves the following:
  • Null Hypothesis (H0): This states that there is no effect of the coatings or soil types on corrosion. It serves as a baseline to compare whether we observe any changes.
  • Alternative Hypothesis (H1): Contrarily, this states that there is a significant effect.
With a chosen significance level (commonly 伪 = 0.05), we aim to verify whether there is enough statistical evidence to reject the null hypothesis. If rejected, it suggests that at least one type of coating or soil affects corrosion more than the others.
F-statistic Calculation
The F-statistic is key to ANOVA as it helps determine whether there are any significant differences between means. It is calculated by comparing the variance between groups to the variance within groups. For our analysis:
  • F for Coatings (F_A): This is calculated by dividing the Mean Square between coatings (MSA) by the Mean Square of Error (MSE). It shows the ratio of variability accounted by coatings to the within-group variability.
  • F for Soil Types (F_B): Similarly, this is calculated by the Mean Square between soil types (MSB) to MSE ratio, indicating if soil type's impact is significant.
We compare these F-values against critical values from F-distribution tables. If the calculated F exceeds the critical F-value, we reject the null hypothesis for that factor, indicating a significant effect on corrosion.
Interpretation of ANOVA Results
Interpreting ANOVA results gives insights into how different factors affect outcomes like corrosion. Upon computing ANOVA, it's key to analyze which factors are statistically significant. Here鈥檚 a simplification: - Significant F-Value: If our F-statistic is larger than the critical value for either coatings or soil types, it highlights that factor's importance in influencing corrosion. - Coefficient Estimates: These values, *

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Most popular questions from this chapter

In an experiment to assess the effect of the angle of pull on the force required to cause separation in electrical connectors, four different angles (factor \(A\) ) were used, and each of a sample of five connectors (factor \(B\) ) was pulled once at each angle ("A Mixed Model Factorial Experiment in Testing Electrical Connectors," Industrial Quality Control, 1960: 12-16). The data appears in the accompanying table. Does the data suggest that true average separation force is affected by the angle of pull? State and test the appropriate hypotheses at level .01 by first constructing an ANOVA table \((\mathrm{SST}=396.13, \mathrm{SSA}=58.16\), and \(\mathrm{SSB}=246.97)\).

The article "Adiabatic Humidification of Air with Water in a Packed Tower" (Chem. Eng. Prog., 1952: 362-370) reports data on gas film heat transfer coefficient (Btu/hr \(\mathrm{ft}^{2}\) on \({ }^{\circ} \mathrm{F}\) ) as a function of gas rate (factor \(A\) ) and liquid rate (factor \(B\) ). \begin{tabular}{ll|cccc} & \(\mathbf{1 ( 1 9 0 )}\) & \(\mathbf{2 ( 2 5 0 )}\) & \(\mathbf{3 ( 3 0 0 )}\) & \(\mathbf{4 ( 4 0 0 )}\) \\ \hline & \(\mathbf{1 ( 2 0 0 )}\) & 200 & 226 & 240 & 261 \\ & \(\mathbf{A}(\mathbf{4 0 0})\) & 278 & 312 & 330 & 381 \\ & \(\mathbf{3 ( 7 0 0 )}\) & 369 & 416 & 462 & 517 \\ & \(\mathbf{4 ( 1 1 0 0 )}\) & 500 & 575 & 645 & 733 \\ \hline \end{tabular} a. After constructing an ANOVA table, test at level .01 both the hypothesis of no gas-rate effect against the appropriate alternative and the hypothesis of no liquid-rate effect against the appropriate alternative. b. Use Tukey's procedure to investigate differences in expected heat transfer coefficient due to different gas rates. c. Repeat part (b) for liquid rates.

A particular county employs three assessors who are responsible for determining the value of residential property in the county. To see whether these assessors differ systematically in their assessments, 5 houses are selected, and each assessor is asked to determine the market value of each house. With factor \(A\) denoting assessors \((I=3)\) and factor \(B\) denoting houses \((J=5)\), suppose \(\mathrm{SSA}=11.7, \mathrm{SSB}=113.5\), and \(\mathrm{SSE}=25.6\) a. Test \(H_{0}: \alpha_{1}=\alpha_{2}=\alpha_{3}=0\) at level \(.05\). ( \(H_{0}\) states that there are no systematic differences among assessors.) b. Explain why a randomized block experiment with only 5 houses was used rather than a one-way ANOVA experiment involving a total of 15 different houses, with each assessor asked to assess 5 different houses (a different group of 5 for each assessor).

In an experiment to assess the effects of curing time (factor \(A\) ) and type of mix (factor \(B\) ) on the compressive strength of hardened cement cubes, three different curing times were used in combination with four different mixes, with three observations obtained for each of the 12 curing time-mix combinations. The resulting sums of squares were computed to be \(\mathrm{SSA}=30,763.0, \mathrm{SSB}=34,185.6, \mathrm{SSE}=97,436.8\), and SST \(=205,966.6\) a. Construct an ANOVA table. b. Test at level .05 the null hypothesis \(H_{\text {aAB }}\) ' all \(\gamma_{i j}\) 's \(=0\) (no interaction of factors) against \(H_{\mathrm{a} / s^{\prime}}\) : at least one \(\gamma_{i j} \neq 0\). c. Test at level \(.05\) the null hypothesis \(H_{0 A}: \alpha_{1}=\alpha_{2}=\) \(\alpha_{3}=0\) (factor \(A\) main effects are absent) against \(H_{\mathrm{ad}}\) : at least one \(\alpha_{i} \neq 0\). d. Test \(H_{0 B}: \beta_{1}=\beta_{2}=\beta_{3}=\beta_{4}=0\) versus \(H_{a B}\) : at least one \(\beta_{j} \neq 0\) using a level \(.05\) test. e. The values of the \(\bar{x}_{i-}\) 's were \(\bar{x}_{1 \mathrm{l}}=4010.88\), \(\bar{x}_{2-}=4029.10\), and \(\bar{x}_{3-}=3960.02\). Use Tukey's procedure to investigate significant differences among the three curing times.

The output of a continuous extruding machine that coats steel pipe with plastic was studied as a function of the thermostat temperature profile \((A\), at three levels), the type of plastic ( \(B\), at three levels), and the speed of the rotating screw that forces the plastic through a tube-forming die \((C\), at three levels). There were two replications ( \(L=2)\) at each combination of levels of the factors, yielding a total of 54 observations on output. The sums of squares were \(\mathrm{SSA}=14,144.44, \quad \mathrm{SSB}=5511.27, \quad \mathrm{SSC}=244,696.39\), \(\mathrm{SSAB}=1069.62, \quad \mathrm{SSAC}=62.67, \quad \mathrm{SSBC}=331.67\), \(\mathrm{SSE}=3127.50\), and \(\mathrm{SST}=270,024.33\). a. Construct the ANOVA table. b. Use appropriate \(F\) tests to show that none of the \(F\) ratios for two- or three-factor interactions is significant at level \(.05\). c. Which main effects appear significant? d. With \(x_{-1 .}=8242, x_{-2}=9732\), and \(x_{-3 .}=11,210\), use Tukey's procedure to identify significant differences among the levels of factor \(C\).
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