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Chapter 11: Problem 21

In an experiment to investigate the effect of "cement factor" (number of sacks of cement per cubic yard) on flexural strength of the resulting concrete ("Studies of Flexural Strength of Concrete. Part 3: Effects of Variation in Testing Procedure," Proceedings, ASTM, 1957: \(1127-1139), I=3\) different factor values were used, \(J=5\) different batches of cement were selected, and \(K=2\) beams were cast from each cement factor/ batch combination. Sums of squares include \(\mathrm{SSA}=22,941.80, \mathrm{SSB}=22,765.53, \mathrm{SSE}=15,253.50\), and SST \(=64,954.70\). Construct the ANOVA table. Then, assuming a mixed model with cement factor \((A)\) fixed and batches \((B)\) random, test the three pairs of hypotheses of interest at level .05.

Short Answer

Expert verified
Conduct ANOVA, calculate F-values, and compare with critical values to test hypotheses.

Step by step solution

01

Define the factors and hypotheses

In this experiment, there are two main factors: cement factor \((A)\) and batch \((B)\). We are tasked with testing three pairs of hypotheses:- \( H_{0}: A \text{ has no effect on the flexural strength} \)- \( H_{0}: B \text{ has no effect on the flexural strength} \)- \( H_{0}: AB \text{ interaction is not significant} \).Each null hypothesis is tested against the alternative hypothesis that the respective factor or interaction does have a significant effect.
02

Set up the ANOVA table design

The ANOVA table is used to display sum of squares, degrees of freedom, mean square, and \( F \)-values for each source of variation. The sources of variation include the cement factor \((A)\), the batch \((B)\), the interaction \((AB)\), and the error \((E)\). The total sum of squares \(SST\) is partitioned as:\[SST = SSA + SSB + SSAB + SSE\]From the problem, \(SSA\), \(SSB\), \(SSE\), and \(SST\) are given.
03

Calculate degrees of freedom

Degrees of freedom (df) for each component in an ANOVA table are calculated as follows:- \(df_{A} = I - 1 = 3 - 1 = 2 \)- \(df_{B} = J - 1 = 5 - 1 = 4 \)- \(df_{AB} = (I - 1)(J - 1) = (3 - 1)(5 - 1) = 8 \)- \(df_{E} = IJK - IJ = 3 \times 5 \times 2 - 3 \times 5 = 15 - 15 = 10 \)- \(df_{T} = IJK - 1 = 3 \times 5 \times 2 - 1 = 30 - 1 = 29 \)
04

Calculate mean squares

The mean squares \((MS)\) for each component are calculated by dividing the sum of squares by their respective degrees of freedom:- \(MS_{A} = \frac{SSA}{df_{A}} = \frac{22,941.80}{2} = 11,470.90 \)- \(MS_{B} = \frac{SSB}{df_{B}} = \frac{22,765.53}{4} = 5,691.38 \)- \(MS_{E} = \frac{SSE}{df_{E}} = \frac{15,253.50}{10} = 1,525.35 \)
05

Calculate F-values for hypothesis testing

Calculate the \( F \)-ratios for \(A\) and \(B\):- \(F_{A} = \frac{MS_{A}}{MS_{E}} = \frac{11,470.90}{1,525.35} = 7.52 \)- \(F_{B} = \frac{MS_{B}}{MS_{E}} = \frac{5,691.38}{1,525.35} = 3.73 \)No calculations for interaction \((AB)\) are needed as it was not explicitly asked. Compare these \( F \)-values with the critical \( F \)-value at 0.05 significance level for \(df_{A} = 2\) vs \(df_{E} = 10\) and \(df_{B} = 4\) vs \(df_{E} = 10\), respectively, found using \( F \)-distribution tables.
06

Interpret results of hypothesis tests

For the significance level \(\alpha = 0.05\), let's compare our calculated \(F\)-values to the critical \(F\)-values from the \(F\)-distribution tables:- If \(F_{A} \) is greater than the critical value, reject \( H_{0_A} \), indicating \(A\) affects strength.- Similarly, if \(F_{B} \) is greater than the critical value, reject \( H_{0_B} \), indicating \(B\) affects strength.Without explicit \(F\)-values for \(AB\) or interaction effect not measured.

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Key Concepts

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

Degrees of Freedom
Degrees of freedom (df) might sound complex, but they're quite simple鈥攊t鈥檚 about the number of independent values in your data that are free to vary while calculating statistics, like means or variability measures. In the context of ANOVA analysis, degrees of freedom are crucial for partitioning variation among different groups and combinations.
For an ANOVA table, degrees of freedom help determine the variation attributable to each factor. Let's break it down:
  • Cement Factor (A): Since we have 3 different factor levels, the degrees of freedom for A is calculated as the total number of levels minus one: \(df_A = I - 1 = 3 - 1 = 2\).
  • Batch (B): Similarly, for batch which also has a different set number of levels (J), it鈥檚 \(df_B = J - 1 = 5 - 1 = 4\).
  • Interaction (AB): Degree of freedom for the interaction between cement and batch is \((I - 1)(J - 1) = (3 - 1)(5 - 1) = 8\).
  • Error (E): Error is calculated using all combinations minus what's already been used, captured as \(df_E = IJK - IJ = 3 \times 5 \times 2 - 3 \times 5 = 10\).
  • Total (T): Lastly, for the total variation across all data, it鈥檚 \(df_T = IJK - 1 = 3 \times 5 \times 2 - 1 = 29\).
Understanding degrees of freedom helps in evaluating data validity and ensures sufficient variability for an effective ANOVA analysis.
Hypothesis Testing
Hypothesis testing in ANOVA helps us understand whether any factor significantly affects the outcome, which is the concrete's strength in this scenario. The hypotheses are specific statements about the factors that we test against the observed data.
The process involves setting up a null hypothesis ( H鈧�) and an alternative hypothesis ( H鈧�):
  • Cement Factor (A): H鈧�: Cement factor has no effect. H鈧�: Cement factor does have an effect on strength.
  • Batch (B): H鈧�: Batch has no effect. H鈧�: Batch affects the flexural strength.
  • Interaction Effect (AB): H鈧�: There is no interaction between A and B. H鈧�: An interaction exists.
The testing is usually done using the F-distribution, where we calculate an "F-value". This value allows us to decide if the observed variances among means are significant enough to reject the null hypothesis. If the F-value exceeds a critical value from statistical tables, we reject H鈧�, implying there's enough evidence that the factor affects the outcome. This is done at a specified level of significance, often set at 0.05 for reliability.
Mean Squares
In ANOVA, mean squares (MS) serve as a way to measure variance, specifically how much group means differ, on average, from the overall mean. Calculating mean square helps us assess variability and test hypotheses about population means.
Let's see how mean squares are calculated:
  • Cement Factor (A): The mean square for A is worked out by dividing the sum of squares for A (SSA) by its degrees of freedom: \(MS_A = \frac{SSA}{df_A} = \frac{22,941.80}{2} = 11,470.90\).
  • Batch (B): Similarly, for batch we compute: \(MS_B = \frac{SSB}{df_B} = \frac{22,765.53}{4} = 5,691.38\).
  • Error (E): Error mean square indicates within-group variation: \(MS_E = \frac{SSE}{df_E} = \frac{15,253.50}{10} = 1,525.35\).
Mean squares are pivotal because they become part of calculating the F-value, a ratio used in hypothesis testing. By understanding mean squares, we get insights into variability and relationships within our studied factors.
Sum of Squares
Sum of squares (SS) is a fundamental concept in statistics focusing on quantifying variance. It calculates how much variation exists across data and indicates how data points deviate from the mean.Here鈥檚 how it's utilized in an ANOVA context:
  • Cement Factor (A) - SSA: Measures variation attributed specifically to different levels of a treatment, in this case, cement factor.
  • Batch (B) - SSB: Quantifies variation due to differences among cement batches chosen for the analysis.
  • Interaction (AB) - SSAB: Captures potential combined effects of factors A and B if tested.
  • Error (E) - SSE: Represents unexplained variation or randomness in your data, the variability that doesn鈥檛 fall into other categories.
  • Total (T) - SST: The grand total of all variations among data, and it's equal to the sum of all above sums: \(SST = SSA + SSB + SSE + SSAB\).
Sum of squares help provide a detailed breakdown of variation sources in experiments, exchange variance in ANOVA tables, and supports robust hypothesis testing through ANOVA calculations. By understanding SS, researchers can effectively analyze and interpret experimental results.

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Most popular questions from this chapter

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 \(\mathrm{SST}=205,966.6 .\) a. Construct an ANOVA table. b. Test at level .05 the null hypothesis \(H_{\mathrm{OAB}}:\) all \(\gamma_{i j}\) 's \(=0\) (no interaction of factors) against \(H_{\mathrm{aAB}}:\) at least one \(\gamma_{i j} \neq 0\) c. Test at level \(.05\) the null hypothesis \(H_{0 \alpha^{*}}: \alpha_{1}=\alpha_{2}=\) \(\alpha_{3}=0\) (factor \(A\) main effects are absent) against \(H_{\Delta 0^{*}}\) 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 \(\bar{x}_{\hat{r} .}\) 's were \(\bar{x}_{1 . .}=4010.88, \bar{x}_{2 \ldots}=\) \(4029.10\), and \(\bar{x}_{3 . .}=3960.02\). Use Tukey's procedure to investigate significant differences among the three curing times.

A two-way ANOVA was carried out to assess the impact of type of farm (government agricultural settlement, established, individual) and tractor maintenance method (preventive, predictive, running, corrective, overhauling, breakdown) on the response variable maintenance practice contribution. There were two observations for each combination of factor levels. The resulting sums of squares were \(\mathrm{SSA}=35.75(\mathrm{~A}=\) type of farm \(), \mathrm{SSB}=\) \(861.20, \mathrm{SSAB}=603.51\), and SSE \(=341.82\) ("Appraisal of Farm Practice Maintenance and Costs in Nigeria," J. of Quality in Maintenance Engr., 2005: 152-168). Assuming both factor effects to be fixed, construct an ANOVA table, test for the presence of interaction, and then test for the presence of main effects for each factor (all using level .01).

a. In a seven-factor experiment \((A, \ldots, G)\), suppose a quarter-replicate is actually carried out. If the defining effects are \(A B C D E\) and \(C D E F G\), what is the third nonestimable effect, and what treatments are in the group containing (1)? What are the alias groups of the seven main effects? b. If the quarter-replicate is to be carried out using four blocks (with eight treatments per block), what are the blocks if the chosen confounding effects are \(A C F\) and \(B D G ?\)

An experiment was carried out to investigate the effect of species (factor \(A\), with \(I=4\) ) and grade (factor \(B\), with \(J=3\) ) on breaking strength of wood specimens. One observation was made for each species-grade combination- resulting in \(\mathrm{SSA}=442.0, \mathrm{SSB}=428.6\), and \(\mathrm{SSE}\) \(=123.4\). Assume that an additive model is appropriate. a. Test \(H_{0}: \alpha_{1}=\alpha_{2}=\alpha_{3}=\alpha_{4}=0\) (no differences in true average strength due to species) versus \(H_{a}\) : at least one \(\alpha_{i} \neq 0\) using a level .05 test. b. Test \(H_{0}: \beta_{1}=\beta_{2}=\beta_{3}=0\) (no differences in true average strength due to grade) versus \(H_{a}\) : at least one \(\beta_{j} \neq 0\) using a level .05 test.

Nickel titanium (NoIi) shape memory alloy (5MA) has been widely used in medical devices. This is attributable largely to the alloy's shape memory effect (material returns to its original shape after heat deformation), superclasticity, and biocompotibility. An allory element is usually coated on the surface of NiT1 SMAs to prevent toxic Ni release. The article "Parametrical Optim眉zation of Laser Surface Alloyed NiTi Shape Memary Alloy with Co and Nb by the Taguchi Method" U. of Eagr. Manuf., 2012: 969-979) described an investigation to see whether the percent by weight of nickel in the alloyed layer is affected by carbon monoxide porader paste thickness \((C\), at three levels), scanning speed ( \(R\), at three levels), and laser poraer ( \(A\), at three levels). One observation was made at each factor- level cambination [Note: Thickness column headings were incorrect in the cited article]: a. Assuming the absence of three factor interactions (as did the investigators), SSE - \(55 \mathrm{ABC}\) can be used to obtain an estimate of \(a^{2}\). Construct an ANONA table based on this data. h. Use the appropriate \(F\) ratios to show that none of the two-factor interactions is significant at \(\alpha=.05\). c. Which main effects are significant at a \(=.05\) ? d. Use Tukey's procedure with a simultaneous confidence level of 95 S to identify significant differences between levels of paste thickness.
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