/*! 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 26 When both factors are random in ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When both factors are random in a two-way ANOVA experiment with \(K\) replications per combination of factor levels, the expected mean squares are \(E(\mathrm{MSE})=\sigma^{2}\), \(E(\mathrm{MSA})=\sigma^{2}+K \sigma_{G}^{2}+J K \sigma_{A}^{2}, E(\mathrm{MSB})=\sigma^{2}+K \sigma_{G}^{2}+\) \(I K \sigma_{B}^{2}\), and \(E(\mathrm{MSAB})=\sigma^{2}+K \sigma_{G^{2}}^{2}\) a. What \(F\) ratio is appropriate for testing \(H_{\alpha^{*}} * \sigma_{G}^{2}=0\) versus \(H_{a}: \sigma_{G}^{2}>0\) ? b. Answer part (a) for testing \(H_{0 \mathrm{~A}}: \sigma_{A}^{2}=0\) versus \(H_{\mathrm{aA}}: \sigma_{A}^{2}>0\) and \(H_{O B}: \sigma_{B}^{2}=0\) versus \(H_{a B^{*}}: \sigma_{B}^{2}>0\).

Short Answer

Expert verified
a) Use \( F = \frac{\mathrm{MSAB}}{\mathrm{MSE}} \); b) Use \( F = \frac{\mathrm{MSA}}{\mathrm{MSAB}} \) and \( F = \frac{\mathrm{MSB}}{\mathrm{MSAB}} \).

Step by step solution

01

Understanding the Question

The problem involves determining the appropriate F-ratio test statistic for hypotheses in a two-way ANOVA with random factors. We are given the expected mean squares (EMS) for the model.
02

Identifying Components for EMS

From the problem, we have:- \(E(\mathrm{MSE}) = \sigma^{2}\)- \(E(\mathrm{MSA}) = \sigma^{2} + K \sigma_{G}^{2} + J K \sigma_{A}^{2}\)- \(E(\mathrm{MSB}) = \sigma^{2} + K \sigma_{G}^{2} + I K \sigma_{B}^{2}\)- \(E(\mathrm{MSAB}) = \sigma^{2} + K \sigma_{G}^{2}\). These expected values reveal how variance components (\(\sigma_{G}^{2}\), \(\sigma_{A}^{2}\), \(\sigma_{B}^{2}\)) contribute to each mean square.
03

Part (a): Selecting F-ratio for \( H_{\alpha^{*}} : \sigma_{G}^{2}=0 \)

To test \( H_{\alpha^{*}} : \sigma_{G}^{2} = 0 \) against \( H_{a} : \sigma_{G}^{2} > 0 \), we observe that both \( \mathrm{MSA} \), \( \mathrm{MSB} \), and \( \mathrm{MSAB} \) include the term \( K \sigma_{G}^{2} \). The only difference in \( \mathrm{MSAB} \) is that it does not have \( \sigma_{A}^{2} \) or \( \sigma_{B}^{2} \). Thus, we use the F-ratio \( F = \frac{\mathrm{MSAB}}{\mathrm{MSE}} \) to test \( \sigma_{G}^{2} = 0 \).
04

Part (b): Selecting F-ratio for \( H_{0 \mathrm{A}}: \sigma_{A}^{2}=0 \)

For \( H_{0 \mathrm{A}} : \sigma_{A}^{2} = 0 \) against \( H_{\mathrm{aA}} : \sigma_{A}^{2} > 0 \), we compare \( \mathrm{MSA} \) with \( \mathrm{MSAB} \) since \(\mathrm{MSAB}\) does not contain \( \sigma_{A}^{2} \). Therefore, \( F = \frac{\mathrm{MSA}}{\mathrm{MSAB}} \) is appropriate for testing \( \sigma_{A}^{2} = 0 \).
05

Part (b): Selecting F-ratio for \( H_{0 \mathrm{B}}: \sigma_{B}^{2}=0 \)

For \( H_{0 \mathrm{B}} : \sigma_{B}^{2} = 0 \) against \( H_{aB^{*}} : \sigma_{B}^{2} > 0 \), compare \( \mathrm{MSB} \) with \( \mathrm{MSAB} \) because \( \mathrm{MSAB} \) doesn't contain \( \sigma_{B}^{2} \). The appropriate F-ratio is \( F = \frac{\mathrm{MSB}}{\mathrm{MSAB}} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

F-ratio
The F-ratio is a key statistical concept used in ANOVA to test specific hypotheses about group variances. In simple terms, the F-ratio is a quotient where the numerator is a mean square from the data related to a particular effect, and the denominator is a mean square that represents error or noise. This ratio helps determine if the variances between groups are significantly different enough to suggest that at least one of the group means is not like the others.

In two-way ANOVA, understanding which components contribute to the numerator and denominator is crucial when forming hypotheses. The F-ratio can be expressed as:
  • \( F = \frac{\text{Mean Square for Effect}}{\text{Mean Square for Error}} \)
For instance, if we test if a variance component \( \sigma_G^2 \) is zero, using \( \mathrm{MSAB} \) for effect and \( \mathrm{MSE} \) for error is appropriate since \( \mathrm{MSAB} \) includes \( K \sigma_G^2 \) but not other components such as \( \sigma_A^2 \) or \( \sigma_B^2 \). This specificity in choosing the right components for numerator and denominator helps in accurate hypothesis testing.
random factors
In a two-way ANOVA, random factors refer to variables where the levels are random samples from a larger population. This means the conclusions from the analysis can be generalized to the whole population rather than just the specific levels studied.

Using random factors implies that we are interested in the variability across the levels rather than the levels themselves. Each random factor contributes to the overall variance, and understanding these contributions is crucial when analyzing variance components.

On the other hand, fixed factors would have specific levels, and analysis would only pertain to those chosen levels. In this context, when both factors in a two-way ANOVA are random, it allows us to pursue a broader inference about population variances, emphasizing the importance of random factor analysis in broader scenarios.
variance components
Variance components in a two-way ANOVA with random factors help us understand how much each factor contributes to the overall variability in the data. These are the specific parts of variance that we denote as \( \sigma^2 \), \( \sigma_G^2 \), \( \sigma_A^2 \), and \( \sigma_B^2 \).

Each component arises from different sources:
  • \( \sigma^2 \) represents the inherent variability within individual measurements.
  • \( \sigma_G^2 \) symbolizes variability due to random interactions or elements present across all treatments.
  • \( \sigma_A^2 \) is the variance due to variations in random factor A.
  • \( \sigma_B^2 \) corresponds to the variance caused by random factor B.
The process of partitioning the total variance into these components plays a pivotal role in understanding where variability comes from, enabling precise interpretation and application of statistical findings.
expected mean squares
Expected mean squares (EMS) are mathematical expressions used to calculate the mean squares for different effects and errors in ANOVA. They help us understand the structure of variance in the data correctly by showing which variance components are expected to be included in each mean square.

The EMS equations for a two-way ANOVA where both factors are random are:
  • \(E(\mathrm{MSE}) = \sigma^{2}\)
  • \(E(\mathrm{MSA}) = \sigma^{2} + K \sigma_{G}^{2} + J K \sigma_{A}^{2}\)
  • \(E(\mathrm{MSB}) = \sigma^{2} + K \sigma_{G}^{2} + I K \sigma_{B}^{2} \)
  • \(E(\mathrm{MSAB}) = \sigma^{2} + K \sigma_{G^{2}}^{2}\)
This accounting of expected variance components in EMS helps derive the correct F-ratios for testing hypotheses about the components' significance. Understanding EMS is vital for correctly interpreting the impact of each factor in ANOVA and for forming informed conclusions about the data's variability.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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 \(.05\). \(\left(H_{0}\right.\) 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 \(\mathrm{SST}=205,966.6 .\) 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.

Use the fact that \(E\left(X_{i j}\right)=\mu \pm \alpha_{i}+\beta_{j}\) with \(\Sigma \alpha_{i}=\Sigma \beta_{j}=0\) to show that \(E\left(\bar{X}_{i}-\bar{X}_{. .}\right)=\alpha_{i}\), so that \(\hat{\alpha}_{i}=\bar{X}_{l}-\bar{X}_{.}\). is an unbiased estimator for \(\alpha_{i}\).

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.

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 Optimüzation 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.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.