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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 \(E(\mathrm{MSAB})=\sigma^{2}+K \sigma_{G}^{2}\) a. What \(F\) ratio is appropriate for testing \(H_{0 G}: \sigma_{G}^{2}=0\) versus \(H_{\mathrm{a} G}: \sigma_{G}^{2}>0\) ? b. Answer part (a) for testing \(H_{0 A}: \sigma_{A}^{2}=0\) versus \(H_{\mathrm{aA}}: \sigma_{A}^{2}>0\) and \(H_{0 B}: \sigma_{B}^{2}=0\) versus \(H_{\mathrm{a} B}: \sigma_{B}^{2}>0\)

Short Answer

Expert verified
a. Use \( F = \frac{\mathrm{MSAB}}{\mathrm{MSE}} \) for \( H_{0 G} \). b. Use \( F = \frac{\mathrm{MSA}}{\mathrm{MSAB}} \) for \( H_{0 A} \) and \( F = \frac{\mathrm{MSB}}{\mathrm{MSAB}} \) for \( H_{0 B} \).

Step by step solution

01

Identify the Hypotheses for Part (a)

In part (a), we are testing if the interaction term variance, denoted by \( \sigma_{G}^{2} \), is zero versus greater than zero. The null hypothesis \( H_{0 G} \) is that \( \sigma_{G}^{2} = 0 \), while the alternative hypothesis \( H_{a G} \) is \( \sigma_{G}^{2} > 0 \).
02

Identify the Mean Squares Relevant to Part (a)

The expected mean squares relevant to testing \( H_{0 G} \) are \( E(\mathrm{MSE}) = \sigma^{2} \) and \( E(\mathrm{MSAB}) = \sigma^{2} + K \sigma_{G}^{2} \).
03

Identify the F-ratio for Part (a)

To test \( H_{0 G}: \sigma_{G}^{2} = 0 \), use the F-ratio:\[F = \frac{\mathrm{MSAB}}{\mathrm{MSE}}\]This ratio tests if the variance due to the interaction term \( \sigma_{G}^{2} \) is significantly greater than zero.
04

Identify the Hypotheses and Mean Squares for Part (b) Testing \(\sigma_{A}^{2}\)

For testing \( H_{0 A}: \sigma_{A}^{2} = 0 \) versus \( H_{aA}: \sigma_{A}^{2} > 0 \), use the expected mean squares:1. \( E(\mathrm{MSA}) = \sigma^{2} + K \sigma_{G}^{2} + J K \sigma_{A}^{2} \)2. \( E(\mathrm{MSAB}) = \sigma^{2} + K \sigma_{G}^{2} \) (serves as denominator)The F-ratio is:\[F = \frac{\mathrm{MSA}}{\mathrm{MSAB}}\]
05

Identify the F-ratio for Part (b) Testing \( \sigma_{B}^{2} \)

For testing \( H_{0 B}: \sigma_{B}^{2} = 0 \) versus \( H_{aB}: \sigma_{B}^{2} > 0 \), use:1. \( E(\mathrm{MSB}) = \sigma^{2} + K \sigma_{G}^{2} + I K \sigma_{B}^{2} \)2. \( E(\mathrm{MSAB}) = \sigma^{2} + K \sigma_{G}^{2} \) (serves as denominator)The F-ratio is:\[F = \frac{\mathrm{MSB}}{\mathrm{MSAB}}\]

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Key Concepts

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

F-ratio
In statistical analysis, especially in Two-Way ANOVA, the F-ratio is a pivotal concept used to compare variances and test hypotheses. It's the ratio between two variance estimates: the mean square for the effect and the mean square for error. The F-ratio is calculated as \( F = \frac{MS_{Effect}}{MS_{Error}} \), where \( MS_{Effect} \) refers to the variance you wish to test (like a factor or interaction), and \( MS_{Error} \) represents the variance that explains the random error.
When conducting a Two-Way ANOVA, the F-ratio helps determine if the observed variances are statistically significant. If the calculated F-ratio is larger than the critical F-value from a statistical table at a given significance level, we can reject the null hypothesis. This implies a significant difference in group means or interactions. Understanding how to calculate and interpret F-ratios is crucial for drawing conclusions in hypothesis testing.
Expected Mean Squares
Expected Mean Squares (EMS) are integral in Two-Way ANOVA, particularly when dealing with random factors. These represent the expected values of the mean square estimates, derived from the levels of random factors in your study. EMS is composed of the estimations of variance components for different sources of variation. For instance, in an experiment, EMS can be represented as \( E(MSA) = \sigma^2 + K\sigma_G^2 + JK\sigma_A^2 \) for factor A. This captures the variance due to random error, variance due to interaction, and variance from factor A.
EMS helps in calculating F-ratios for hypothesis testing and ensures that we account correctly for different sources of variation present, making the results of ANOVA accurate and meaningful. Identifying these expected values guides us in evaluating which effects in the data are systemic and which are random variations.
Random Factors
In Two-Way ANOVA, random factors are those that consist of randomly chosen levels from a broader population. These are key when the analysis needs to generalize beyond specific levels of the factor being tested. Unlike fixed factors, random factors allow inferences about a wider spectrum of possible levels. This is vital when the objective extends beyond just comparing the treatments present in the experiment. Consider a scenario where schools are randomized in an educational study; the specific schools tested are not of primary interest, but rather the population they represent.
Random factors necessarily adjust the calculation of expected mean squares and influence the F-ratio. Understanding their correct application is essential, as they help in quantifying the variability attributable to the wider universe of data from which the sample is drawn, simplifying the explanation and generalization of findings.
Hypothesis Testing
Hypothesis Testing in Two-Way ANOVA involves systematically evaluating whether there are significant effects present due to specific factors or their interaction. Generally, the hypotheses focus on whether variance attributed to a factor (or interaction) equals zero.
For example, a hypothesis like \( H_{0G}: \sigma_G^2 = 0 \) vs. \( H_{aG}: \sigma_G^2 > 0 \) tests if there’s no interaction effect, suggesting all group means are equal despite random variances. If hypothesis testing concludes with rejecting \( H_0 \), it signifies noteworthy differences likely due to the factor under study.
  • Define null and alternative hypotheses.
  • Calculate the relevant mean squares and F-ratios.
  • Compare calculated F to the critical F-value.
  • Make a decision: reject or fail to reject the null hypothesis.
Hypothesis testing is about making decisions based on the calculated F-ratios, fulfilling a central role in the analytical framework of ANOVA.

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Most popular questions from this chapter

Four laboratories \((1-4)\) are randomly selected from a large population, and each is asked to make three determinations of the percentage of methyl alcohol in specimens of a compound taken from a single batch. Based on the accompanying data, are differences among laboratories a source of variation in the percentage of methyl alcohol? State and test the relevant hypotheses using significance level .05. \(\begin{array}{llll}1: & 85.06 & 85.25 & 84.87 \\ 2: & 84.99 & 84.28 & 84.88 \\\ 3: & 84.48 & 84.72 & 85.10 \\ 4: & 84.10 & 84.55 & 84.05\end{array}\)

In single-factor ANOVA, suppose the \(x_{i j}\) 's are "coded" by \(y_{i j}=c x_{i j}+d\). How does the value of the \(F\) statistic computed from the \(y_{i j}\) 's compare to the value computed from the \(x_{i j}\) 's? Justify your assertion.

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.

The article "Computer-Assisted Instruction Augmented with Planned Teacher/Student Contacts" (J. Exper. Ed., Winter 1980-1981: 120-126) compared five different methods for teaching descriptive statistics. The five methods were traditional lecture and discussion (L/D), programmed textbook instruction (R), programmed text with lectures (R/L), computer instruction (C), and computer instruction with lectures (C/L). Forty-five students were randomly assigned, 9 to each method. After completing the course, the students took a 1-h exam. In addition, a 10-minute retention test was administered 6 weeks later. Summary quantities are given. $$ \begin{array}{lcccc} & {\text { Exam }} & {\text { Retention Test }} \\ \text { Method } & \overline{\boldsymbol{x}}_{\boldsymbol{i}} \cdot & \boldsymbol{s}_{\boldsymbol{i}} & \overline{\boldsymbol{x}}_{\boldsymbol{i}} \cdot & \boldsymbol{s}_{\boldsymbol{i}} \\ \hline \text { L/D } & 29.3 & 4.99 & 30.20 & 3.82 \\ \text { R } & 28.0 & 5.33 & 28.80 & 5.26 \\ \text { R/L } & 30.2 & 3.33 & 26.20 & 4.66 \\ \text { C } & 32.4 & 2.94 & 31.10 & 4.91 \\ \text { C/L } & 34.2 & 2.74 & 30.20 & 3.53 \end{array} $$ The grand mean for the exam was \(30.82\), and the grand mean for the retention test was \(29.30\). a. Does the data suggest that there is a difference among the five teaching methods with respect to true mean exam score? Use \(\alpha=.05\). b. Using a \(.05\) significance level, test the null hypothesis of no difference among the true mean retention test scores for the five different teaching methods.

a. Show that a constant \(d\) can be added to (or subtracted from) each \(x_{i j}\) without affecting any of the ANOVA sums of squares. b. Suppose that each \(x_{i j}\) is multiplied by a nonzero constant \(c\). How does this affect the ANOVA sums of squares? How does this affect the values of the \(F\) statistics \(F_{A}\) and \(F_{B}\) ? What effect does "coding" the data by \(y_{i j}=c x_{i j}+d\) have on the conclusions resulting from the ANOVA procedures?

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