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Chapter 10: Problem 10

In single-factor ANOVA with \(I\) treatments and \(J\) observations per treatment, let \(\mu=(1 / I) \sum \mu_{i}\). a. Express \(E\left(\bar{X}_{\underline{.}}\right)\) in terms of \(\mu\). [Hint: \(\bar{X}_{. .}=(1 / I) \sum \bar{X}_{i \cdot}\) ] b. Compute \(E\left(\bar{X}_{i .}^{2}\right)\). [Hint: For any rv \(Y, E\left(Y^{2}\right)=V(Y)+\) \(\left.[E(Y)]^{2} .\right]\) c. Compute \(E\left(\bar{X}_{. .}^{2}\right)\). d. Compute \(E(\mathrm{SSTr})\) and then show that $$ E(\mathrm{MSTr})=\sigma^{2}+\frac{J}{I-1} \sum\left(\mu_{i}-\mu\right)^{2} $$ e. Using the result of part (d), what is E(MSTr) when \(H_{0}\) is true? When \(H_{0}\) is false, how does \(E(\mathrm{MSTr})\) compare to \(\sigma^{2} ?\)

Short Answer

Expert verified
a. \( E(\bar{X}_{..}) = \mu \); b. \( E(\bar{X}_{i.}^{2}) = \frac{\sigma^{2}}{J} + \mu_{i}^{2} \); c. \( E(\bar{X}_{..}^{2}) = \frac{\sigma^{2}}{IJ} + \mu^{2} \); d. show result; e. \( E(MSTr) = \sigma^{2} \) if \( H_{0} \) true; else \( > \sigma^{2} \).

Step by step solution

01

Expressing E(X虅..) in terms of 渭

Given that \( \bar{X}_{..} = (1 / I) \sum \bar{X}_{i.} \), we need to express \( E(\bar{X}_{..}) \) in terms of \( \mu \). We have:\[ E(\bar{X}_{..}) = E\left(\frac{1}{I} \sum \bar{X}_{i.}\right) = \frac{1}{I} \sum E(\bar{X}_{i.}) \]The expected value \( E(\bar{X}_{i.}) \) is \( \mu_{i} \), so:\[ E(\bar{X}_{..}) = \frac{1}{I} \sum \mu_{i} = \mu \]
02

Computing E(X虅i.^2)

We want to compute \( E(\bar{X}_{i.}^{2}) \). Using the hint, for any random variable \( Y \), we have:\[ E(Y^{2}) = V(Y) + [E(Y)]^{2} \]Substitute \( Y = \bar{X}_{i.} \), so:\[ E(\bar{X}_{i.}^{2}) = V(\bar{X}_{i.}) + [E(\bar{X}_{i.})]^{2} \]The variance \( V(\bar{X}_{i.}) = \frac{\sigma^{2}}{J} \), hence:\[ E(\bar{X}_{i.}^{2}) = \frac{\sigma^{2}}{J} + \mu_{i}^{2} \]
03

Computing E(X虅..^2)

Now, we compute \( E(\bar{X}_{..}^{2}) \). Since \( \bar{X}_{..} = (1 / I) \sum \bar{X}_{i.} \):\[ E(\bar{X}_{..}^{2}) = V(\bar{X}_{..}) + [E(\bar{X}_{..})]^{2} \]We previously found \( E(\bar{X}_{..}) = \mu \), and \( V(\bar{X}_{..}) = \frac{\sigma^{2}}{IJ} \):\[ E(\bar{X}_{..}^{2}) = \frac{\sigma^{2}}{IJ} + \mu^{2} \]
04

Computing E(SSTr)

The Sum of Squares for Treatments (SSTr) is given by:\[ SSTr = J \sum (\bar{X}_{i.} - \bar{X}_{..})^{2} \]We compute \( E(SSTr) \):\[ E(SSTr) = J \sum (V(\bar{X}_{i.}) + [E(\bar{X}_{i.}) - E(\bar{X}_{..})]^{2}) \]Substitute known values:\[ E(SSTr) = J \sum \left(\frac{\sigma^{2}}{J} + (\mu_{i} - \mu)^{2}\right) = \sum \sigma^{2} + J \sum (\mu_{i} - \mu)^{2} \]
05

Showing E(MSTr) formula

The Mean Squares for Treatment (MSTr) is \( MSTr = \frac{SSTr}{I-1} \). Thus:\[ E(MSTr) = \frac{1}{I-1} E(SSTr) = \frac{1}{I-1} \left(I\sigma^{2} + J \sum (\mu_{i} - \mu)^{2}\right) \]This simplifies to:\[ E(MSTr) = \sigma^{2} + \frac{J}{I-1} \sum (\mu_{i} - \mu)^{2} \]
06

Impact of Hypotheses on E(MSTr)

If \( H_{0} \) is true, all \( \mu_{i} = \mu \) making the second term zero:\[ E(MSTr) = \sigma^{2} \]If \( H_{0} \) is false, the second term is positive, so \( E(MSTr) > \sigma^{2} \).

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Key Concepts

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

Single-Factor ANOVA
ANOVA, which stands for Analysis of Variance, is a statistical method used to analyze differences among group means. In single-factor ANOVA, we focus on one independent variable or factor that has different levels, often known as treatments. The essence of this method is to determine if there is a significant difference in treatment means. Single-factor ANOVA is a powerful tool for comparing more than two means at the same time, unlike a t-test, which only compares two means. It allows researchers to see if any of the treatment groups differ significantly from each other.
  • Treatments: The different levels or categories of the factor being studied.
  • Observations: The data or responses collected under each treatment.
Understanding single-factor ANOVA equips students with tools to explore variations within and between groups, establishing a foundation for more complex analyses.
Treatment Means
In the context of ANOVA, treatment means refer to the average outcomes measured within each treatment group, also called treatment levels. For instance, if you have three different fertilizers used on plants, the treatment means are the average plants' growth results for each fertilizer. Calculating these means is the first step in determining if there are significant differences between different treatments.
  • Calculated as sum of all observations in a group divided by the number of observations in that group.
  • Plays a crucial role in forming hypotheses about overall mean differences.
Comparing treatment means can unveil insightful variations affected by different treatment conditions, thus it forms a crucial aspect of ANOVA.
Expected Value
Expected value, often denoted as \( E(X) \), is a significant concept in statistics that represents the average or mean value of a random variable in a probability distribution. In single-factor ANOVA, the expected value helps to predict the mean response over all treatments when factors do not affect the response variable.
  • It can be thought of as the center of mass of a distribution.
  • In ANOVA, the expected value of the overall mean is typically denoted as \( \mu \).
  • The formula for expected value is \( E(\bar{X}_{..}) = \mu \).
Using expected values allows researchers to make informed predictions and decisions based on anticipated outcomes over repeated trials.
Variance
Variance measures how far a set of numbers are spread out from their average value. In ANOVA, variance is a crucial measure to determine how much treatments differ from one another and if there's a significant divergence in the data. Variance is represented among the treatments and within treatments to analyze the distribution of the data.
  • Within-group variance: Variation due to differences within treatment groups.
  • Between-group variance: Variation due to differences between treatment means.
  • It is calculated as \( V(Y) = E(Y^2) - [E(Y)]^2 \).
Understanding how variance operates in the context of ANOVA can identify the consistency or variability of treatment effects, aiding accurate results interpretation.
Hypothesis Testing
Hypothesis testing in ANOVA involves formulating and testing assumptions about treatment means. The null hypothesis \( H_{0} \), state that all treatment means are equal, thus any observed variance is due to random sampling error. The alternative hypothesis \( H_{a} \) suggests that at least one treatment mean is different.
  • Null Hypothesis (\(H_{0}\)): \( \mu_{1} = \mu_{2} = \ldots = \mu_{I} \)
  • Alternative Hypothesis (\(H_{a}\)): Not all \( \mu_{i} \) are equal
  • This testing helps to identify if the variance between groups is greater than what could be expected by chance.
  • If \( H_{0} \) is true, the Mean Squares for Treatment (MSTr) equals \( \sigma^{2} \), otherwise it is greater.
Hypothesis testing is the cornerstone of statistical analysis in ANOVA, driving decisions regarding the effects and differences in treatments.

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Most popular questions from this chapter

The article "Computer-Assisted Instruction Augmented with Planned Teacher/Student Contacts" (J. Exp. Educ., 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 \((\mathrm{R})\), 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-hour exam. In addition, a 10-minute retention test was administered 6 weeks later. Summary quantities are given. $$ \begin{array}{lcccc} & {}{}{\text { Exam }} && {}{}{\text { Retention Test }} \\ \hline \text { Method } & \overline{\boldsymbol{x}}_{\boldsymbol{i} \cdot} & \boldsymbol{s}_{i} & \overline{\boldsymbol{x}}_{i \cdot} & \boldsymbol{s}_{i} \\\ \hline \mathrm{L} / \mathrm{D} & 29.3 & 4.99 & 30.20 & 3.82 \\ \mathrm{R} & 28.0 & 5.33 & 28.80 & 5.26 \\ \mathrm{R} / \mathrm{L} & 30.2 & 3.33 & 26.20 & 4.66 \\ \mathrm{C} & 32.4 & 2.94 & 31.10 & 4.91 \\ \mathrm{C} / \mathrm{L} & 34.2 & 2.74 & 30.20 & 3.53 \\ \hline \end{array} $$ 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.

The article "Origin of Precambrian Iron Formations" (Econ. Geology, 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.

In a study to assess the effects of malaria infection on mosquito hosts ("Plasmodium Cynomolgi: Effects of Malaria Infection on Laboratory Flight Performance of Anopheles Stephensi Mosquitos," Experimental Parasitology, 1977: 397-404), mosquitos were fed on either infective or noninfective rhesus monkeys. Subsequently the distance they flew during a 24-hour period was measured using a flight mill. The mosquitos were divided into four groups of eight mosquitos each: infective rhesus and sporozites present (IRS), infective rhesus and oocysts present (IRD), infective rhesus and no infection developed (IRN), and noninfective (C). The summary data values are \(\bar{x}_{1}=4.39\left(\right.\) IRS), \(\bar{x}_{2}=4.52\) (IRD), \(\bar{x}_{3 .}=5.49(\mathrm{IRN}), \bar{x}_{4}=6.36(\mathrm{C}), \bar{x}_{. .}=5.19\), and \(\sum \sum x_{i j}^{2}=\) 911.91. Use the ANOVA \(F\) test at level .05 to decide whether there are any differences between true average flight times for the four treatments.

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 \(x_{i j} \mathrm{~s}\) ? Justify your assertion.

Four types of mortars-ordinary cement mortar (OCM), polymer impregnated mortar (PIM), resin mortar (RM), and polymer cement mortar (PCM)-were subjected to a compression test to measure strength (MPa). Three strength observations for each mortar type are given in the article. "Polymer Mortar Composite Matrices for MaintenanceFree Highly Durable Ferrocement" (J. Ferrocement, 1984: 337-345) and are reproduced here. Construct an ANOVA table. Using a \(.05\) significance level, determine whether the data suggests that the true mean strength is not the same for all four mortar types. If you determine that the true mean strengths are not all equal, use Tukey's method to identify the significant differences. $$ \begin{array}{lrrr} O C M & 32.15 & 35.53 & 34.20 \\ \text { PIM } & 126.32 & 126.80 & 134.79 \\ R M & 117.91 & 115.02 & 114.58 \\ \text { PCM } & 29.09 & 30.87 & 29.80 \end{array} $$
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