/*! 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 55 Consider the ANOVA with \(a=2\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the ANOVA with \(a=2\) treatments. Show that the \(M S_{E}\) in this analysis is equal to the pooled variance estimate used in the two-sample \(t\) -test.

Short Answer

Expert verified
The MSE in ANOVA with two treatments is equal to the pooled variance estimate in the two-sample t-test.

Step by step solution

01

Understand the problem

We need to show that the mean square error (MSE) in a two-treatment ANOVA is equal to the pooled variance estimate used in the two-sample \( t \)-test.
02

Define MSE in ANOVA

In ANOVA, the MSE is given by \( MSE = \frac{SS_E}{df_E} \) where \( SS_E \) is the sum of squares for error and \( df_E \) is the degrees of freedom for error.
03

Define Pooled Variance in t-test

In a two-sample \( t \)-test, the pooled variance estimate is \( s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} \) where \( s_1^2 \) and \( s_2^2 \) are the sample variances, and \( n_1 \) and \( n_2 \) are the sample sizes.
04

Connect ANOVA and t-test

Notice that in ANOVA, with two treatments, \( a = 2 \), the error sum of squares \( SS_E \) is composed of the differences within each treatment group, just as the numerator in the pooled variance estimate. The degrees of freedom for error \( df_E = n_1 + n_2 - 2 \).
05

Show MSE and Pooled Variance are Equivalent

Since in both the ANOVA and the t-test, the expressions for the error (or within-group) sum of squares and degrees of freedom are identical, the resulting MSE is equal to the pooled variance estimate, i.e., \( MSE = s_p^2 \).

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Key Concepts

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

Mean Square Error (MSE) in ANOVA
The mean square error, often abbreviated as MSE, is a crucial concept in the analysis of variance (ANOVA). It represents the average squared discrepancies between observed values and their corresponding group means. In a two-treatment ANOVA, MSE helps us understand how much of the variation in data can be attributed to differences within groups, rather than between them. In mathematical terms, MSE is calculated as:\[ MSE = \frac{SS_E}{df_E} \]Here, \(SS_E\) stands for the sum of squares for error and \(df_E\) denotes the degrees of freedom for error.
  • \(SS_E\) captures the total variability within groups.
  • \(df_E\) is determined by the total number of observations minus the number of groups.
Understanding MSE in ANOVA is critical as it provides a baseline against which treatment effects can be compared. It indicates how consistent your measures are and how tightly your data clusters around the group means.
Two-Sample T-Test
The two-sample t-test is a statistical procedure used to determine whether there is a significant difference between the means of two independent groups. This test is commonly applied when comparing the means of two samples and is ideal when the data is normally distributed and variances are equal or nearly equal. Here's a breakdown of the two-sample t-test:- It evaluates the null hypothesis, which states that there is no difference between the population means of the two groups.- The test computes a t-statistic, which is used to determine the probability of observing the data if the null hypothesis is true.A critical component of the two-sample t-test is the pooled variance estimate, given by:\[ s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} \]This pooled variance \(s_p^2\) integrates the two groups' sample variances \(s_1^2\) and \(s_2^2\), reflecting the shared variance based on the assumption that both samples are from populations with the same variance. This characteristic is what links it closely to the MSE in ANOVA.
Pooled Variance
Pooled variance is a concept used to combine variances from two or more groups, assuming the samples come from populations with equal variances. It provides a more reliable estimate of variance across the groups, as it considers sample size and variance from each group.Here's how pooled variance is calculated:- Factor in each group's variance and degrees of freedom.- Combine them using the formula: \[ s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} \]- \(n_1\) and \(n_2\) are the number of observations in each group.- \(s_1^2\) and \(s_2^2\) are the sample variances.Pooled variance is a cornerstone in both the two-sample t-test and ANOVA with two treatments. In both cases, it represents how much random error is present within groups, thus enabling comparison of means with a single measure of variability. By understanding this shared variance, analysts can draw more accurate conclusions about the differences between group means.

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Most popular questions from this chapter

Consider testing the equality of the means of two normal populations where the variances are unknown but are assumed equal. The appropriate test procedure is the two-sample \(t\) -test. Show that the two-sample \(t\) -test is equivalent to the single-factor analysis of variance \(F\) -test.

Suppose that four normal populations have common variance \(\sigma^{2}=25\) and means \(\mu_{1}=50, \mu_{2}=60, \mu_{3}=50,\) and \(\mu_{4}=60 .\) How many observations should be taken on each population so that the probability of rejecting the hypothesis of equality of means is at least \(0.90 ?\) Use \(\alpha=0.05 .\)

An article in Lubrication Engineering (December 1990) described the results of an experiment designed to investigate the effects of carbon material properties on the progression of blisters on carbon face seals. The carbon face seals are used extensively in equipment such as air turbine starters. Five different carbon materials were tested, and the surface roughness was measured. The data are as follows: $$\begin{array}{llllll}\hline {\begin{array}{l}\text { Carbon } \\\\\text { Material } \\\\\text { Type }\end{array}} & {\text { Surface Roughness }} \\\\\hline \text { EC10 } & 0.50 & 0.55 & 0.55 & 0.36 & \\\\\text { EC10A } & 0.31 & 0.07 & 0.25 & 0.18 & 0.56 & 0.20 \\\\\text { EC4 } & 0.20 & 0.28 & 0.12 & & \\\\\text { ECl } & 0.10 & 0.16 & & & \\\\\hline\end{array}$$ (a) Does carbon material type have an effect on mean surface roughness? Use \(\alpha=0.05\). (b) Find the residuals for this experiment. Does a normal probability plot of the residuals indicate any problem with the normality assumption? (c) Plot the residuals versus \(\hat{y}_{i j}\). Comment on the plot. (d) Find a \(95 \%\) confidence interval on the difference between the mean surface roughness between the EC 10 and the EC 1 carbon grades. (e) Apply the Fisher LSD method to this experiment. Summarize your conclusions regarding the effect of material type on surface roughness.

Consider the random-effect model for the single-factor completely randomized design. Show that a \(100(1-\alpha) \%\) confidence interval on the ratio of variance components \(\sigma_{\tau}^{2} / \sigma^{2}\) is given by $$L \leq \frac{\sigma_{\tau}^{2}}{\sigma^{2}} \leq U$$ where $$L=\frac{1}{n}\left[\frac{M S_{\text {Treatments }}}{M S_{E}} \times\left(\frac{1}{f_{\alpha / 2, a-1, N-a}}\right)-1\right]$$ and$$U=\frac{1}{n}\left[\frac{M S_{\text {Treatments }}}{M S_{E}} \times\left(\frac{1}{f_{1-\alpha / 2, a-1, Na}}\right)-1\right]$$

An article in the Journal of Agricultural Engineering Research (Vol. \(52,1992,\) pp. \(53-76\) ) described an experiment to investigate the effect of drying temperature of wheat grain on the baking quality of bread. Three temperature levels were used, and the response variable measured was the volume of the loaf of bread produced. The data are as follows: $$\begin{array}{clllll}\hline \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) &{\text { Volume }(\mathrm{CC})} \\\\\hline 70.0 & 1245 & 1235 & 1285 & 1245 & 1235 \\\75.0 & 1235 & 1240 & 1200 & 1220 & 1210 \\\80.0 & 1225 & 1200 & 1170 & 1155 & 1095 \\\\\hline\end{array}$$ (a) Does drying temperature affect mean bread volume? Use $$\alpha=0.01$$ (b) Find the \(P\) -value for this test. (c) Use the Fisher LSD method to determine which means are different. (d) Analyze the residuals from this experiment and comment on model adequacy.

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