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Given the following ANOVA output, answer the questions that follow. \(\begin{array}{lrrrrr}\text { Source } & \text { df } & \text { SS } & \text { MS } & F & P \\ \text { Factor A } & 2 & 156 & 78 & 0.39 & 0.679 \\ \text { Factor B } & 2 & 132 & 66 & 0.33 & 0.720 \\ \text { Interaction } & 4 & 311 & 78 & 0.39 & 0.813 \\ \text { Error } & 27 & 5354 & 198 & & \\ \text { Total } & 35 & 5952 & & & \end{array}\) (a) Is there evidence of an interaction effect? Why or why not? (b) Based on the \(P\) -value, is there evidence of a difference in the means from factor A? Based on the \(P\) -value, is there evidence of a difference in the means from factor \(\mathrm{B} ?\) (c) What is the mean square error?

Short Answer

Expert verified
No interaction effect (P=0.813). No difference in means for Factor A (P=0.679). No difference in means for Factor B (P=0.720). The mean square error is 198.

Step by step solution

01

Identify Interaction Effect from P-value

To determine if there is evidence of an interaction effect, examine the P-value for the Interaction row. If the P-value is less than 0.05, there is evidence of an interaction effect.
02

Interpretation of the Interaction Effect

The given P-value for the Interaction effect is 0.813, which is greater than 0.05. Therefore, there is no evidence of an interaction effect.
03

Determine Evidence from Factor A Using P-value

To check if there is evidence for Factor A, look at the P-value in the Factor A row. If this P-value is less than 0.05, there is evidence of a difference in means.
04

Interpretation for Factor A

The given P-value for Factor A is 0.679, which is greater than 0.05. Therefore, there is no evidence of a difference in means for Factor A.
05

Determine Evidence from Factor B Using P-value

Similarly, evaluate Factor B by examining its P-value. If the P-value is less than 0.05, then there is evidence of a difference in means.
06

Interpretation for Factor B

The P-value for Factor B is 0.720, which is also greater than 0.05. Thus, there is no evidence of a difference in means for Factor B.
07

Mean Square Error Calculation

The mean square error can be found in the Error row's MS column.. Here, the mean square error is clearly given as 198.

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Key Concepts

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

interaction effect
In ANOVA (Analysis of Variance), the interaction effect investigates whether two factors interact with each other to affect the outcome variable. Simply put, it asks if the effect of one factor depends on the level of another factor.

In our given ANOVA table, we check the interaction effect by looking at the P-value in the 'Interaction' row. To determine if the interaction is significant, the P-value should be less than 0.05. If it is, this indicates there is a statistically significant interaction effect between the factors.

For our exercise, the provided P-value for the interaction effect is 0.813, which is well above 0.05. This means there is no statistically significant interaction effect. In other words, the combined influence of Factor A and Factor B on the outcome is not greater than would be expected based on their individual influences alone.
P-value interpretation
P-value is a crucial element in statistics for determining the significance of results. In the context of ANOVA, the P-value helps us conclude whether the observed differences among group means are statistically significant.

When the P-value is less than the chosen significance level (commonly 0.05), this indicates strong evidence against the null hypothesis, suggesting that at least one group's mean is different.

For Factor A, the P-value is 0.679. Since this is greater than 0.05, it suggests there is no significant difference in means among the groups defined by Factor A.

Similarly, Factor B has a P-value of 0.720. Again, this is more than 0.05, indicating no significant difference in means for the groups defined by Factor B either.

This interpretation helps us understand if the factors individually have a significant impact on the outcome or not. In our example, both Factor A and B do not exhibit significant effects.
mean square error
Mean Square Error (MSE) is an important measure in ANOVA. It estimates the variance within the groups, acting as an indicator of the variability in individual observations that is not explained by the factors.

MSE is calculated as the sum of squares due to error (SS Error) divided by its degrees of freedom (df Error). It is found in the 'Error' row of the ANOVA table under the 'MS' column. In our example, the MSE is 198.

Here's why MSE is critical: it provides a baseline for comparing the variance explained by the factors against the unexplained variance. A lower MSE implies that the model (including factors and their levels) explains much of the variability in the response. Conversely, a high MSE suggests there's considerable unexplained variability.

Knowing MSE helps us understand the overall quality of the model used in ANOVA. Be sure to always note the MSE when interpreting ANOVA results, as it plays a pivotal role in understanding the data's underlying variation.

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Most popular questions from this chapter

In Problems 5 and \(6,\) assume that the data come from populations that are normally distributed with the same variance $$ \begin{array}{cccc} \text { Block } & \text { Treatment 1 } & \text { Treatment 2 } & \text { Treatment 3 } \\ \hline \mathbf{1} & 9.7 & 8.4 & 8.8 \\ \hline \mathbf{2} & 10.4 & 8.9 & 8.5 \\ \hline \mathbf{3} & 10.5 & 9.3 & 9.0 \\ \hline \mathbf{4} & 10.7 & 10.5 & 9.3 \\ \hline \mathbf{5} & 11.1 & 10.7 & 10.3 \\ \hline \end{array} $$ (a) Test \(H_{0}: \mu_{1}=\mu_{2}=\mu_{3}\) against \(H_{1}:\) at least one of the means is different, where \(\mu_{1}\) is the mean for treatment \(1,\) and so on, at the \(\alpha=0.05\) level of significance. (b) If the null hypothesis from part (a) was rejected, use Tukey's test to determine which pairwise means differ using a familywise error rate of \(\alpha=0.05\). (c) Draw boxplots of the data for each treatment using the same scale to support the analytical results obtained in parts (a) and (b).

The data in the table represent the number of corn plants in randomly sampled rows (a 17 -foot by 5 -inch strip ) for various types of plot. An agricultural researcher wants to know whether the mean number of plants for each plot type are equal. $$\begin{array}{lcccccc}\text { Plot Type } & {\text { Number of Plants }} \\\\\hline \text { Sludge plot } & 25 & 27 & 33 & 30 & 28 & 27 \\\\\hline \text { Spring disc } & 32 & 30 & 33 & 35 & 34 & 34 \\\\\hline \text { No till } & 30 & 26 & 29 & 32 & 25 & 29\end{array}$$ (a) Write the null and alternative hypotheses. (b) State the requirements that must be satisfied to use the oneway ANOVA procedure. (c) Use the following partial Minitab output to test the hypothesis of equal means at the \(\alpha=0.05\) level of significance $$\begin{aligned}&\text { One-way ANOVA: Sludge Plot, Spring Disc, No Till }\\\&\begin{array}{lrrrrr}\text { Source } & \text { df } & \text { SS } & \text { MS } & F & P \\\\\text { Factor } & 2 & 84.11 & 42.06 & 7.10 & 0.007 \\\\\text { Error } & 15 & 88.83 & 5.92 & & \\\\\text { Total } & 17 & 172.94 & & &\end{array}\end{aligned}$$ (d) Shown are side-by-side boxplots of each type of plot. Do these boxplots support the results obtained in part (c)? (e) Verify that the \(F\) -test statistic is 7.10 . (f) Verify the residuals are normally distributed.

Researchers wanted to determine if the psychological profile of healthy children was different than for children suffering from recurrent abdominal pain (RAP) or recurring headaches. A total of 210 children and adolescents were studied and their psychological profiles were graded according to the Child Behavior Checklist \(4-18\) (CBCL). Children were stratified in two age groups: 4 to 11 years and 12 to 18 years. The results of the study are summarized in the following table: $$\begin{array}{lccc} & n & \text { Sample Mean } & \text { Sample Variance } \\\\\hline \text { Control group } & 70 & 11.7 & 21.6 \\\\\hline \text { RAP } & 70 & 9.0 & 13.0 \\\\\hline \text { Headache } & 70 & 12.4 & 8.4\end{array}$$ (a) Compute the sample standard deviations for each group. (b) What sampling method was used for each treatment group? Why? (c) Use a two sample \(t\) -test for independent samples to determine if there is a significant difference in mean \(\mathrm{CBCL}\) scores between the control group and the RAP group (assume that both samples are simple random samples). (d) Is it necessary to check the normality assumption to answer part (c)? Explain. (e) Use the one-way ANOVA procedure with \(\alpha=0.05\) to determine if the mean CBCL scores are different for the three treatment groups. (f) Based on your results from parts (c) and (e), can you determine if there is a significant difference between the mean scores of the RAP group and the headache group? Explain.

A medical researcher wanted to determine the effectiveness of coagulants on the healing rate of a razor cut on lab mice. Because healing rates of mice vary from mouse to mouse, the researcher decided to block by mouse. First, the researcher gave each mouse a local anesthesia and then made a 5 -mm incision that was \(2 \mathrm{~mm}\) deep on each mouse. He randomly selected one of the three treatments and recorded the time it took for the wound to stop bleeding (in minutes). He repeated this process two more times on each mouse and obtained the results shown. $$ \begin{array}{cccc} \text { Mouse } & \text { No Drug } & \begin{array}{l} \text { Experimental } \\ \text { Drug 1 } \end{array} & \begin{array}{l} \text { Experimental } \\ \text { Drug 2 } \end{array} \\ \hline \mathbf{1} & 3.2 & 3.4 & 3.4 \\ \hline \mathbf{2} & 4.8 & 4.4 & 3.4 \\ \hline \mathbf{3} & 6.6 & 5.9 & 5.4 \\ \hline \mathbf{4} & 6.5 & 6.3 & 5.2 \\ \hline \mathbf{5} & 6.4 & 6.3 & 6.1 \\ \hline \end{array} $$ (a) Explain how each mouse forms a block. Explain how blocking might reduce variability of time to heal. (b) Normal probability plots for each treatment indicate that the requirement of normality is satisfied. Verify that the requirement of equal population variances for each treatment is satisfied. (c) Is there sufficient evidence that the mean healing time is different among the three treatments at the \(\alpha=0.05\) level of significance? (d) If the null hypothesis from part (c) was rejected, use Tukey's test to determine which pairwise means differ using a familywise error rate of \(\alpha=0.05 .\) (e) Based on your results for part (d), what do you conclude?

An automotive engineer wanted to determine whether the octane of gasoline used in a car increases gas mileage. Recognizing that car and driver are variables that affect gas mileage, he selected six different brands of car and assigned a driver to each car, so he blocked by car type and driver. For each car (and driver), the researcher randomly selected a number from 1 to 3 , with 1 representing 87 -octane gasoline, 2 representing 89 -octane gasoline, and 3 representing 92 -octane gasoline. Then 5 gallons of the gasoline selected was placed in the car. The car was driven around a closed track at 40 miles per hour until the car ran out of gas. The number of miles driven was recorded and then divided by 5 to obtain the miles per gallon. He obtained the following results: $$ \begin{array}{lccc} & \mathbf{8 7} \text { Octane } & \mathbf{8 9} \text { Octane } & \mathbf{9 2} \text { Octane } \\ \hline \text { Chevrolet Impala } & 28.3 & 28.4 & 28.7 \\ \hline \text { Chrysler } \mathbf{3 0 0 M} & 27.1 & 26.9 & 27.2 \\ \hline \text { Ford Taurus } & 26.4 & 26.1 & 26.8 \\ \hline \text { Lincoln LS } & 26.1 & 26.4 & 27.3 \\ \hline \text { Toyota Camry } & 28.4 & 28.9 & 29.1 \\ \hline \text { Volvo S60 } & 25.3 & 25.1 & 25.8 \end{array} $$ (a) Normal probability plots for each treatment indicate that the requirement of normality is satisfied. Verify that the requirement of equal population variances for each treatment is satisfied. (b) Explain the role blocking plays in reducing the variability of gas mileage. (c) Is there sufficient evidence that the mean miles per gallon are different among the three octane levels at the \(\alpha=0.05\) level of significance? (d) If the null hypothesis from part (c) was rejected, use Tukey's test to determine which pairwise means differ using a familywise error rate of \(\alpha=0.05\)

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