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Chapter 13: Problem 13

Given the following ANOVA output, answer the questions that follow. \(\begin{array}{lrrrrr}\text { Source } & \text { df } & \text { SS } & \text { MS } & F & P \\ \text { Factor A } & 2 & 2269.8 & 1134.9 & 35.63 & 0.000 \\\ \text { Factor B } & 2 & 115.2 & 57.6 & 1.81 & 0.183 \\ \text { Interaction } & 4 & 1694.8 & 423.7 & 13.30 & 0.000 \\ \text { Error } & 27 & 860.0 & 31.9 & & \\ \text { Total } & 35 & 4939.8 & & & \end{array}\) (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
There is an interaction effect. Evidence shows differences in means for Factor A but not for Factor B. The mean square error is 31.9.

Step by step solution

01

Determine Interaction Effect

Look at the P-value for the 'Interaction' row. The P-value is given as 0.000. Since this value is less than 0.05, there is evidence of a significant interaction effect.
02

Analyze Factor A

Examine the P-value for 'Factor A'. The P-value for Factor A is 0.000. Since this value is less than 0.05, there is evidence of a significant difference in the means from Factor A.
03

Analyze Factor B

Check the P-value for 'Factor B'. The P-value for Factor B is 0.183. Since this value is greater than 0.05, there is no evidence of a significant difference in the means from Factor B.
04

Find Mean Square Error

Locate the mean square value in the 'Error' row. The mean square error (MSE) is the value under the 'MS' column for the 'Error' row, which is 31.9.

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Key Concepts

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

P-value interpretation
Understanding the P-value is vital when analyzing the results from ANOVA. A P-value helps us determine whether the observed data deviates significantly from what is expected under the null hypothesis. In simple terms, it tells us if the effect we are seeing is likely due to chance or if it is statistically significant.

The common threshold for significance is 0.05. If the P-value is less than 0.05, we reject the null hypothesis and conclude that there is evidence of a significant effect.
  • For Factor A in the given output, the P-value is 0.000. This is less than 0.05, indicating a significant difference in the means from Factor A.
  • For Factor B, the P-value is 0.183. Since this is greater than 0.05, there is no significant difference in the means from Factor B.
  • The interaction term also has a P-value of 0.000, indicating a significant interaction effect.
Interpreting these P-values allows us to understand whether our factors and their interactions have significant impacts.

Interaction effect
An interaction effect in ANOVA occurs when the effect of one factor depends on the level of another factor. This means the combined effect of two factors is different from their individual effects.

In the provided example, we look at the 'Interaction' row and note the P-value is 0.000. This value is much less than 0.05, so we conclude that there is a significant interaction effect between Factor A and Factor B.
  • An interaction plot can be helpful to visualize this effect. The lines will show whether they cross or have different slopes, indicating that the effect of one factor changes depending on the level of the other factor.
  • Understanding interaction effects is important for comprehensive data analysis, as it provides insights into how different variables work together.
Overall, a significant interaction effect means that the impact of one factor is influenced by the presence or level of another factor.

Mean square error
Mean square error (MSE) is a measure used to quantify the average of the squared differences between observed and estimated values. In ANOVA, MSE helps indicate the variance within the groups.

In the ANOVA table, you can find the MSE under the 'Error' row in the 'MS' column. In this example, the MSE is 31.9.
  • MSE is calculated by dividing the sum of squares for errors (SS) by the degrees of freedom (df) for errors. In this case, the sum of squares for errors (860.0) is divided by the degrees of freedom (27), resulting in 31.9.
  • MSE provides an estimate of the variance within the groups being compared, which is essential for understanding the overall variability in the data.
By analyzing the MSE, researchers can understand how much of the observed variance is due to random error, aiding in more accurate and reliable interpretations of the results.

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Most popular questions from this chapter

Researchers Olivia Carrillo-Gamboa and Richard Gunst presented the following data in their article, "Measurement-Error-Model Collinearities." The data represent the compressive strength (in pounds per square inch) of a random sample of concrete 2 days, 7 days, and 28 days after pouring. $$ \begin{array}{cccc} \text { Concrete } & \text { 2 Days } & \text { 7 Days } & \text { 28 Days } \\\ \hline \mathbf{1} & 2830 & 3505 & 4470 \\ \hline \mathbf{2} & 3295 & 3430 & 4740 \\ \hline \mathbf{3} & 2710 & 3670 & 5115 \\ \hline \mathbf{4} & 2855 & 3355 & 4880 \\ \hline \mathbf{5} & 2980 & 3985 & 4445 \\ \hline \mathbf{6} & 3065 & 3630 & 4080 \\ \hline \mathbf{7} & 3765 & 4570 & 5390 \end{array} $$ (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) Is there sufficient evidence that the mean strength is different among the three days at the \(\alpha=0.05\) level of significance? (c) If the null hypothesis from part (b) was rejected, use Tukey's test to determine which pairwise means differ using a familywise error rate of \(\alpha=0.05 .\)

Putting It Together: Time to Complete a Degree A researcher wanted to determine if the mean time to complete a bachelor's degree was different depending on the selectivity of the first institution of higher education that was attended. The following data represent a random sample of 12 th-graders who earned their degree within eight years. Probability plots indicate that the data for each treatment level are normally distributed. $$ \begin{array}{ccccc} \text { Highly } & & & \text { Not } \\ \text { Selective } & \text { Selective } & \text { Nonselective } & \text { Open-door } & \text { Rated } \\ \hline 2.5 & 4.6 & 4.7 & 5.9 & 5.1 \\ \hline 5.1 & 4.3 & 2.3 & 4.2 & 4.3 \\ \hline 4.3 & 4.4 & 4.3 & 6.4 & 4.4 \\ \hline 4.8 & 4.0 & 4.2 & 5.6 & 3.8 \\ \hline 5.5 & 4.1 & 5.1 & 6.0 & 4.7 \\ \hline 2.6 & 3.1 & 4.7 & 5.0 & 4.5 \\ \hline 4.1 & 3.8 & 4.2 & 7.3 & 5.5 \\ \hline 3.4 & 5.2 & 5.7 & 5.6 & 5.5 \\ \hline 4.3 & 4.5 & 2.3 & 6.9 & 4.6 \\ \hline 3.9 & 4.0 & 4.5 & 5.1 & 4.1 \\ \hline \end{array} $$ (a) What type of observational study was conducted? What is the response variable? (b) Find the sample mean for each treatment level. (c) Find the sample standard deviation for each treatment level. Using the general rule presented in this chapter, does it appear that the population variances are the same? (d) Use the time to degree completion for students first attending highly selective institutions to construct a \(95 \%\) confidence interval estimate for the population mean. (e) How many pairwise comparisons are possible among the treatment levels? (f) Consider the null hypothesis \(H_{0}: \mu_{1}=\mu_{2}=\mu_{3}=\mu_{4}=\mu_{5}\). If we test this hypothesis using \(t\) -tests for each pair of treatments, use your answer from part (e) to compute the probability of making a Type I error, assuming that each test uses an \(\alpha=0.05\) level of significance. (g) Use the one-way ANOVA procedure to determine if there is a difference in the mean time to degree completion for the different types of initial institutions. If the null hypothesis is rejected, use Tukey's test to determine which pairwise differences are significant using a familywise error rate of \(\alpha=0.05\)

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?

The variability among the sample means is called _____ sample variability, and the variability of each sample is the _____ sample variability.

Researchers at NASA wanted to determine the effects of space flight on a rat's daily consumption of water. The following data represent the water consumption (in milliliters per day) at lift-off minus \(1,\) return plus \(1,\) and 1 month after return for six rats sent to space on the Spacelab Sciences 1 flight. $$ \begin{array}{cccc} \text { Rat } & \begin{array}{l} \text { Lift-Off } \\ \text { Minus 1 } \end{array} & \begin{array}{l} \text { Return } \\ \text { Plus 1 } \end{array} & \begin{array}{l} \text { Return } \\ \text { Plus 1 Month } \end{array} \\ \hline \mathbf{1} & 18.5 & 32 & 30 \\ \hline \mathbf{2} & 17.5 & 18 & 34 \\ \hline \mathbf{3} & 28.0 & 31 & 39 \\ \hline \mathbf{4} & 28.5 & 29 & 44 \\ \hline \mathbf{5} & 31.0 & 48 & 54 \\ \hline \mathbf{6} & 22.5 & 25 & 32 \\ \hline \end{array} $$ (a) What is the response variable in this study? What is the treatment? How many levels does it have? (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 water consumption is different for the three days 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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