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Chapter 11: Problem 62

Suppose you were to conduct a two-factor factorial experiment, factor \(\mathrm{A}\) at four levels and factor \(\mathrm{B}\) at two levels, with \(r\) replications per treatment. a. How many treatments are involved in the experiment? b. How many observations are involved? c. List the sources of variation and their respective degrees of freedom.

Short Answer

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Question: In a two-factor factorial experiment, factor A has 4 levels, factor B has 2 levels, and there are r replications per treatment. Find the following: a. The number of treatments involved in the experiment. b. The number of observations involved. c. The sources of variation and their respective degrees of freedom. Answer: a. There are 8 treatments involved in the experiment. b. There are a total of 8r observations in the experiment. c. The sources of variation and their respective degrees of freedom are: - Factor A: 3 degrees of freedom - Factor B: 1 degree of freedom - Interaction (AB): 3 degrees of freedom - Error: (8r - 8) degrees of freedom

Step by step solution

01

a. Number of treatments

To find the number of treatments in the experiment, we need to multiply the number of levels for both factors (factor A and factor B). Since factor A has 4 levels and factor B has 2 levels, there are 4 * 2 = 8 treatments in the experiment.
02

b. Number of observations

To find the total number of observations, we need to multiply the number of treatments by the number of replications per treatment (r). Since there are 8 treatments and r replications per treatment, there are a total of 8 * r = 8r observations in the experiment.
03

c. Sources of variation and degrees of freedom

There are four sources of variation in this experiment: Factor A, Factor B, Interaction (AB), and Error. The degrees of freedom for each source of variation are calculated as follows: 1. Factor A: Degrees of Freedom for Factor A = (Levels of factor A - 1) = (4-1) = 3 2. Factor B: Degrees of Freedom for Factor B = (Levels of factor B - 1) = (2-1) = 1 3. Interaction (AB): Degrees of Freedom for Interaction = (Degree of freedom for Factor A) * (Degree of freedom for Factor B) = 3 * 1 = 3 4. Error: Degrees of Freedom for Error = Total number of observations - Number of treatments = 8r - 8 So, the sources of variation and their respective degrees of freedom are: - Factor A: 3 degrees of freedom - Factor B: 1 degree of freedom - Interaction (AB): 3 degrees of freedom - Error: (8r - 8) degrees of freedom

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Most popular questions from this chapter

A study was conducted to determine the effect of two factors on terrain visualization training for soldiers. \({ }^{4}\) During the training programs, participants viewed contour maps of various terrains and then were permitted to view a computer reconstruction of the terrain as it would appear from a specified angle. The two factors investigated in the experiment were the participants' spatial abilities (abilities to visualize in three dimensions) and the viewing procedures (active or passive). Active participation permitted participants to view the computer-generated reconstructions of the terrain from any and all angles. Passive participation gave the participants a set of preselected reconstructions of the terrain. Participants were tested according to spatial ability, and from the test scores 20 were categorized as possessing high spatial ability, 20 medium, and 20 low. Then 10 participants within each of these groups were assigned to each of the two training modes, active or passive. The accompanying tables are the ANOVA table computed the researchers and the table of the treatment means. $$ \begin{array}{lcc} & {\text { Training Condition }} \\ \hline \text { Spatial Ability } & \text { Active } & \text { Passive } \\ \hline \text { High } & 17.895 & 9.508 \\ \text { Medium } & 5.031 & 5.648 \\ \text { Low } & 1.728 & 1.610 \\ \hline \end{array} $$ a. Explain how the authors arrived at the degrees of freedom shown in the ANOVA table. b. Are the \(F\) -values correct? c. Interpret the test results. What are their practical implications? d. Use Table 6 in Appendix I to approximate the \(p\) values for the \(F\) statistics shown in the ANOVAtable.

A randomized block design was used to compare the means of three treatments within six blocks. Construct an ANOVA table showing the sources of variation and their respective degrees of freedom.

The data that follow are observations collected from an experiment that compared four treatments, \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D},\) within each of three blocks, using a randomized block design. $$ \begin{array}{lrrrrrr} &&&{\text { Treatment }} \\ \hline \text { Block } & \text { A } & \text { B } & \text { C } & \text { D } & \text { Total } \\ \hline 1 & 6 & 10 & 8 & 9 & 33 \\ 2 & 4 & 9 & 5 & 7 & 25 \\ 3 & 12 & 15 & 14 & 14 & 55 \\ \hline \text { Total } & 22 & 34 & 27 & 30 & 113 \end{array} $$ a. Do the data present sufficient evidence to indicate differences among the treatment means? Test using $$ \alpha=.05 . $$ b. Do the data present sufficient evidence to indicate differences among the block means? Test using \(\alpha=.05 .\) c. Rank the four treatment means using Tukey's method of paired comparisons with \(\alpha=.01\) d. Find a \(95 \%\) confidence interval for the difference in means for treatments \(\mathrm{A}\) and \(\mathrm{B}\). e. Does it appear that the use of a randomized block design for this experiment was justified? Explain.

An independent random sampling design was used to compare the means of six treatments based on samples of four observations per treatment. The pooled estimator of \(\sigma^{2}\) is \(9.12,\) and the sample means follow: \(\bar{x}_{1}=101.6 \quad \bar{x}_{2}=98.4 \quad \bar{x}_{3}=112.3\) \(\bar{x}_{4}=92.9 \quad \bar{x}_{5}=104.2 \quad \bar{x}_{6}=113.8\) a. Give the value of \(\omega\) that you would use to make pairwise comparisons of the treatment means for $$ \alpha=.05 . $$ b. Rank the treatment means using pairwise comparisons.

A nationa home builder wants to compare the prices per 1,000 board feet of standard or better grade Douglas fir framing lumber. He randomly selects five suppliers in each of the four states where the builder is planning to begin construction. The prices are given in the table. $$ \begin{array}{rrrr} && {\text { State }} \\ \hline 1 & 2 & 3 & 4 \\ \hline \$ 241 & \$ 216 & \$ 230 & \$ 245 \\ 235 & 220 & 225 & 250 \\ 238 & 205 & 235 & 238 \\ 247 & 213 & 228 & 255 \\ 250 & 220 & 240 & 255 \end{array} $$ a. What type of experimental design has been used? b. Construct the analysis of variance table for this data. c. Do the data provide sufficient evidence to indicate that the average price per 1000 board feet of Douglas fir differs among the four states? Test using \(\alpha=.05\)
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