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Suppose you were to conduct a two-factor factorial experiment, factor A at four levels and factor \(\mathrm{B}\) at five levels, with three 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

Expert verified
So, the total number of treatments in the experiment is 4 x 5 = 20 treatments.

Step by step solution

01

a. Number of treatments

In a two-factor factorial experiment, the total number of treatments (combinations of factor levels) is the product of the number of levels of both factors. For factor A with 4 levels and factor B with 5 levels, there are (4 levels in A) x (5 levels in B) treatments.

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Most popular questions from this chapter

An experiment was conducted to compare the effects of four different chemicals, \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D},\) in producing water resistance in textiles. A strip of material, randomly selected from a bolt, was cut into four pieces, and the four pieces were randomly assigned to receive one of the four chemicals, \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) or \(\mathrm{D} .\) This process was replicated three times, thus producing a randomized block design. The design, with moistureresistance measurements, is as shown in the figure (low readings indicate low moisture penetration). Analyze the experiment using a method appropriate for this randomized block design. Identify the blocks and treatments, and investigate any possible differences in treatment means. If any differences exist, use an appropriate method to specifically identify where the differences lie. What are the practical implications for the chemical producers? Has blocking been effective in this experiment? Present your results in the form of a report. $$$ \begin{aligned} &\text { Blocks (bolt samples) }\\\ &\begin{array}{|c|c|c|} \hline 1 & 2 & 3 \\ \hline \mathrm{C} & \mathrm{D} & \mathrm{B} \\ 9.9 & 13.4 & 12.7 \\ \mathrm{A} & \mathrm{B} & \mathrm{D} \\ 10.1 & 12.9 & 12.9 \\ \mathrm{B} & \mathrm{A} & \mathrm{C} \\ 11.4 & 12.2 & 11.4 \\ \mathrm{D}_{2} & \mathrm{C} & \mathrm{A} \\ 12.1 & 12.3 & 11.9 \end{array} \end{aligned} $$

The partially completed ANOVA table for a randomized block design is presented here: $$ \begin{array}{lcl} \text { Source } & d f & \text { SS } & \text { MS } \quad F \\ \hline \text { Treatments } & 4 & 14.2 & \\ \text { Blocks } & & 18.9 & \\ \text { Error } & 24 & & \\ \hline \text { Total } & 34 & 41.9 & \end{array} $$ a. How many blocks are involved in the design? b. How many observations are in each treatment total? c. How many observations are in each block total? d. Fill in the blanks in the ANOVA table. e. Do the data present sufficient evidence to indicate differences among the treatment means? Test using \(\alpha=.05\) f. Do the data present sufficient evidence to indicate differences among the block means? Test using \(\alpha=.05\)

Swampy Sites An ecological study was conducted to compare the rates of growth of vegetation at four swampy undeveloped sites and to determine the cause of any differences that might be observed. Part of the study involved measuring the leaf lengths of a particular plant species on a preselected date in May. Six plants were randomly selected at each of the four sites to be used in the comparison. The data in the table are the mean leaf length per plant (in centimeters) for a random sample of ten leaves per plant. The MINITAB analysis of variance computer printout for these data is also provided. $$ \begin{array}{lllllll} \text { Location } & {\text { Mean Leaf Length (cm) }} \\ \hline 1 && 5.7 & 6.3 & 6.1 & 6.0 & 5.8 & 6.2 \\ 2 && 6.2 & 5.3 & 5.7 & 6.0 & 5.2 & 5.5 \\ 3 && 5.4 & 5.0 & 6.0 & 5.6 & 4.9 & 5.2 \\ 4 && 3.7 & 3.2 & 3.9 & 4.0 & 3.5 & 3.6 \end{array} $$ a. You will recall that the test and estimation procedures for an analysis of variance require that the observations be selected from normally distributed (at least, roughly so) populations. Why might you feel reasonably confident that your data satisfy this assumption? b. Do the data provide sufficient evidence to indicate a difference in mean leaf length among the four locations? What is the \(p\) -value for the test? c. Suppose, prior to seeing the data, you decided to compare the mean leaf lengths of locations 1 and \(4 .\) Test the null hypothesis \(\mu_{1}=\mu_{4}\) against the alternative \(\mu_{1} \neq \mu_{4}\) d. Refer to part c. Construct a \(99 \%\) confidence interval for \(\left(\mu_{1}-\mu_{4}\right)\) e. Rather than use an analysis of variance \(F\) -test, it would seem simpler to examine one's data, select the two locations that have the smallest and largest sample mean lengths, and then compare these two means using a Student's \(t\) -test. If there is evidence to indicate a difference in these means, there is clearly evidence of a difference among the four. (If you were to use this logic, there would be no need for the analysis of variance \(F\) -test.) Explain why this procedure is invalid.

Refer to Exercise \(11.28 .\) Find a \(95 \%\) confidence interval for the difference between a pair of treatment means \(\mathrm{A}\) and \(\mathrm{B}\) if \(\bar{x}_{\mathrm{A}}=21.9\) and \(\bar{x}_{\mathrm{B}}=24.2\).

Suppose you wish to compare the means of four populations based on independent random samples, each of which contains six observations. Insert, in an ANOVA table, the sources of variation and their respective degrees of freedom.

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