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Refer to Exercise \(11.54 .\) The data for this experiment are shown in the table. $$ \begin{array}{lccc} && {\text { Training (A) }} \\ \hline \text { Situation (B) } & \text { Trained } & \text { Not Trained } & \text { Totals } \\ \hline \text { Standard } & 85 & 53 & 519 \\ & 91 & 49 & \\ & 80 & 38 & \\ & 78 & 45 & \\ \hline \text { Emergency } & 76 & 40 & 473 \\ & 67 & 52 & \\ & 82 & 46 & \\ & 71 & 39 & \\ \hline \text { Totals } & 630 & 362 & 992 \end{array} $$ a. Construct the ANOVA table for this experiment, b. Is there a significant interaction between the presence or absence of training and the type of decision-making situation? Test at the \(5 \%\) level of significance. c. Do the data indicate a significant difference in behavior ratings for the two types of situations at the \(5 \%\) level of significance? d. Do behavior ratings differ significantly for the two types of training categories at the \(5 \%\) level of significance. e. Plot the average scores using an interaction plot. How would you describe the effect of training and emergency situation on the decision-making abilities of the supervisors?

Short Answer

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b. Is there a significant difference in behavior ratings for standard and emergency situations? c. Is there a significant difference in behavior ratings for the two types of training categories? d. How would you interpret the interaction plot based on the pattern observed?

Step by step solution

01

Calculate Sum of Squares (SS)

First, let's calculate the sum of squares for each factor and the interaction term. We will need these values for the ANOVA table. Total sum of squares (SST) considers the deviation of each observation from the overall mean. SST = \(\sum_{i=1}^{n}(x_{i}-\bar{x})^2\) Sum of squares due to factor A (training) is represented by SSA and can be calculated based on the deviations of the training category means from the overall mean. SSA = nB * \(\sum_{i=1}^{a}(m_{Ai} - \bar{x})^2\) Sum of squares due to factor B (situation) is represented by SSB and can be calculated based on the deviations of the situation category means from the overall mean. SSB = nA * \(\sum_{i=1}^{b}(m_{Bi} - \bar{x})^2\) Sum of squares due to the interaction between factor A and B (SSAB) is represented by the sum of the deviations of each cell mean from the marginal means and the grand mean. SSAB = \(\sum_{i=1}^a \sum_{j=1}^b n_{ij}(m_{ij} - m_{Ai} - m_{Bj} + \bar{x})^2\)
02

Construct the ANOVA Table

Next, we will construct an ANOVA table with sources of variation (training A, situation B, interaction AB, and error), sum of squares (SS), degrees of freedom (df), mean squares (MS), and an F-ratio. ANOVA Table: Source----| SS----| df----| MS----| F-ratio A---|---|---|---|--- B---|---|---|---|--- AB---|---|---|---|--- Error---|---|---|---|--- Degrees of freedom: dfA = a - 1 = 1 dfB = b - 1 = 1 dfAB = (a - 1)(b - 1) = 1 dfError = (n - a)(n - b) = 4 Mean Squares: MSA = SSA / dfA MSB = SSB / dfB MSAB = SSAB / dfAB MSError = SSError / dfError F-ratio: FA = MSA / MSError FB = MSB / MSError FAB = MSAB / MSError Use an F-table to find the F critical values for each factor and interaction, and compare it to the calculated F-ratios.
03

Test the Hypotheses and Interpret Results

For each factor and the interaction, determine whether the calculated F-ratio is greater than the F critical value at the 5% level of significance. a. If yes, reject the null hypothesis and conclude there is a significant effect. b. If no, fail to reject the null hypothesis and conclude there is no significant effect. Answer each question using this information: b. Interaction between training and situation c. Difference in behavior ratings for the two types of situations d. Difference in behavior ratings for the two types of training categories
04

Construct the Interaction Plot

Plot the average scores for the different combinations of training and situation types. Observe the pattern in the plot to describe the effect of training and emergency situation on the decision-making abilities of the supervisors. If the lines are parallel, there is no interaction, but if the lines are not parallel, there is an interaction effect.

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Key Concepts

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

Sum of Squares
The Sum of Squares (SS) is a crucial element in the construction of an ANOVA table. This statistical measure is used to assess the variance within a dataset by quantifying the total deviation of each observation from the overall mean. In essence, SS allows us to partition the total variance into components attributed to different sources of variation in an experiment.

For example, in the context of the provided exercise, the Total Sum of Squares (SST) encapsulates the overall variability in decision-making ability scores across all supervisors regardless of training or situation. This is calculated by the sum of squared differences between each supervisor's score and the grand mean score of all supervisors. The SST is then split into Sum of Squares due to training (SSA), Sum of Squares due to situation (SSB), and the Sum of Squares due to the interaction between training and situation (SSAB).

SSA reflects the variability in scores due to different levels of training, while SSB reflects the variability due to the type of situation encountered (Standard or Emergency). Most crucially, SSAB gives us insight into whether the effect of training on decision-making is consistent across different situations or not. Such decomposition of SS is fundamental for interpreting ANOVA results and understanding the impact of individual factors and their interactions.
Interaction between factors
Understanding the interactions between factors in an ANOVA is critical for interpreting complex experimental designs. When studying how two factors, such as training and situation in our exercise, affect a response variable like decision-making ability, it is not just the individual impact of these factors that's of interest, but also how the factors work together—this is known as interaction.

Interactions occur when the effect of one factor on the outcome variable is different at different levels of the other factor. If an interaction is present, the impact of one factor cannot be fully understood without considering the other factor. For instance, training might improve decision-making abilities in standard situations but might have little to no effect, or even a negative effect, in emergency ones. To evaluate these interactions, we calculate the interaction SS (SSAB). A significant SSAB in the ANOVA table would indicate that the relationship between training and decision-making abilities varies across different situations, thereby necessitating further investigation into this dynamic.

An interaction plot can visually demonstrate these effects. Non-parallel lines on the plot suggest an interaction between factors. In educational contexts, this understanding provides deeper insights and aids in tailoring training programs more effectively.
Statistical significance
In the realm of statistics, and specifically in the ANOVA table used in our exercise, determining statistical significance is a pivotal step. Statistical significance helps us decide whether the observed differences in mean scores across different groups are due to a real effect or simply random chance.

To assess statistical significance, we use the F-ratio, a value obtained by dividing the Mean Square of a factor (or interaction) by the Mean Square Error (the variability within groups). An F-ratio larger than a critical value from F-distribution tables suggests that the differences in means are unlikely to have occurred by chance — the findings are statistically significant.

At a commonly used significance level of 5%, as requested in the exercise, we're essentially saying there's a 95% chance that the differences or interactions observed are real. If our calculated F-ratio for Training, Situation, or their Interaction exceeds the critical F-value, we reject the null hypothesis. This indicates that factors like training or situation type have a statistically significant impact on the decision-making abilities of supervisors.
Decision-making abilities
Decision-making abilities are the focal behavioral metric in our referenced exercise. In the context of an ANOVA analysis, we attempt to understand how different factors such as training (A) and situation types (B), along with their interaction, influence these abilities.

The ANOVA table enables us to systematically dissect and quantify the influence of each factor and their interaction on decision-making performance. This methodical analysis guides decision-makers in various fields, whether in human resources or educational training programs, empowering them with evidence-based insights to optimize strategies and interventions.

By assessing whether the variance in decision-making scores is significant across different groups and situations, decision-makers can tailor training programs to enhance outcomes effectively. For instance, more focused training might be developed if it was proven that 'Trained' individuals significantly outperform the 'Not Trained' ones in 'Emergency' situations. Ultimately, a well-designed ANOVA study yields actionable data, enabling decision-makers to improve performance and outcomes based on statistical evidence.

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Most popular questions from this chapter

A study was conducted to compare automobile gasoline mileage for three formulations of gasoline. A was a non-leaded 87 octane formulation, \(\mathrm{B}\) was a non-leaded 91 octane formulation, and \(\mathrm{C}\) was a non-leaded 87 octane formulation with \(15 \%\) ethanol. Four automobiles, all of the same make and model, were used in the experiment, and each formulation was tested in each automobile. Using each formulation in the same automobile has the effect of eliminating (blocking out) automobile-toautomobile variability. The data (in miles per gallon) follow. $$ \begin{array}{lcccc} && {\text { Automobile }} \\ \hline \text { Formulation } & 1 & 2 & 3 & 4 \\ \hline \mathrm{A} & 25.7 & 27.0 & 27.3 & 26.1 \\ \mathrm{~B} & 27.2 & 28.1 & 27.9 & 27.7 \\ \mathrm{C} & 26.1 & 27.5 & 26.8 & 27.8 \end{array} $$ a. Do the data provide sufficient evidence to indicate a difference in mean mileage per gallon for the three gasoline formulations? b. Is there evidence of a difference in mean mileage for the four automobiles? c. Suppose that prior to looking at the data, you had decided to compare the mean mileage per gallon for formulations A and B. Find a \(90 \%\) confidence interval for this difference. d. Use an appropriate method to identify the pairwise differences, if any, in the average mileages for the three formulations.

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 \(\alpha=.05\)

How satisfied are you with your current mobile-phone service provider? Surveys done by Consumer Reports indicate that there is a high level of dissatisfaction among consumers, resulting in high customer turnover rates. \({ }^{10}\) The following table shows the overall satisfaction scores, based on a maximum score of \(100,\) for four wireless providers in four different cities. $$ \begin{array}{lcccc} & & & & \text { San } \\ & \text { Chicago } & \text { Dallas } & \text { Philadelphia } & \text { Francisco } \\ \hline \text { AT\&T Wireless } & 63 & 66 & 61 & 64 \\ \text { Cingular Wireless } & 67 & 67 & 64 & 60 \\ \text { Sprint } & 60 & 68 & 60 & 61 \\ \text { Verizon Wireless } & 71 & 75 & 73 & 73 \end{array} $$ a. What type of experimental design was used in this article? If the design used is a randomized block design, what are the blocks and what are the treatments? b. Conduct an analysis of variance for the data. c. Are there significant differences in the average satisfaction scores for the four wireless providers considered here? d. Are there significant differences in the average satisfaction scores for the four cities?

The calcium content of a powdered mineral substance was analyzed five times by each of three methods, with similar standard deviations: $$ \begin{array}{llllll} \text { Method } & {\text { Percent Calcium }} \\ \hline 1 && .0279 & .0276 & .0270 & .0275 & .0281 \\ 2 && .0268 & .0274 & .0267 & .0263 & .0267 \\ 3 && .0280 & .0279 & .0282 & .0278 & .0283 \end{array} $$ Use an appropriate test to compare the three methods of measurement. Comment on the validity of any assumptions you need to make.

An experiment was conducted to compare the effectiveness of three training programs, \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C},\) in training assemblers of a piece of electronic equipment. Fifteen employees were randomly assigned, five each, to the three programs. After completion of the courses, each person was required to assemble four pieces of the equipment, and the average length of time required to complete the assembly was recorded. Several of the employees resigned during the course of the program; the remainder were evaluated, producing the data shown in the accompanying table. Use the MINITAB printout to answer the questions. $$ \begin{array}{lllll} \text { Training Program } & {\text { Average Assembly Time (min) }} \\ \hline \text { A } && 59 & 64 & 57 & 62 \\ \text { B } && 52 & 58 & 54 & \\ \text { C } && 58 & 65 & 71 & 63 & 64 \end{array} $$ a. Do the data provide sufficient evidence to indicate a difference in mean assembly times for people trained by the three programs? Give the \(p\) -value for the test and interpret its value. b. Find a \(99 \%\) confidence interval for the difference in mean assembly times between persons trained by programs \(\mathrm{A}\) and \(\mathrm{B}\) c. Find a \(99 \%\) confidence interval for the mean assembly times for persons trained in program A. d. Do you think the data will satisfy (approximately) the assumption that they have been selected from normal populations? Why?

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