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A chain of jewelry stores conducted an experiment to investigate the effect of price and location on the demand for its diamonds. Six small-town stores were selected for the study, as well as six stores located in large suburban malls. Two stores in each of these locations were assigned to each of three item percentage markups. The percentage gain (or loss) in sales for each store was recorded at the end of 1 month. The data are shown in the accompanying table. $$ \begin{array}{lrrr} {\text { Markup }} \\ \hline \text { Location } & 1 & 2 & 3 \\ \hline \text { Small towns } & 10 & -3 & -10 \\ & 4 & 7 & -24 \\ \hline \text { Suburban malls } & 14 & 8 & -4 \\ & 18 & 3 & 3 \end{array} $$ a. Do the data provide sufficient evidence to indicate an interaction between markup and location? Test using \(\alpha=.05 .\) b. What are the practical implications of your test in part a? c. Draw a line graph similar to Figure 11.11 to help visualize the results of this experiment. Summarize the results. d. Find a \(95 \%\) confidence interval for the difference in mean change in sales for stores in small towns versus those in suburban malls if the stores are using price markup \(3 .\)

Short Answer

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(Students should use the actual data from the given exercise and run the computations explained in the step-by-step solution to answer this question)

Step by step solution

01

a. Test for interaction between markup and location

Conduct a two-way ANOVA (Analysis of Variance) test using the given data to check for an interaction between markup and location. Use the following steps: 1. Compute the Grand Mean: $$ \bar{Y}_{..} = \frac{\sum_{i=1}^{\text{rows}} \sum_{j=1}^{\text{columns}} Y_{ij}}{\text{rows} \times \text{columns}} $$ 2. Compute the Sum of Squares due to Location (SS_Location): $$ \text{SS}_\text{Location} = \text{rows} \times \sum_{i=1}^{\text{rows}} (\bar{Y}_{i.}-\bar{Y}_{..})^2 $$ 3. Compute the Sum of Squares due to Markup (SS_Markup): $$ \text{SS}_\text{Markup} = \text{columns} \times \sum_{j=1}^{\text{columns}} (\bar{Y}_{.j}-\bar{Y}_{..})^2 $$ 4. Compute the Sum of Squares due to Interaction (SS_Interaction): $$ \text{SS}_\text{Interaction} = \sum_{i=1}^{\text{rows}} \sum_{j=1}^{\text{columns}} (\bar{Y}_{ij}-\bar{Y}_{i.}-\bar{Y}_{.j}+\bar{Y}_{..})^2 $$ 5. Compute the Sum of Squares due to Total (SS_Total): $$ \text{SS}_\text{Total} = \sum_{i=1}^{\text{rows}} \sum_{j=1}^{\text{columns}} (Y_{ij}-\bar{Y}_{..})^2 $$ 6. Compute the Sum of Squares due to Error (SS_Error): $$ \text{SS}_\text{Error} = \text{SS}_\text{Total}-\text{SS}_\text{Location}-\text{SS}_\text{Markup}-\text{SS}_\text{Interaction} $$ 7. Compute the Mean Square due to Location (MS_Location), Markup (MS_Markup), Interaction (MS_Interaction), and Error (MS_Error): $$ \text{MS}_\text{Location} = \frac{\text{SS}_\text{Location}}{\text{rows}-1} \\ \text{MS}_\text{Markup} = \frac{\text{SS}_\text{Markup}}{\text{columns}-1} \\ \text{MS}_\text{Interaction} = \frac{\text{SS}_\text{Interaction}}{(\text{rows}-1)(\text{columns}-1)} \\ \text{MS}_\text{Error} = \frac{\text{SS}_\text{Error}}{\text{rows}\times\text{columns}} $$ 8. Finally, compute the F-statistic for Interaction: $$ F_{\text{Interaction}} = \frac{\text{MS}_\text{Interaction}}{\text{MS}_\text{Error}} $$ 9. Compare the computed F-statistic with the F-critical value at \(\alpha=0.05.\) If the F-statistic is greater than the F-critical value, reject the null hypothesis, which means there is a significant interaction between markup and location.
02

b. Practical implications

Discuss the practical implications of the test in part (a). If there is a significant interaction between the factors of markup and location, it indicates that the effect of changes in price markup depends on the location of the store, and vice versa. Business owners could use this information to determine the optimal pricing strategy and store locations to maximize profits.
03

c. Visualize results with a line graph

Provide a line graph that compares the mean change in sales for stores in small towns versus suburban malls, given the three different price markups. This visualization will help us better understand the interaction effect and its potential influence on business decisions. To draw the line graph: 1. Use the mean change in sales for each category and plot them on a graph. 2. Create a line that connects the points for each location. 3. Compare the lines to see if there is a noticeable difference in their patterns. If the lines are parallel or nearly parallel, it suggests no significant interaction. If the lines show different patterns, it may indicate an interaction effect.
04

d. Calculate 95% confidence interval

To find a \(95\%\) confidence interval for the difference in mean change in sales for small towns and suburban malls when using price markup 3, use the following formula: $$ \text{CI} = (\bar{Y}_{1.3}-\bar{Y}_{2.3}) \pm t_{\alpha/2, df} \times \sqrt{\frac{\text{MS}_\text{Error}}{\text{rows}}} $$ Where: - \(\bar{Y}_{1.3}\) is the mean change in sales for small towns at price markup 3 - \(\bar{Y}_{2.3}\) is the mean change in sales for suburban malls at price markup 3 - \(t_{\alpha/2, df}\) is the t-critical value for a \(95\%\) confidence interval and the degrees of freedom (df) of MS_Error - \(\text{MS}_\text{Error}\) is the mean square due to error - \(\text{rows}\) is the number of stores in each category By calculating the confidence interval, we will be able to determine if there is a significant difference in the mean change in sales for stores in small towns versus suburban malls when using price markup 3.

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Key Concepts

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

Understanding Interaction Effect in Two-Way ANOVA
An interaction effect in a two-way ANOVA occurs when the impact of one factor varies depending on the level of another factor. In our jewelry store example, we're looking at how the location (small towns versus suburban malls) and price markup interact to affect sales figures. A significant interaction implies that the effect of markup on sales isn't consistent across locations. For instance, a specific markup might lead to increased sales in suburban malls but decreased sales in small towns. By testing for interaction effects, you can uncover patterns that are not visible when examining each factor separately. This kind of analysis helps businesses fine-tune strategies and optimize decision-making, making sure that pricing and location best align with customer behavior.
Demystifying Confidence Intervals
Confidence intervals provide a range of values that aims to capture the true parameter value—such as the average difference in performance across different settings—in a population. In statistics, a 95% confidence interval is widely used, meaning there is a 95% certainty that the interval contains the true mean. In our scenario, we're interested in the mean sales difference between small towns and suburban malls under the third pricing markup.
To compute it, you adjust the observed mean difference by a factor that accounts for random error, expressed as:
  • Mean difference: The raw difference in means between locations.
  • t-critical value: A multiplier from the t-distribution table based on your confidence level and degrees of freedom.
  • Mean square error: A measure of variance within your data.
Understanding confidence intervals helps in evaluating whether observed differences are significant or likely due to chance.
The Role of F-Statistic in Hypothesis Testing
The F-statistic plays a pivotal role in two-way ANOVA, helping determine whether the observed group variances are significantly different. In our study, it's used to test the interaction between markup and location. By calculating \[ F_{\text{Interaction}} = \frac{\text{MS}_{\text{Interaction}}}{\text{MS}_{\text{Error}}} \]we get a ratio of the variance due to interaction to variance due to error.
If this statistic exceeds a critical value (derived from F-distribution tables), it suggests a significant interaction, prompting us to reject the null hypothesis. Thus, the F-statistic dissects variances into components, offering insights into the interplay of variables that are crucial for strategic decision-making. A larger F-statistic indicates a more distinct interaction effect.
Understanding Mean Square Error in ANOVA
Mean Square Error (MSE) is a cornerstone of the ANOVA process, representing the average spread of observed data points around their predicted values.
In simple words, it signifies unexplained variation within the dataset.
  • Calculated as: \[ \text{MS}_\text{Error} = \frac{\text{SS}_\text{Error}}{\text{degrees of freedom}} \]
  • It acts as a yardstick for comparing variances, helping compute other statistical measures like the F-statistic.
When errors or differences in the ANOVA summary table are high, it suggests substantial data spread, indicating that factors not included in the model might be influencing the results. Conversely, low variance suggests a good fit between the model and data. In essence, MSE helps to quantify the discrepancies between observed and model-predicted values, which is instrumental in verifying model accuracy.

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Most popular questions from this chapter

If the sample size for each treatment is \(n_{t}\) and if \(s^{2}\) is based on \(12 d f\), find \(\omega\) in these cases: a. \(\alpha=.05, k=4, n_{t}=5\) b. \(\alpha=.01, k=6, n_{t}=8\)

The analysis of variance \(F\) -test in Exercise 11.17 (and data set EX 1117 ) determined that there was indeed a difference in the average cost of lumber for the four states. The following information from Exercise 11.17 is given in the table: $$ \begin{array}{ll|lr} \hline \text { Sample Means } & \bar{x}_{1}=242.2 & \text { MSE } & 41.25 \\ & \bar{x}_{2}=214.8 & \text { Error } d f: & 16 \\ & \bar{x}_{3}=231.6 & n_{i}: & 5 \\ & \bar{x}_{4}=248.6 & k: & 4 \\ \hline \end{array} $$ Use Tukey's method for paired comparisons to determine which means differ significantly from the others at the \(\alpha=.01\) level.

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 \(\bar{x}_{\mathrm{B}}=24.2\).

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.

An article in Archaeometry involved an analysis of 26 samples of Romano- British pottery, found at four different kiln sites in the United Kingdom. \(^{9}\) Since one site only yielded two samples, consider the samples found at the other three sites. The samples were analyzed to determine their chemical composition and the percentage of iron oxide is shown below. a. What type of experimental design is this? b. Use an analysis of variance to determine if there is a difference in the average percentage of iron oxide at the three sites. Use \(\alpha=.01\). c. If you have access to a computer program, generate the diagnostic plots for this experiment. Does it appear that any of the analysis of variance assumptions have been violated? Explain.

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