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Exercise 10.40 examined an advertisement for Albertsons, a supermarket chain in the western United States. The advertiser claims that Albertsons has consistently had lower prices than four other full-service supermarkets. As part of a survey conducted by an "independent market basket price-checking company," the average weekly total based on the prices of approximately 95 items is given for five different supermarket chains recorded during 4 consecutive weeks. $$ \begin{array}{llrlll} & \text { Albertsons } & \text { Ralphs } & \text { Vons } & \text { Alpha Beta } & \text { Lucky } \\ \hline \text { Week 1 } & \$ 254.26 & \$ 256.03 & \$ 267.92 & \$ 260.71 & \$ 258.84 \\ \text { Week 2 } & 240.62 & 255.65 & 251.55 & 251.80 & 242.14 \\ \text { Week 3 } & 231.90 & 255.12 & 245.89 & 246.77 & 246.80 \\ \text { Week 4 } & 234.13 & 261.18 & 254.12 & 249.45 & 248.99 \end{array} $$ a. What type of design has been used in this experiment? b. Conduct an analysis of variance for the data. c. Is there sufficient evidence to indicate that there is a difference in the average weekly totals for the five supermarkets? Use \(\alpha=.05\) d. Use Tukey's method for paired comparisons to determine which of the means are significantly different from each other. Use \(\alpha=.05 .\)

Short Answer

Expert verified
Answer: To determine if there is a significant difference among the supermarkets, perform an analysis of variance (ANOVA) and compare the F-statistic for supermarkets with the critical F-value at a significance level of 0.05. If the F-statistic is greater than the critical F-value, there is a significant difference. Additionally, utilize Tukey's method for paired comparisons to identify which means are significantly different from each other.

Step by step solution

01

a. Identify the design type

This experiment follows a two-factor factorial design, with factors being the supermarkets and the weeks, both having fixed levels.
02

b. Conduct an ANOVA

First, let's calculate the overall mean, the mean price of each supermarket, and the mean price of each week, as well as the sums of squares for the supermarkets (SS_supermarkets), weeks (SS_weeks), and the error (SS_error). $$ \bar{X} = \frac{\sum_{i=1}^5 \sum_{j=1}^4 X_{ij}}{20} $$ $$ \bar{X_i} = \frac{\sum_{j=1}^4 X_{ij}}{4}, \bar{X_j} = \frac{\sum_{i=1}^5 X_{ij}}{5} $$ $$ SS_{supermarkets} = 4 \sum_{i=1}^5 (\bar{X_i} - \bar{X})^2 $$ $$ SS_{weeks} = 5 \sum_{j=1}^4 (\bar{X_j} - \bar{X})^2 $$ $$ SS_{error} = \sum_{i=1}^5 \sum_{j=1}^4 (X_{ij} - \bar{X_i} - \bar{X_j} + \bar{X})^2 $$ Calculate the mean squares by dividing the sums of squares by their respective degrees of freedom: $$ MS_{supermarkets} = \frac{SS_{supermarkets}}{4-1}, MS_{weeks} = \frac{SS_{weeks}}{4-1}, MS_{error} = \frac{SS_{error}}{(4-1)(5-1)} $$ Now, we can calculate the F-statistics for the supermarkets and weeks: $$ F_{supermarkets} = \frac{MS_{supermarkets}}{MS_{error}}, F_{weeks} = \frac{MS_{weeks}}{MS_{error}} $$
03

c. Test for sufficient evidence

To test for sufficient evidence of a difference in average weekly totals among the supermarkets, compare the calculated F-statistic for supermarkets (\(F_{supermarkets}\)) with the critical F-value, \(F_{3,12;0.05}\), at a significance level of 0.05. If \(F_{supermarkets} > F_{3,12;0.05}\), we can conclude there is sufficient evidence to indicate a difference in average weekly totals for the five supermarkets.
04

d. Tukey's method for paired comparisons

To use Tukey's method, first, we calculate the differences in the mean prices between each pair of supermarkets: $$ \Delta_{ij} = \bar{X}_i - \bar{X}_j $$ Next, we compare these differences with the Tukey's critical value, \(q_{3,12;\alpha}\), multiplied by the square root of the mean square error (MSE) divided by the number of observations per cell (in this case, weeks). If the absolute difference in means, \(|\Delta_{ij}|\), is greater than the critical value times the standard error, we can conclude that there is a significant difference between the means of supermarkets i and j: $$ |\Delta_{ij}| > q_{3,12;0.05} \times \sqrt{\frac{MS_{error}}{4}} $$

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Key Concepts

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

Two-Factor Factorial Design
In the context of statistical experiments, a two-factor factorial design is a type of experimental structure that allows researchers to observe the effects of two different factors simultaneously. Here, the two factors are the supermarket chains and the weeks. This design is particularly useful when wanting to understand how multiple factors interact with each other.
  • The benefits of a two-factor factorial design include efficiency and a better understanding of interaction effects.
  • Each factor is considered to have fixed levels, meaning the levels are pre-set and not randomly selected.
  • This design provides more information than conducting separate one-factor experiments, as it helps in exploring interaction between the factors.
These interactions can reveal complexities in data that single-factor designs may overlook, offering a comprehensive view of the factors' combined effects on the outcome. It is especially important when the factors are likely to influence each other, as in the case of supermarket pricing across different weeks.
Tukey's Method
Tukey's method, or Tukey's Honest Significant Difference test, is a statistical tool used for making pairwise comparisons while controlling for family-wise error rates. It is particularly useful following an ANOVA when you've determined there is a significant difference in averages among groups.
  • Tukey's method is essential for identifying where differences lie between group means.
  • The process involves calculating the difference between all pairs of group means.
  • It compares each of these differences against a critical value that accounts for variations across the dataset.
If the absolute mean difference between any two groups exceeds the critical value, it indicates a significant difference. This method is robust across various scenarios, ensuring you have clear results on which groups differ significantly. This is valuable in contexts such as evaluating supermarket price differences, where identifying specific price disparities between chains is crucial.
F-Statistics
The F-statistic is a ratio used in ANOVA to determine if there are any statistically significant differences between the means of several groups. The F-statistic serves as a test for comparing variance among group means relative to variance within the groups.
  • Calculation involves dividing the mean square for the factor by the mean square for error.
  • A higher F-statistic suggests a greater degree of variance among group means compared to within-group variance.
  • If the F-statistic is greater than the critical F-value from the F-distribution table, you can conclude there is a significant difference among the means.
This comparison allows researchers to determine whether the observed differences are due to actual group effects or are simply random variations. In the case of supermarket prices over different weeks, a significant F-statistic would indicate that the differences in weekly prices are not due to random chance, affirming the effect of the supermarkets on the pricing structure.

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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.

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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.

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