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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.

Short Answer

Expert verified
Answer: The pairs of states with significant differences in the average cost of lumber at the α = 0.01 level are (state 1, state 2), (state 2, state 3), (state 2, state 4), and (state 3, state 4).

Step by step solution

01

Calculate the Critical Value for Tukey's Test

We have to find the critical value for Tukey's test using the Studentized range statistic table. We do so by looking for the parameters: α = 0.01, k (number of groups) = 4, and error degrees of freedom = 16. Checking a Studentized range statistic table, we find the critical value: 4.98
02

Calculate the Differences Between All Pairs of Sample Means

Find the differences between all possible pairs of sample means: 1. \(|\bar{x}_{1} - \bar{x}_{2}| = |242.2 - 214.8| = 27.4\) 2. \(|\bar{x}_{1} - \bar{x}_{3}| = |242.2 - 231.6| = 10.6\) 3. \(|\bar{x}_{1} - \bar{x}_{4}| = |242.2 - 248.6| = 6.4\) 4. \(|\bar{x}_{2} - \bar{x}_{3}| = |214.8 - 231.6| = 16.8\) 5. \(|\bar{x}_{2} - \bar{x}_{4}| = |214.8 - 248.6| = 33.8\) 6. \(|\bar{x}_{3} - \bar{x}_{4}| = |231.6 - 248.6| = 17.0\)
03

Compare the Absolute Differences to the Critical Value

Now we must compare the absolute differences to the critical value we found earlier (4.98). To do this, we should use the mean square error (MSE = 41.25), the number of observations per group (nᵢ = 5), and the critical value (4.98) to calculate the least significant difference (LSD) between the means: LSD = Critical Value × \(\sqrt{\frac{MSE}{n_i}}\) = 4.98 × \(\sqrt{\frac{41.25}{5}} \approx 11.61\) Now, we compare the absolute differences calculated in Step 2: 1. \(27.4 > 11.61\) → Significant difference between states 1 and 2 2. \(10.6 < 11.61\) → No significant difference between states 1 and 3 3. \(6.4 < 11.61\) → No significant difference between states 1 and 4 4. \(16.8 > 11.61\) → Significant difference between states 2 and 3 5. \(33.8 > 11.61\) → Significant difference between states 2 and 4 6. \(17.0 > 11.61\) → Significant difference between states 3 and 4 With Tukey's paired comparisons method, we can conclude that there are significant differences between the means at the α = 0.01 level for the following pairs of states: (state 1, state 2), (state 2, state 3), (state 2, state 4), and (state 3, state 4).

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Key Concepts

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

Tukey's Test
Tukey's Test is a statistical method designed to find out which specific group means are different after conducting an analysis of variance (ANOVA). When we have several means to compare, such as the average cost of lumber across different states, Tukey’s Test helps identify exactly where the differences lie. It is especially useful when we have more than two samples to compare, ensuring that we catch significant differences without making a common statistical mistake called Type I error - mistakenly identifying a non-significant effect as significant contrary to the findings due to too many comparisons.
Tukey's Test operates under the assumption that the data is normally distributed and that the variances are homogeneous, meaning variations within these different groups should be similar. By comparing every possible pair of means, it helps pinpoint where specific differences indicate significant variations. If the absolute difference between any two means exceeds the critical value we calculate, that tells us there's a statistically significant difference between those two means.
Therefore, after performing an ANOVA and finding an overall significant result, you would apply Tukey's Test to determine exactly which groups are different from one another. It's like figuring out which cities have notably differing average costs, beyond just knowing there is variation without identifying which ones.
Significant Difference
A significant difference refers to a difference between group means that is strong enough to be unlikely due to chance alone. In the context of statistical testing, especially Tukey's Test, we are looking to identify pairs of group means that are notably different when accounting for sampling error.
The notion of significance in this setting is usually validated through a critical threshold or value. Once the calculated differences in means (as outlined in various steps, like between different states) are compared to this threshold, if they exceed it, that signals a significant difference. Essentially, we are finding confirmation through statistical evidence that the variations observed in data from different groups (say, from different states in our example) are likely due to true differences rather than random variability.
Such assessments allow researchers and stakeholders to make informed decisions because they can ascertain that variations are substantial and not merely noise. This is especially important in practical fields such as economics and resource management where understanding statistical significance can influence policy and financial decisions dramatically.
Critical Value
The critical value in Tukey's Test is a number that helps us determine whether the differences between means are statistically significant. This value acts kind of like a benchmark, and it is derived from the Studentized range statistic specially constructed for comparing multiple groups.
The critical value depends on several factors:
  • The selected significance level (\(\alpha = 0.01\) in our example), ensuring we have high confidence in our findings.
  • The number of groups being compared (\(k\)), as the more groups you compare, the higher the probability of variance.
  • The error degrees of freedom, which accounts for the data's variability unrelated to the testing groups.
For our case, the critical value was calculated as 4.98 because these parameters influenced the level we considered enough to assume true significance in mean differences, leading to conclusions drawn that only the largest differences are signal-worthy. Understanding and finding this critical value is core to executing Tukey’s test accurately.
Mean Square Error
The Mean Square Error (MSE) is a core component in ANOVA and Tucker's Test calculations. It essentially measures the average of the squares of errors—that is, the average squared differences between what was estimated and the actual observed data points.
In simpler terms, it's a measure of the variance or dispersion in the data. A small mean square error indicates your predictions are close to the observed results, meaning less unexplained variability. Think of it like understanding how tightly data points cluster around a mean; closer means more consistency between observations.
In Tukey's Test, the MSE becomes part of calculating the least significant difference (LSD) for paired comparisons. With our example MSE = 41.25, it guided us through assessing the precise threshold between statistical significance and mere fluctuation. Thus, MSE not only plays a role in helping determine the size of the critical value needed for significant differences but also enhances how confidently one can interpret the differences found in group means. Knowing the MSE is essential because it anchors our understanding of what variations are normal and which ones truly stand out.

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Most popular questions from this chapter

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.

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 \(3 .\)

An experiment was conducted to compare the glare characteristics of four types of automobile rearview mirrors. Forty drivers were randomly selected to participate in the experiment. Each driver was exposed to the glare produced by a headlight located 30 feet behind the rear window of the experimental automobile. The driver then rated the glare produced by the rearview mirror on a scale of 1 (low) to 10 (high). Each of the four mirrors was tested by each driver; the mirrors were assigned to a driver in random order. An analysis of variance of the data produced this ANOVA table: $$ \begin{array}{lcc} \text { Source } & d f & \text { SS } & \text { MS } \\ \hline \text { Mirrors } & 46.98 & \\ \text { Drivers } & & 8.42 \\ \text { Error } & & & \\ \hline \text { Total } & 638.61 & \end{array} $$ a. Fill in the blanks in the ANOVA table. b. Do the data present sufficient evidence to indicate differences in the mean glare ratings of the four rearview mirrors? Calculate the approximate \(p\) -value and use it to make your decision. c. Do the data present sufficient evidence to indicate that the level of glare perceived by the drivers varied from driver to driver? Use the \(p\) -value approach. d. Based on the results of part b, what are the practical implications of this experiment for the manufacturers of the rearview mirrors?

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

Four chemical plants, producing the same product and owned by the same company, discharge effluents into streams in the vicinity of their locations. To check on the extent of the pollution created by the effluents and to determine whether this varies from plant to plant, the company collected random samples of liquid waste, five specimens for each of the four plants. The data are shown in the table: $$ \begin{array}{llllll} \text { Plant } & {\text { Polluting Effluents }} {\text { (Ib/gal of waste) }} \\ \hline \text { A } && 1.65 & 1.72 & 1.50 & 1.37 & 1.60 \\ \text { B } & &1.70 & 1.85 & 1.46 & 2.05 & 1.80 \\ \text { C } && 1.40 & 1.75 & 1.38 & 1.65 & 1.55 \\ \text { D } && 2.10 & 1.95 & 1.65 & 1.88 & 2.00 \end{array} $$ a. Do the data provide sufficient evidence to indicate a difference in the mean amounts of effluents discharged by the four plants? b. If the maximum mean discharge of effluents is 1.5 lb/gal, do the data provide sufficient evidence to indicate that the limit is exceeded at plant \(\mathrm{A} ?\) c. Estimate the difference in the mean discharge of effluents between plants \(\mathrm{A}\) and \(\mathrm{D},\) using a \(95 \%\) confidence interval.

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