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Chapter 11: Problem 71

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.

Short Answer

Expert verified
#Answer# The experimental design used is a one-way (or one-factor) ANOVA design, and the main factor being studied is the kiln site.

Step by step solution

01

(Step 1: Identify the type of experimental design)

This is a one-way (or one-factor) ANOVA design, as we are studying the effect of a single factor (kiln site) on the percentage of iron oxide in the samples.
02

(Step 2: Calculate summary statistics)

Since we don't have the actual data values, we will write a generic guideline to perform the calculations. 1. For each site, calculate the number of samples (n), the mean (\(\bar{x}\)), and the variance (s^2) of the percentage of iron oxide. 2. Calculate the overall mean (\(\bar{x}_{G}\)) of the percentage of iron oxide for all samples combined.
03

(Step 3: Calculate the ANOVA's three sum of squares)

1. Calculate the sum of squares between groups (SSB) using the formula: \(SSB = \sum_{i=1}^{3} n_i (\bar{x}_i - \bar{x}_G)^2\) 2. Calculate the sum of squares within groups (SSW) using the formula: \(SSW = \sum_{i=1}^{3} (n_i - 1)s_i^2\) 3. Calculate the total sum of squares (SST) using the formula: \(SST = SSB + SSW\)
04

(Step 4: Calculate the F statistic)

1. Calculate the degrees of freedom for the numerator (dfb) using: \(dfb = k - 1\), where k is the number of groups (k = 3 in this case). 2. Calculate the degrees of freedom for the denominator (dfw) using: \(dfw = N - k\), where N is the total number of samples. 3. Calculate the mean square between groups (MSB) using: \(MSB = \frac{SSB}{dfb}\) 4. Calculate the mean square within groups (MSW) using: \(MSW = \frac{SSW}{dfw}\) 5. Calculate the F statistic using: \(F = \frac{MSB}{MSW}\)
05

(Step 5: Determine if there is a significant difference between the three sites)

1. Identify the critical F value (\(F_{\alpha, dfb, dfw}\)) from an F-distribution table or using a statistical software, where \(\alpha = 0.01\), dfb and dfw are calculated in step 4. 2. Compare the calculated F statistic to the critical F value: If \(F > F_{\alpha, dfb, dfw}\), then reject the null hypothesis and conclude that there is a significant difference in the average percentage of iron oxide at the three sites. Otherwise, do not reject the null hypothesis, and there is no significant difference.
06

(Step 6: Generating diagnostic plots and checking ANOVA assumptions)

1. Generate diagnostic plots using statistical software. Common plots include the residual plot, normal probability plot (QQ plot), and boxplot. 2. Assess if the assumptions have been violated by examining the plots: - Homogeneity of variances: The residual plot should show no clear pattern or systematic deviation from zero. - Normality: The QQ plot should be roughly linear. - Independence: Not possible to assess from the plots. We assume that the samples are independent. If any of these assumptions appear to be violated, consider transforming the data or using non-parametric methods as an alternative analysis method.

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Key Concepts

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

Experimental Design
In research and data analysis, experimental design is crucial as it outlines the framework for collecting and analyzing data. In the given exercise, we are dealing with a one-way ANOVA experimental design. This design is used when the focus is on understanding the impact of a single factor on a particular outcome. Here, the single factor is the kiln site, and the outcome is the percentage of iron oxide in pottery samples. This means we want to see if different kiln sites result in different average percentages of iron oxide. By using a structured approach like this, researchers can derive meaningful insights and make statistically sound conclusions. The one-way ANOVA helps in isolating and understanding these specific effects, which in turn, aids in effectively comparing groups.
One-way ANOVA
One-way ANOVA is a statistical method used to compare the means of three or more groups to find out if at least one group mean is different from the others. In this exercise, the one-way ANOVA is applied to test if there's a difference in the mean percentage of iron oxide across pottery samples from three different kiln sites. This type of ANOVA is powerful when dealing with more than two groups and helps in identifying overall variations without conducting multiple t-tests, which could increase the risk of errors.

The steps involve calculating various sums of squares: between groups (SSB), within groups (SSW), and the total sum of squares (SST). These calculations help in determining the variance within and between the groups. The F statistic, calculated as the ratio of mean square between groups to mean square within groups, indicates whether the observed variance among means is significant.
ANOVA Assumptions
Conducting a one-way ANOVA requires certain assumptions to be fulfilled for accurate results. The three major assumptions are:

  • Independence: The samples must be independent of each other, meaning the data from one group should not affect the data from another.
  • Normality: The data in each group should be approximately normally distributed. This can be assessed using a normal probability (QQ) plot, which should appear roughly linear if this assumption holds.
  • Homogeneity of Variance: The variance among the groups should be similar, meaning each group should have a roughly equal spread of data. A residual plot can help determine if this assumption is met, which should ideally show no discernible pattern.
Ensuring these assumptions are met helps in validating the results of the ANOVA analysis. If any of these assumptions are violated, alternatives such as data transformation or non-parametric tests might be considered.
Chemical Composition Analysis
Chemical composition analysis in archaeology helps in understanding the materials and processes used in ancient times. For this exercise, it involves analyzing the percentage of iron oxide in pottery samples, which can indicate various aspects like the source of materials or the conditions of the kiln sites.

This analysis provides insights into the technological capabilities and choices at different kiln sites. Variations in chemical composition can be linked to differences in geological sources or different firing temperatures and methods. By comparing iron oxide percentages, researchers can infer differences in the pottery-making techniques or the origin of raw materials. Furthermore, such compositional analysis is crucial in provenance studies, helping to trace back cultural exchanges and influences in historical contexts.

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Most popular questions from this chapter

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.

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.

Water samples were taken at four different locations in a river to determine whether the quantity of dissolved oxygen, a measure of water pollution, varied from one location to another. Locations 1 and 2 were selected above an industrial plant, one near the shore and the other in midstream; location 3 was adjacent to the industrial water discharge for the plant; and location 4 was slightly downriver in midstream. Five water specimens were randomly selected at each location, but one specimen, corresponding to location \(4,\) was lost in the laboratory. The data and a MINITAB analysis of variance computer printout are provided here (the greater the pollution, the lower the dissolved oxygen readings). $$ \begin{array}{llllll} \text { Location } && {\text { Mean Dissolved }} {\text { Oxygen Content }} \\\ \hline 1 &&& 5.9 & 6.1 & 6.3 & 6.1 & 6.0 \\ 2 &&& 6.3 & 6.6 & 6.4 & 6.4 & 6.5 \\ 3 &&& 4.8 & 4.3 & 5.0 & 4.7 & 5.1 \\ 4 &&& 6.0 & 6.2 & 6.1 & 5.8 & \end{array} $$ a. Do the data provide sufficient evidence to indicate a difference in the mean dissolved oxygen contents for the four locations? b. Compare the mean dissolved oxygen content in midstream above the plant with the mean content adjacent to the plant (location 2 versus location 3 ). Use a \(95 \%\) confidence interval.

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?

Refer to Exercise \(11.63 .\) The means of all observations, at the factor A levels \(\mathrm{A}_{1}\) and \(\mathrm{A}_{2}\) are \(\bar{x}_{1}=3.7\) and \(\bar{x}_{2}=1.4,\) respectively. Find a \(95 \%\) confidence interval for the difference in mean response for factor levels \(\mathrm{A}_{1}\) and \(\mathrm{A}_{2}\)
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