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Chapter 11: Problem 26

Samples of six different brands of diet/imitation margarine were analyzed to determine the level of physiologically active polyunsaturated fatty acids (PAPFUA, in percentages), resulting in the following data: \(\begin{array}{llllll}\text { Imperial } & 14.1 & 13.6 & 14.4 & 14.3 & \\\ \text { Parkay } & 12.8 & 12.5 & 13.4 & 13.0 & 12.3 \\ \text { Blue Bonnet } & 13.5 & 13.4 & 14.1 & 14.3 & \\ \text { Chiffon } & 13.2 & 12.7 & 12.6 & 13.9 & \\ \text { Mazola } & 16.8 & 17.2 & 16.4 & 17.3 & 18.0 \\ \text { Fleischmann's } & 18.1 & 17.2 & 18.7 & 18.4 & \end{array}\) (The preceding numbers are fictitious, but the sample means agree with data reported in the January 1975 issue of Consumer Reports.) a. Use ANOVA to test for differences among the true average PAPFUA percentages for the different brands. b. Compute CIs for all \(\left(\mu_{i}-\mu_{j}\right)\) 's. c. Mazola and Fleischmann's are corn-based, whereas the others are soybean- based. Compute a CI for $$ \frac{\mu_{1}+\mu_{2}+\mu_{3}+\mu_{4}}{4}-\frac{\mu_{5}+\mu_{6}}{2} $$

Short Answer

Expert verified
Perform ANOVA, calculate means, F-statistic, and confidence intervals for brand comparisons.

Step by step solution

01

Formulate the Problem

We need to use ANOVA to test if there are differences in the mean levels of PAPFUA among the six margarine brands. ANOVA helps us determine if at least one of the group means is significantly different from the others.
02

Calculate Group Means

Compute the mean PAPFUA for each brand. - Imperial: \((14.1 + 13.6 + 14.4 + 14.3)/4 = 14.1\)- Parkay: \((12.8 + 12.5 + 13.4 + 13.0 + 12.3)/5 = 12.8\)- Blue Bonnet: \((13.5 + 13.4 + 14.1 + 14.3)/4 = 13.825\)- Chiffon: \((13.2 + 12.7 + 12.6 + 13.9)/4 = 13.1\)- Mazola: \((16.8 + 17.2 + 16.4 + 17.3 + 18.0)/5 = 17.14\)- Fleischmann's: \((18.1 + 17.2 + 18.7 + 18.4)/4 = 18.1\)
03

Calculate Overall Mean

Find the overall mean of all samples:\[ \text{Overall Mean} = \frac{(14.1 + 13.6 + 14.4 + 14.3 + 12.8 + 12.5 + 13.4 + 13.0 + 12.3 + 13.5 + 13.4 + 14.1 + 14.3 + 13.2 + 12.7 + 12.6 + 13.9 + 16.8 + 17.2 + 16.4 + 17.3 + 18.0 + 18.1 + 17.2 + 18.7 + 18.4)}{26} \approx 15.11 \]
04

Compute ANOVA Summary Statistics

Calculate the 'between-group' and 'within-group' sum of squares. 1. SST (total sum of squares): \[ SST = \sum (X_{ij} - \text{Overall Mean})^2 \]2. SSB (between sum of squares):\[ SSB = n_i \sum (\bar{X}_{i} - \text{Overall Mean})^2 \]3. SSE (within sum of squares):\[ SSE = \sum \sum (X_{ij} - \bar{X}_{i})^2 \]Calculate each component using data and plug into an ANOVA table.
05

Conduct ANOVA Test

1. Calculate the F-statistic: \[ F = \frac{SSB/df_B}{SSE/df_E} \]2. Compare the calculated F-value to the critical F-value from the F-distribution table based on chosen significance level (e.g., 0.05) to determine if the null hypothesis can be rejected.
06

Construct Confidence Intervals for Mean Differences

Compute confidence intervals for each pairwise mean difference using the formula:\[ \bar{X}_i - \bar{X}_j \pm t_{\alpha/2, df} \cdot \sqrt{\frac{s^2}{n_i}+\frac{s^2}{n_j}} \]where \(s^2\) is the pooled variance estimate and \(t\) is the critical value from the t-distribution.
07

Calculate CI for the Combination of Means

Given the formula for the combination of means:\[ \frac{\mu_1+\mu_2+\mu_3+\mu_4}{4}-\frac{\mu_5+\mu_6}{2} \]Calculate the mean difference and use similar variance and degrees of freedom as in Step 6 to construct the confidence interval.

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Key Concepts

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

Confidence Interval
A confidence interval is a range of values that is used to estimate the true value of a population parameter. In the context of ANOVA (Analysis of Variance), confidence intervals help to determine the mean differences between groups. When we calculate a confidence interval for a mean difference, we want to understand with a certain level of confidence, usually 95%, how different the means are between two groups.

To construct a confidence interval for the differences in means \(\mu_i - \mu_j\), we use sample estimates and consider the variability in the data. A common formula is:
  • \( \bar{X}_i - \bar{X}_j \pm t_{\alpha/2, df} \cdot \sqrt{\frac{s^2}{n_i}+\frac{s^2}{n_j}} \)
Where:
  • \( \bar{X}_i \) and \( \bar{X}_j \) are the sample means of the two groups,
  • \( t_{\alpha/2, df} \) is the critical value from the t-distribution that corresponds to the desired confidence level,
  • \( s^2 \) is the pooled estimate of variance, and
  • \( n_i \) and \( n_j \) are the respective sample sizes for each group.
By calculating this interval, we can determine whether the true mean difference is significant or not, which informs us about differences in the margarine brands' PAPFUA levels.
Mean Differences
Mean differences are a crucial part of interpreting ANOVA results. This process involves comparing the average values, or means, of samples from different groups to assess if those group means are statistically different from each other. ANOVA aims to test if at least one group mean varies significantly among multiple tested groups.

For instance, consider multiple margarine brands and their PAPFUA levels. By examining the mean differences, we focus on pairwise comparisons of brands like 'Imperial vs Parkay' or 'Mazola vs Fleischmann鈥檚' to see if certain brands have significantly different levels of PAPFUA.

These mean differences are calculated by subtracting one group mean from another (i.e., \( \bar{X}_i - \bar{X}_j \)). The magnitude of these differences, when paired with statistical estimates from ANOVA, indicate potential areas where significant differences exist, warranting further investigation. Understanding these differences helps consumer reports and customers make informed choices about nutritional content.
Sum of Squares
The concept of Sum of Squares is foundational in understanding the variance components in ANOVA. It is essentially a measure of the total variability in the data.

There are different types of sum of squares used in ANOVA:
  • Total Sum of Squares (SST): This measures the overall variability in the data by comparing each observation to the grand mean. It quantifies the total deviation observed.
  • Between Group Sum of Squares (SSB): This measures the variability due to differences among group means. It helps determine if there are actual differences between the groups or if observed differences are due to random chance.
  • Within Group Sum of Squares (SSE): This calculates the variability within each group, showing how much of the total variability is due to differences within individual samples from the same group.
These sums of squares are used to calculate the F-statistic in ANOVA, which helps determine if the observed between-group variation is greater than what we would expect by chance, given the within-group variability. Understanding these concepts is key to interpreting ANOVA results effectively.

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Most popular questions from this chapter

The article "Towards Improving the Properties of Plaster Moulds and Castings" (J. Engrg. Manuf., 1991: 265-269) describes several ANOVAs carried out to study how the amount of carbon fiber and sand additions affect various characteristics of the molding process. Here we give data on casting hardness and on wet-mold strength. $$ \begin{array}{llll} \hline \begin{array}{l} \text { Sand } \\ \begin{array}{l} \text { Addition } \\ (\%) \end{array} & \begin{array}{l} \text { Carbon } \\ \text { Fiber } \\ \text { Addition } \\ (\%) \end{array} & & \begin{array}{l} \text { Casting } \\ \text { Hardness } \end{array} & \begin{array}{l} \text { Wet- } \\ \text { Mold } \\ \text { Strength } \end{array} \\ \hline 0 & 0 & 61.0 & 34.0 \\ 0 & 0 & 63.0 & 16.0 \\ 15 & 0 & 67.0 & 36.0 \\ 15 & 0 & 69.0 & 19.0 \\ 30 & 0 & 65.0 & 28.0 \\ 30 & 0 & 74.0 & 17.0 \\ 0 & .25 & 69.0 & 49.0 \\ 0 & .25 & 69.0 & 48.0 \\ 15 & .25 & 69.0 & 43.0 \\ 15 & .25 & 74.0 & 29.0 \\ 30 & .25 & 74.0 & 31.0 \\ 30 & .25 & 72.0 & 24.0 \\ 0 & .50 & 67.0 & 55.0 \\ 0 & .50 & 69.0 & 60.0 \\ 15 & .50 & 69.0 & 45.0 \\ 15 & .50 & 74.0 & 43.0 \\ 30 & .50 & 74.0 & 22.0 \\ 30 & .50 & 74.0 & 48.0 \end{array} \end{array} $$ a. An ANOVA for wet-mold strength gives SSSand \(=705\), SSFiber \(=1278\), SSE \(=843\), and SST \(=3105 .\) Test for the presence of any effects using \(\alpha=.05\). b. Carry out an ANOVA on the casting hardness observations using \(\alpha=.05\). c. Make an interaction plot with sand percentage on the horizontal axis, and discuss the results of part (b) in terms of what the plot shows.

In an experiment to assess the effects of curing time (factor \(A\) ) and type of \(\operatorname{mix}(\) factor \(B)\) on the compressive strength of hardened cement cubes, three different curing times were used in combination with four different mixes, with three observations obtained for each of the 12 curing time-mix combinations. The resulting sums of squares were computed to be SSA \(=30,763.0\), SSB \(=34,185.6, \quad\) SSE \(=97,436.8, \quad\) and \(\quad\) SST \(=205,966.6\). a. Construct an ANOVA table. b. Test at level \(.05\) the null hypothesis \(H_{0 A B}\) : all \(\gamma_{i j}\) 's \(=0\) (no interaction of factors) against \(H_{0 A B}\) : at least one \(\gamma_{i j} \neq 0\). c. Test at level \(.05\) the null hypothesis \(H_{0 A}: \alpha_{1}=\) \(\alpha_{2}=\alpha_{3}=0\) (factor \(A\) main effects are absent) against \(H_{\mathrm{aA}}\) : at least one \(\alpha_{i} \neq 0\). d. Test \(H_{0 B}: \beta_{1}=\beta_{2}=\beta_{3}=\beta_{4}=0\) versus \(H_{\mathrm{aB}}\) : at least one \(\beta_{j} \neq 0\) using a level \(.05\) test. e. The values of the \(\bar{x}_{i . .}\) 's were \(\bar{x}_{1 . .}=\) \(4010.88, \bar{x}_{2} .=4029.10\), and \(\bar{x}_{3} \ldots=3960.02\). Use Tukey's procedure to investigate significant differences among the three curing times.

Four different coatings are being considered for corrosion protection of metal pipe. The pipe will be buried in three different types of soil. To investigate whether the amount of corrosion depends either on the coating or on the type of soil, 12 pieces of pipe are selected. Each piece is coated with one of the four coatings and buried in one of the three types of soil for a fixed time, after which the amount of corrosion (depth of maximum pits, in \(.0001\) in.) is determined. The depths are shown in this table: $$ \begin{aligned} &\text { Soil Type (B) }\\\ &\begin{array}{l|lll} & \mathbf{1} & \mathbf{2} & \mathbf{3} \\ \hline \mathbf{1} & 64 & 49 & 50 \\ \mathbf{2} & 53 & 51 & 48 \\ \mathbf{3} & 47 & 45 & 50 \\ \mathbf{4} & 51 & 43 & 52 \\ \hline \end{array} \end{aligned} $$ a. Assuming the validity of the additive model, carry out the ANOVA analysis using an ANOVA table to see whether the amount of corrosion depends on either the type of coating used or the type of soil. Use \(\alpha=.05\). b. Compute \(\hat{\mu}, \hat{\alpha}_{1}, \hat{\alpha}_{2}, \hat{\alpha}_{3}, \hat{\alpha}_{4}, \hat{\beta}_{1}, \hat{\beta}_{2}\), and \(\hat{\beta}_{3}\)

The number of miles of useful tread wear (in 1000's) was determined for tires of each of five different makes of subcompact car (factor \(A\), with \(I=5\) ) in combination with each of four different brands of radial tires (factor \(B\), with \(J=4\) ), resulting in \(I J=20\) observations. The values \(\mathrm{SSA}=30.6, \mathrm{SSB}=44.1\), and \(\mathrm{SSE}=\) \(59.2\) were then computed. Assume that an additive model is appropriate. a. Test \(H_{0}: \alpha_{1}=\alpha_{2}=\alpha_{3}=\alpha_{4}=\alpha_{5}=0\) (no differences in true average tire lifetime due to makes of cars) versus \(H_{\mathrm{a}}\) : at least one \(\alpha_{i} \neq 0\) using a level \(.05\) test. b. \(H_{0}: \beta_{1}=\beta_{2}=\beta_{3}=\beta_{4}=0\) (no differences in true average tire lifetime due to brands of tires) versus \(H_{\mathrm{a}}\) : at least one \(\beta_{j} \neq 0\) using a level \(.05\) test.

The strength of concrete used in commercial construction tends to vary from one batch to another. Consequently, small test cylinders of concrete sampled from a batch are "cured" for periods up to about 28 days in temperature- and moisture-controlled environments before strength measurements are made. Concrete is then "bought and sold on the basis of strength test cylinders" (ASTM C 31 Standard Test Method for Making and Curing Concrete Test Specimens in the Field). The accompanying data resulted from an experiment carried out to compare three different curing methods with respect to compressive strength (MPa). Analyze this data. $$ \begin{array}{lccc} \hline \text { Batch } & \text { Method A } & \text { Method B } & \text { Method C } \\ \hline 1 & 30.7 & 33.7 & 30.5 \\ 2 & 29.1 & 30.6 & 32.6 \\ 3 & 30.0 & 32.2 & 30.5 \\ 4 & 31.9 & 34.6 & 33.5 \\ 5 & 30.5 & 33.0 & 32.4 \\ 6 & 26.9 & 29.3 & 27.8 \\ 7 & 28.2 & 28.4 & 30.7 \\ 8 & 32.4 & 32.4 & 33.6 \\ 9 & 26.6 & 29.5 & 29.2 \\ 10 & 28.6 & 29.4 & 33.2 \end{array} $$
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