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Chapter 10: Problem 3

The accompanying summary data on skeletal-muscle CS activity (nmol/min/mg) appeared in the article "Impact of Lifelong Sedentary Behavior on Mitochondrial Function of Mice Skeletal Muscle" (J. of Gerontology, 2009: 927-939): \begin{tabular}{lccc} \hline & & Old & Old \\ & Young & Sedentary & Active \\ \hline Sample size & 10 & 8 & 10 \\ Sample mean & \(46.68\) & \(47.71\) & \(58.24\) \\ Sample sd & \(7.16\) & \(5.59\) & \(8.43\) \\ \hline \end{tabular} Carry out a test to decide whether true average activity differs for the three groups. If appropriate, investigate differences amongst the means with a multiple comparisons method.

Short Answer

Expert verified
Perform ANOVA to test for differences among groups; if significant, use Tukey's HSD to find specific group differences.

Step by step solution

01

Formulate the Hypotheses

We are testing if the true average CS activity differs among the three groups: Young, Old Sedentary, and Old Active. Formulate the null hypothesis as \( H_0: \mu_1 = \mu_2 = \mu_3 \), stating there is no difference between group means. The alternative hypothesis \( H_a \) is that at least one group mean differs.
02

Choose the Test Method

Since we have more than two groups to compare, we use ANOVA (Analysis of Variance) to test our hypotheses. ANOVA is appropriate here to compare the means of more than two groups.
03

Calculate ANOVA Test Statistic

Using the sample means, sample sizes, and standard deviations provided, we compute the total sum of squares (SST), the between-group sum of squares (SSB), and the within-group sum of squares (SSW). Then, calculate the F-statistic using the formula: \[ F = \frac{\text{SSB} / (k-1)}{\text{SSW} / (N-k)} \]where \( k \) is the number of groups and \( N \) is the total number of observations.
04

Determine the Critical Value and Decision Rule

Find the critical value of \( F \) from the F-distribution table for the significance level (e.g., \( \alpha = 0.05 \)), with \( k-1 \) and \( N-k \) degrees of freedom. If the computed \( F \)-statistic is greater than the critical \( F \), we reject the null hypothesis.
05

Perform Post-Hoc Tests

If the ANOVA results are significant, proceed with a post-hoc test such as Tukey's HSD to identify which group means significantly differ from others. Calculate the differences between group means and compare with a critical difference value from the Tukey's table.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a method used to determine if there is enough evidence to reject a null hypothesis in favor of an alternative one. It involves several key steps:
  • Formulating the null and alternative hypotheses.
  • Deciding on a significance level (often 0.05).
  • Choosing the appropriate test based on the data characteristics.
  • Calculating a test statistic from the sample data.
  • Making a decision about the hypotheses based on the test statistic and significance level.
In this case, the null hypothesis (\( H_0: \mu_1 = \mu_2 = \mu_3 \)) states that the average CS activity is the same across all groups. The alternative hypothesis (\( H_a \)) suggests that at least one group differs. The choice of ANOVA helps us assess these hypotheses by comparing group variances.
F-statistic
The F-statistic is a critical component of the ANOVA process. It is used to determine if there are significant differences between group means in a dataset. To compute the F-statistic, you first need to calculate the sums of squares:
  • Total sum of squares (SST), which measures overall variation in the data.
  • Between-group sum of squares (SSB), capturing variation due to differences between group means.
  • Within-group sum of squares (SSW), reflecting variation within individual groups.
Once you have these, the F-statistic calculation is:\[F = \frac{\text{SSB} / (k-1)}{\text{SSW} / (N-k)}\]where \(k\) is the number of groups and \(N\) is the total number of observations. A higher F-value indicates that the group means differ more than expected by chance, prompting rejection of the null hypothesis.
Post-Hoc Tests
Post-hoc tests come into play after an ANOVA test indicates significant differences among group means. They help pinpoint which specific means are different. Commonly used methods include Tukey's Honestly Significant Difference (HSD) test and Bonferroni correction. These tests control for Type I errors that can occur from conducting multiple comparisons.
  • Tukey鈥檚 HSD: Compares all possible pairs of group means while controlling for Type I error, giving you a critical difference value to determine significance between each pair.
  • Bonferroni correction: Adjusts the significance level based on the number of comparisons, offering a more conservative approach to identify differences.
By applying these tests, you ensure that your results are statistically reliable and pertinent in understanding the nuances between group means.
Mean Comparison
Mean comparison is the core purpose of performing ANOVA and subsequent post-hoc tests. It involves evaluating the average values across different groups to see if they differ from each other. In our skeletal muscle study, we are comparing the mean CS activity of three groups: Young, Old Sedentary, and Old Active.
  • Sample Means: These are the average values within each group, serving as initial points of comparison.
  • Variations in Means: Detected using statistical methods, indicating potential differences in group dynamics or characteristics.
Through these comparisons, insights can be drawn about factors possibly affecting CS activity, aiding in the broader understanding of how lifestyle choices influence physiological functions.

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Most popular questions from this chapter

Four laboratories (1-4) are randomly selected from a large population, and each is asked to make three determinations of the percentage of methyl alcohol in specimens of a compound taken from a single batch. Based on the accompanying data, are differences among laboratories a source of variation in the percentage of methyl alcohol? State and test the relevant hypotheses using significance level .05. \(\begin{array}{llll}\text { 1: } & 85.06 & 85.25 & 84.87 \\ \text { 2: } & 84.99 & 84.28 & 84.88 \\ \text { 3: } & 84.48 & 84.72 & 85.10 \\ \text { 4: } & 84.10 & 84.55 & 84.05\end{array}\)

When sample sizes are equal \(\left(J_{j}=J\right)\), the parameters \(\alpha_{1}, \alpha_{2}, \ldots \alpha_{1}\) of the altemative parameterization are restricted by \(\Sigma \alpha_{i}=0\). For unequal sample sizes, the most natural restriction is \(\sum J_{i} \alpha_{i}=0\). Use this to show that $$ E(\mathrm{MSTr})=\sigma^{2}+\frac{1}{I-1} \sum J_{i} \alpha_{i}^{2} $$ What is \(E\) (MSTr) when \(H_{0}\) is true? [This expectation is correct if \(\sum J_{j} \alpha_{j}=0\) is replaced by the restriction \(\sum \alpha_{i}=0\) (or any other single linear restriction on the \(\alpha_{i}\) 's used to reduce the model to \(I\) independent parameters), but \(\sum J \alpha_{i}=0\) simplifies the algebra and yields natural estimates for the model parameters (in particular, \(\hat{\alpha}_{i}=\bar{x}_{j}-\bar{x}\)..).]

Although tea is the world's most widely consumed beverage after water, little is known about its nutritional value. Folacin is the only B vitamin present in any significant amount in tea, and recent advances in assay methods have made accurate determination of folacin content feasible. Consider the accompanying data on folacin content for randomly selected specimens of the four leading brands of green tea. \(\begin{array}{llllllll}\text { 1: } & 7.9 & 6.2 & 6.6 & 8.6 & 8.9 & 10.1 & 9.6 \\ \text { 2: } & 5.7 & 7.5 & 9.8 & 6.1 & 8.4 & & \\ \text { 3: } & 6.8 & 7.5 & 5.0 & 7.4 & 5.3 & 6.1 & \\ \text { 4: } & 6.4 & 7.1 & 7.9 & 4.5 & 5.0 & 4.0 & \end{array}\) (Data is based on "Folacin Content of Tea," J. of the Amer. Dietetic Assoc., 1983: 627-632.) Does this data suggest that true average folacin content is the same for all brands? a. Carry out a test using \(\alpha=.05\) via the \(P\)-value method. b. Assess the plausibility of any assumptions required for your analysis in part (a). c. Perform a multiple comparisons analysis to identify significant differences among brands.

Consider a single-factor ANOVA experiment in which \(I=3, J=5, \bar{x}_{1 .}=10, \bar{x}_{2 .}=12\), and \(\bar{x}_{3}=20\). Find a value of SSE for which \(f>F_{03,2,12}\), so that \(H_{0}: \mu_{1}=\mu_{2}=\mu_{3}\) is rejected, yet when Tukey's procedure is applied none of the \(\mu_{i}\) 's can be said to differ significantly from one another.

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 $$ \frac{\left(\mu_{1}+\mu_{2}+\mu_{3}+\mu_{4}\right)}{4}-\frac{\left(\mu_{5}+\mu_{6}\right)}{2} $$ [Hint: Modify the expression for \(V(\hat{\theta})\) that led to \((10.5)\) in the previous section.]
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