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Chapter 13: Problem 16

The comparisonwise error rate, denoted \(\alpha_{c}\), is the probability of making a Type I error when comparing two means. It is related to the familywise error rate, \(\alpha\), through the formula \(1-\alpha=\left(1-\alpha_{c}\right)^{k},\) where \(k\) is the number of means being compared. (a) If the familywise error rate is \(\alpha=0.05\) and \(k=3\) means are being compared, what is the comparisonwise error rate? (b) If the familywise error rate is \(\alpha=0.05\) and \(k=5\) means are being compared, what is the comparisonwise error rate? (c) Based on the results of parts (a) and (b), what happens to the comparisonwise error rate as the number of means compared increases?

Short Answer

Expert verified
For (a), \(\alpha_c\)\(\approx 0.01686\). For (b), \(\alpha_c\)\(\approx 0.01048\). As \(k\) increases, \(\alpha_c\) decreases.

Step by step solution

01

Understand the Formula

The relationship between the familywise error rate (\(\alpha\)) and the comparisonwise error rate (\(\alpha_c\)) is given by:\[1 - \alpha = (1 - \alpha_c)^k\]where \(k\) is the number of means being compared.
02

Rearrange the Formula for Comparisonwise Error Rate

Solve for \(\alpha_c\) by rearranging the equation. Take the \(k\)-th root on both sides: \[ (1 - \alpha)^{1/k} = 1 - \alpha_c \]Then solve for \(\alpha_c\): \[\alpha_c = 1 - (1 - \alpha)^{1/k} \]
03

Calculate \(\alpha_c\) for \(k=3\)

Use the given values: \(\alpha = 0.05\) and \(k = 3\). Plug into the formula: \[\alpha_c = 1 - (1 - 0.05)^{1/3}\]Calculate: \[(1 - 0.05)^{1/3} = 0.95^{1/3} \approx 0.98314\]Thus, \[\alpha_c = 1 - 0.98314 \approx 0.01686\]
04

Calculate \(\alpha_c\) for \(k=5\)

Use the given values: \(\alpha = 0.05\) and \(k = 5\). Plug into the formula: \[\alpha_c = 1 - (1 - 0.05)^{1/5}\]Calculate: \[(1 - 0.05)^{1/5} = 0.95^{1/5} \approx 0.98952\]Thus, \[\alpha_c = 1 - 0.98952 \approx 0.01048\]
05

Comparison of Results

Observe the comparisonwise error rates calculated for parts (a) and (b). Notice that \(\alpha_c\) decreases as \(k\) increases from 3 to 5.
06

Conclusion

As the number of means being compared (\(k\)) increases, the comparisonwise error rate (\(\alpha_c\)) decreases.

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Key Concepts

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

Comparisonwise Error Rate
Understanding the comparisonwise error rate (\(\alpha_c\)) is crucial when analyzing statistical tests. This rate represents the probability of making a Type I error when comparing two means. The formula connecting the familywise error rate (\(\alpha\)) with \(\alpha_c\) is given by: \(1 - \alpha = (1 - \alpha_c)^k\).
Here, \(k\) stands for the number of comparisons being made.
To find the comparisonwise error rate, the formula can be rearranged to: \(\alpha_c = 1 - (1 - \alpha)^{1/\{k\}}\). This rearrangement lets us isolate \(\alpha_c\) to calculate it for any given \(\alpha\) and \(k\). After simplifying, you'll see that the value of \(\alpha_c\) decreases when \(k\) increases.
Type I Error
A Type I error occurs when a true null hypothesis is rejected. It signifies a false positive, where you detect an effect that isn't actually there. The probability of making a Type I error is denoted by \(\alpha\) and is also referred to as the significance level of a test.
In statistical comparisons, controlling for Type I errors is essential because they can lead to incorrect conclusions.
By maintaining \(\alpha\) at a low level (e.g., 0.05), we limit the likelihood of these false positives. However, as the number of comparisons increases, the cumulative chance of making at least one Type I error also increases.
This leads to concepts like the familywise error rate (\(\alpha\)) and comparisonwise error rate (\(\alpha_c\) ).
Multiple Comparisons Correction
When performing multiple comparisons, the risk of encountering Type I errors across tests increases. This is why multiple comparisons correction is vital. The goal is to adjust the error rates to control for the increased risk.
The familywise error rate (\(\alpha\)) considers the probability of making one or more Type I errors across all comparisons. By applying corrections, such as the Bonferroni correction, the significance level for each individual test is adjusted to \(\alpha/k\), where \(k\) is the number of comparisons. This reduces the \(\alpha_c\), keeping the overall error rate lower.
Methods like Tukey's HSD and Holm鈥檚 step-down procedure are also used to address multiple comparison issues. These methods ensure that the conclusions drawn from multiple tests are robust and reliable.

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Most popular questions from this chapter

Determine the F-test statistic based on the given summary statistics. $$ \begin{array}{cccc} \text { Population } & \text { Sample Size } & \text { Sample Mean } & \text { Sample Variance } \\ \hline 1 & 10 & 40 & 48 \\ \hline 2 & 10 & 42 & 31 \\ \hline 3 & 10 & 44 & 25 \end{array} $$

The variability among the sample means is called _____ sample variability, and the variability of each sample is the _____ sample variability.

Given the following ANOVA output, answer the questions that follow. \(\begin{array}{lrrrrr}\text { Source } & \text { df } & \text { SS } & \text { MS } & F & P \\ \text { Factor A } & 1 & 531.2 & 531.2 & 11.73 & 0.003 \\\ \text { Factor B } & 2 & 3018.0 & 1509.0 & 33.33 & 0.000 \\ \text { Interaction } & 2 & 16.3 & 8.2 & 0.18 & 0.836 \\ \text { Error } & 18 & 814.9 & 45.3 & & \end{array}\) (a) Is there evidence of an interaction effect? Why or why not? (b) Based on the \(P\) -value, is there evidence of a difference in the means from factor A? Based on the \(P\) -value, is there evidence of a difference in the means from factor \(\mathrm{B} ?\) (c) What is the mean square error?

The following data are taken from four different populations that are known to be normally distributed, with equal population variances based on independent simple random samples. $$ \begin{array}{cccc} \text { Sample 1 } & \text { Sample 2 } & \text { Sample 3 } & \text { Sample 4 } \\ \hline 110 & 138 & 98 & 130 \\ \hline 85 & 140 & 100 & 116 \\ \hline 83 & 130 & 94 & 157 \\ \hline 95 & 115 & 110 & 137 \\ \hline 103 & 101 & 104 & 144 \\ \hline 105 & 130 & 118 & 124 \\ \hline 107 & 123 & 102 & 139 \\ \hline \end{array} $$ (a) Test the hypothesis that each sample comes from a population with the same mean at the \(\alpha=0.05\) level of significance. That is, test \(H_{0}: \mu_{1}=\mu_{2}=\mu_{3}=\mu_{4}\). (b) If you rejected the null hypothesis in part (a), use Tukey's test to determine which pairwise means differ using a familywise error rate of \(\alpha=0.05\). (c) Draw boxplots of each set of sample data to support your results from parts (a) and (b).

Given the following ANOVA output, answer the questions that follow: $$ \begin{aligned} &\text { Analysis of Variance for Response }\\\ &\begin{array}{lrrrrr} \text { Source } & \text { df } & \text { SS } & \text { MS } & F & P \\ \text { Block } & 6 & 1712.37 & 285.39 & 134.20 & 0.000 \\ \text { Treatment } & 3 & 2.27 & 0.76 & 0.36 & 0.786 \\ \text { Error } & 18 & 38.28 & 2.13 & & \\ \text { Total } & 27 & 1752.91 & & & \end{array} \end{aligned} $$ (a) The researcher wants to test \(H_{0}: \mu_{1}=\mu_{2}=\mu_{3}=\mu_{4}\) against \(H_{1}:\) at least one of the means is different. Based on the ANOVA table, what should the researcher conclude? (b) What is the mean square due to error? (c) Explain why it is not necessary to use Tukey's test on these data.
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