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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

Putting It Together: Time to Complete a Degree A researcher wanted to determine if the mean time to complete a bachelor's degree was different depending on the selectivity of the first institution of higher education that was attended. The following data represent a random sample of 12 th-graders who earned their degree within eight years. Probability plots indicate that the data for each treatment level are normally distributed. $$ \begin{array}{ccccc} \text { Highly } & & & \text { Not } \\ \text { Selective } & \text { Selective } & \text { Nonselective } & \text { Open-door } & \text { Rated } \\ \hline 2.5 & 4.6 & 4.7 & 5.9 & 5.1 \\ \hline 5.1 & 4.3 & 2.3 & 4.2 & 4.3 \\ \hline 4.3 & 4.4 & 4.3 & 6.4 & 4.4 \\ \hline 4.8 & 4.0 & 4.2 & 5.6 & 3.8 \\ \hline 5.5 & 4.1 & 5.1 & 6.0 & 4.7 \\ \hline 2.6 & 3.1 & 4.7 & 5.0 & 4.5 \\ \hline 4.1 & 3.8 & 4.2 & 7.3 & 5.5 \\ \hline 3.4 & 5.2 & 5.7 & 5.6 & 5.5 \\ \hline 4.3 & 4.5 & 2.3 & 6.9 & 4.6 \\ \hline 3.9 & 4.0 & 4.5 & 5.1 & 4.1 \\ \hline \end{array} $$ (a) What type of observational study was conducted? What is the response variable? (b) Find the sample mean for each treatment level. (c) Find the sample standard deviation for each treatment level. Using the general rule presented in this chapter, does it appear that the population variances are the same? (d) Use the time to degree completion for students first attending highly selective institutions to construct a \(95 \%\) confidence interval estimate for the population mean. (e) How many pairwise comparisons are possible among the treatment levels? (f) Consider the null hypothesis \(H_{0}: \mu_{1}=\mu_{2}=\mu_{3}=\mu_{4}=\mu_{5}\). If we test this hypothesis using \(t\) -tests for each pair of treatments, use your answer from part (e) to compute the probability of making a Type I error, assuming that each test uses an \(\alpha=0.05\) level of significance. (g) Use the one-way ANOVA procedure to determine if there is a difference in the mean time to degree completion for the different types of initial institutions. If the null hypothesis is rejected, use Tukey's test to determine which pairwise differences are significant using a familywise error rate of \(\alpha=0.05\)

The following data represent the number of fish species living in various Andirondack Lakes and the \(\mathrm{pH}\) of the lakes. From chemistry, we know \(\mathrm{pH}\) is a measure of the acidity or basicity of a solution. Solutions with \(\mathrm{pH}\) less than 7 are said to be acidic. As pH increases, the solution is said to be less acidic. $$\begin{array}{lc|lc}\text { pH } & \text { Species } & \text { pH } & \text { Species } \\\\\hline 4.6 & 0 & 5.8 & 8 \\\\\hline 4.7 & 0 & 6 & 3 \\\\\hline 4.8 & 0 & 6.1 & 4 \\\\\hline 5 & 0 & 6.2 & 9 \\\\\hline 5 & 2 & 6.25 & 9 \\\\\hline 5.2 & 2 & 6.3 & 2 \\\\\hline 5.2 & 1 & 6.3 & 4 \\\\\hline 5.25 & 0 & 6.3 & 9 \\\\\hline 5.3 & 1 & 6.4 & 5 \\\\\hline 5.35 & 1 & 6.7 & 6 \\\\\hline 5.5 & 5 & 6.7 & 8 \\\\\hline 5.7 & 4 & 6.7 & 8 \\\\\hline 5.75 & 3 & 6.8 & 10\end{array}$$ (a) Draw a scatter diagram of the data treating \(\mathrm{pH}\) as the explanatory variable. (b) Determine the linear correlation coefficient between \(\mathrm{pH}\) and number of fish species. (c) Does a linear relation exist between \(\mathrm{pH}\) and number of fish species? (d) Find the least-squares regression line treating \(\mathrm{pH}\) as the explanatory variable. (e) Interpret the slope. (f) Is it reasonable to interpret the intercept? Explain. (g) What proportion of the variability in number of fish species is explained by \(\mathrm{pH} ?\) (h) Is the number of fish species in the lake whose \(\mathrm{pH}\) is 5.5 above or below average? Explain. (i) In part (g), you found the proportion of variability in number of fish species that is explained by the variability in \(\mathrm{pH}\). Can you think of other variables that might also explain the variability in the number of fish species?

Discrimination To determine if there is gender and/or race discrimination in car buying, Ian Ayres put together a team of fifteen white males, five white females, four black males, and seven black females who were each asked to obtain an initial offer price from the dealer on a certain model car. The 31 individuals were made to appear as similar as possible to account for other variables that may play a role in the offer price of a car. The following data are based on the results in the article and represent the profit on the initial price offered by the dealer. Ayres wanted to determine if the profit based on the initial offer differed among the four groups. $$\begin{array}{cc|c|c|c}\text { White Male } & \text { Black Male } & \text { White Female } & \text { Black Female } \\\\\hline 1300 & 853 & 1241 & 951 & 1899 \\\\\hline 646 & 727 & 1824 & 954 & 2053 \\\\\hline 951 & 559 & 1616 & 754 & 1943 \\\\\hline 794 & 429 & 1537 & 706 & 2168 \\\\\hline 661 & 1181 & & 596 & 2325 \\\\\hline 824 & 853 & & & 1982 \\\\\hline 1038 & 877 & & & 1780 \\\\\hline 754 & & & &\end{array}$$ (a) What is the response variable in this study? Is it qualitative or quantitative? (b) State the null and alternative hypotheses. (c) A normal probability plot of each group suggests the data come from a population that is normally distributed. Verify the requirement of equal variances is satisfied. (d) Test the hypothesis stated in part (b). (e) Draw side-by-side boxplots of the four groups to support the analytic results of part (d). (f) What do the results of the analysis suggest? (g) Because the group of black males has a small sample size, the normality requirement is best verified by assessing the normality of the residuals. Verify the normality requirement by drawing a normal probability plot of the residuals.

Suppose that there is sufficient evidence to reject \(H_{0}: \mu_{1}=\mu_{2}=\mu_{3}\) using a one-way ANOVA. The mean square error from ANOVA is determined to be \(26.2 .\) The sample means are \(\bar{x}_{1}=9.5, \bar{x}_{2}=9.1, \bar{x}_{3}=18.1,\) with \(n_{1}=n_{2}=n_{3}=5 .\) Use Tukey's test to determine which pairwise means are significantly different using a familywise error rate of \(\alpha=0.05\)

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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