/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Suppose there is sufficient evid... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 4

Suppose there is sufficient evidence to reject \(H_{0}: \mu_{1}=\mu_{2}=\mu_{3}=\mu_{4}\) using a one-way ANOVA. The mean square error from ANOVA is determined to be \(26.2 .\) The sample means are \(\bar{x}_{1}=42.6, \bar{x}_{2}=49.1, \bar{x}_{3}=46.8, \bar{x}_{4}=63.7,\) with \(n_{1}=n_{2}=n_{3}=n_{4}=6 .\) Use Tukey's test to determine which pairwise means are significantly different using a familywise error rate of \(\alpha=0.05 .\)

Short Answer

Expert verified
Pairs \(\bar{x}_{1} \& \bar{x}_{4}, \bar{x}_{2} \& \bar{x}_{4}, \bar{x}_{3} \& \bar{x}_{4}\) are significantly different.

Step by step solution

01

Identify the sample means and sample sizes

Extract the sample means and sample sizes from the given data. The sample means are \(\bar{x}_{1}=42.6, \bar{x}_{2}=49.1, \bar{x}_{3}=46.8,\bar{x}_{4}=63.7\). The sample sizes are \(n_{1}=n_{2}=n_{3}=n_{4}=6\).
02

Determine the mean square error

From the given data, the mean square error (MSE) is determined to be 26.2.
03

Calculate the critical value (q-value)

Using the Tukey's test table for \(\alpha=0.05\), degrees of freedom for the error term (df = N - k = 24 - 4 = 20), and number of groups (k = 4), find the critical value, \(q_{\text{crit}}\). This involves looking up the values from Tukey's table or using statistical software, which gives \(q_{\text{crit}} \approx 3.96\).
04

Calculate the standard error

The standard error (SE) is calculated as: \[\text{SE} = \sqrt{\frac{\text{MSE}}{n}} = \sqrt{\frac{26.2}{6}} \approx 2.09 \].
05

Calculate the minimum significant difference (MSD)

The formula for MSD is: \[\text{MSD} = q_{\text{crit}} \times \text{SE} \approx 3.96 \times 2.09 \approx 8.28 \].
06

Compare pairwise mean differences with MSD

Calculate the pairwise differences and compare each against the MSD. Pairwise differences: \(|\bar{x}_{1} - \bar{x}_{2}| = |42.6 - 49.1| = 6.5 \) \(|\bar{x}_{1} - \bar{x}_{3}| = |42.6 - 46.8| = 4.2 \) \(|\bar{x}_{1} - \bar{x}_{4}| = |42.6 - 63.7| = 21.1 \) \(|\bar{x}_{2} - \bar{x}_{3}| = |49.1 - 46.8| = 2.3 \) \(|\bar{x}_{2} - \bar{x}_{4}| = |49.1 - 63.7| = 14.6 \) \(|\bar{x}_{3} - \bar{x}_{4}| = |46.8 - 63.7| = 16.9 \). Compare each with MSD = 8.28.
07

Determine significant differences

If each pairwise difference is greater than MSD, it is significant: Pair \(|\bar{x}_{1} - \bar{x}_{4}|\rightarrow 21.1 > 8.28\) - Significant Pair \(|\bar{x}_{2} - \bar{x}_{4}|\rightarrow 14.6 > 8.28\) - Significant Pair \(|\bar{x}_{3} - \bar{x}_{4}|\rightarrow 16.9 > 8.28\) - Significant. Other pairs are not significant.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

One-Way ANOVA
One-way ANOVA, or Analysis of Variance, is a statistical technique used to compare the means of three or more groups to identify if there is a significant difference among them. It is particularly useful when you want to test hypotheses about the means of several populations. The null hypothesis (\(H_0\)) in a one-way ANOVA states that all group means are equal (e.g., \(\bar{x}_1 = \bar{x}_2 = \bar{x}_3 = \bar{x}_4\)). The alternative hypothesis (\(H_1\)) is that at least one group mean is different. The one-way ANOVA test involves several steps: Calculating the group means and the grand mean. Estimating the variances within each group and between groups. Computing the F-statistic to determine if the observed variances are significantly different. If the F-statistic is large enough, you reject the null hypothesis, concluding that not all group means are equal. This test gets its name because it considers only one independent variable (or factor) with multiple levels (or groups). When the null hypothesis is rejected, post hoc tests like Tukey's test are needed to find out which specific means are significantly different from each other.
Mean Square Error (MSE)
The Mean Square Error (MSE) is a measure of variation within the groups being compared in an ANOVA test. It represents how much the individual observations vary around their respective group means. The formula for calculating the MSE is: \[ \text{MSE} = \frac{\text{Sum of Squares Within}}{df_{within}} \] Here, \( df_{within} \) denotes the degrees of freedom within groups, typically calculated as the total number of observations minus the number of groups. In the context of the provided problem, the MSE is given as 26.2. This value is used to compute the standard error and subsequently determine the minimum significant difference for Tukey's test. A lower MSE means that the data points are closely packed around the mean, indicating less randomness. Understanding MSE is crucial because it forms the foundation for further calculations in various statistical tests, including ANOVA and post hoc tests like Tukey's.
Pairwise Mean Differences
Pairwise mean differences are evaluated to determine which pairs of group means are significantly different from each other. This comparison helps in understanding the specific relationships between each pair of groups. In Tukey's test, the process is as follows: Calculate the differences between each pair of group means. Compare these differences with a threshold called the Minimum Significant Difference (MSD). The MSD is derived using the critical value from Tukey's test and the standard error. The formula for the standard error (SE) is: \[ \text{SE} = \sqrt{\frac{\text{MSE}}{n}} \] where \(n\) is the sample size for each group. Then, the formula for MSD is: \[ \text{MSD} = q_{\text{crit}} \times SE \] If the absolute value of any pairwise mean difference exceeds the MSD, that pair is considered significantly different. In the exercise, significant differences were found in pairs \( |\bar{x}_1 - \bar{x}_4| \), \( |\bar{x}_2 - \bar{x}_4| \), and \( |\bar{x}_3 - \bar{x}_4| \). Non-significant pairs do not exceed the MSD and indicate that those group means are not different enough to be statistically significant. This analytical approach is vital for interpreting the relationships between multiple groups effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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.

Researchers wanted to determine if the psychological profile of healthy children was different than for children suffering from recurrent abdominal pain (RAP) or recurring headaches. A total of 210 children and adolescents were studied and their psychological profiles were graded according to the Child Behavior Checklist \(4-18\) (CBCL). Children were stratified in two age groups: 4 to 11 years and 12 to 18 years. The results of the study are summarized in the following table: $$\begin{array}{lccc} & n & \text { Sample Mean } & \text { Sample Variance } \\\\\hline \text { Control group } & 70 & 11.7 & 21.6 \\\\\hline \text { RAP } & 70 & 9.0 & 13.0 \\\\\hline \text { Headache } & 70 & 12.4 & 8.4\end{array}$$ (a) Compute the sample standard deviations for each group. (b) What sampling method was used for each treatment group? Why? (c) Use a two sample \(t\) -test for independent samples to determine if there is a significant difference in mean \(\mathrm{CBCL}\) scores between the control group and the RAP group (assume that both samples are simple random samples). (d) Is it necessary to check the normality assumption to answer part (c)? Explain. (e) Use the one-way ANOVA procedure with \(\alpha=0.05\) to determine if the mean CBCL scores are different for the three treatment groups. (f) Based on your results from parts (c) and (e), can you determine if there is a significant difference between the mean scores of the RAP group and the headache group? Explain.

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 \(\alpha=0.05\)

David Elkington, the Chief Executive Officer (CEO) of InsideSales.com, wondered whether the cycles of the moon affect sales. He got the idea from his father-in-law, who said that emergency rooms are crowded when the moon is full. There are eight phases of the moon: New Moon, Waning Crescent, Third Quarter, Waning Gibbous, Full, Waxing Gibbous, First Quarter, and Waxing Crescent. Explain how to obtain and analyze data to determine whether moon cycles affect sales.

A medical researcher wanted to determine the effectiveness of coagulants on the healing rate of a razor cut on lab mice. Because healing rates of mice vary from mouse to mouse, the researcher decided to block by mouse. First, the researcher gave each mouse a local anesthesia and then made a 5 -mm incision that was \(2 \mathrm{~mm}\) deep on each mouse. He randomly selected one of the three treatments and recorded the time it took for the wound to stop bleeding (in minutes). He repeated this process two more times on each mouse and obtained the results shown. $$ \begin{array}{cccc} \text { Mouse } & \text { No Drug } & \begin{array}{l} \text { Experimental } \\ \text { Drug 1 } \end{array} & \begin{array}{l} \text { Experimental } \\ \text { Drug 2 } \end{array} \\ \hline \mathbf{1} & 3.2 & 3.4 & 3.4 \\ \hline \mathbf{2} & 4.8 & 4.4 & 3.4 \\ \hline \mathbf{3} & 6.6 & 5.9 & 5.4 \\ \hline \mathbf{4} & 6.5 & 6.3 & 5.2 \\ \hline \mathbf{5} & 6.4 & 6.3 & 6.1 \\ \hline \end{array} $$ (a) Explain how each mouse forms a block. Explain how blocking might reduce variability of time to heal. (b) Normal probability plots for each treatment indicate that the requirement of normality is satisfied. Verify that the requirement of equal population variances for each treatment is satisfied. (c) Is there sufficient evidence that the mean healing time is different among the three treatments at the \(\alpha=0.05\) level of significance? (d) If the null hypothesis from part (c) was rejected, use Tukey's test to determine which pairwise means differ using a familywise error rate of \(\alpha=0.05 .\) (e) Based on your results for part (d), what do you conclude?
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.