/*! 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 15 The paper "Trends in Blood Lead ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 15

The paper "Trends in Blood Lead Levels and Blood Lead Testing among U.S. Children Aged 1 to 5 Years" (Pediatrics [2009]: e376-e385) gave data on blood lead levels (in \(\mu \mathrm{g} / \mathrm{dL}\) ) for samples of children living in homes that had been classified either at low, medium, or high risk of lead exposure based on when the home was constructed. After using a multiple comparison procedure, the authors reported the following: 1\. The difference in mean blood lead level between low-risk housing and medium-risk housing was significant. 2\. The difference in mean blood lead level between low-risk housing and high- risk housing was significant. 3\. The difference in mean blood lead level between medium-risk housing and high-risk housing was significant. Which of the following sets of \(\mathrm{T}-\mathrm{K}\) intervals \((\mathrm{Set} 1,2\), or 3) is consistent with the authors' conclusions? Explain your choice. \(\mu_{L}=\) mean blood lead level for children living in low-risk housing \(\mu_{M}=\) mean blood lead level for children living in medium-risk housing \(\mu_{H}=\) mean blood lead level for children living in high-risk housing $$ \begin{array}{lccc} \text { Difference } & \text { Set 1 } & \text { Set 2 } & \text { Set } 3 \\ \hline \mu_{L}-\mu_{M} & (-0.6,0.1) & (-0.6,-0.1) & (-0.6,-0.1) \\ \mu_{L}-\mu_{H} & (-1.5,-0.6) & (-1.5,-0.6) & (-1.5,-0.6) \\ \mu_{M}-\mu_{H} & (-0.9,-0.3) & (-0.9,0.3) & (-0.9,-0.3) \\ \hline \end{array} $$

Short Answer

Expert verified
The sets of T-K intervals that are consistent with the authors' conclusions are Set 1 and Set 3 because all the intervals in these sets do not contain 0, indicating that there are significant differences among all pairs of low-risk housing, medium-risk housing, and high-risk housing, as the authors concluded.

Step by step solution

01

Checking Set 1

For the set 1, all intervals for \(\mu_{L}-\mu_{M}\), \(\mu_{L}-\mu_{H}\), and \(\mu_{M}-\mu_{H}\), do not contain 0. Therefore, all are showing significant differences, which corresponds to the authors' conclusions.
02

Checking Set 2

For the set 2, the interval for \(\mu_{M}-\mu_{H}\) contains 0. This means that there is a possibility that the difference between the means of medium and high risk housing could be 0 (i.e., no difference). This contradicts the authors' conclusion, so set 2 is not consistent with the authors' conclusions.
03

Checking Set 3

For the set 3, all intervals for \(\mu_{L}-\mu_{M}\), \(\mu_{L}-\mu_{H}\), and \(\mu_{M}-\mu_{H}\) do not contain 0. Therefore, all are showing significant differences, which corresponds to the authors' conclusions.

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.

Multiple Comparison Procedure
In statistics, when researchers conduct multiple comparisons, they are comparing several different groups or conditions to understand if differences exist between them. A multiple comparison procedure involves conducting several statistical tests simultaneously. In the context of the given exercise, the authors tested differences in mean blood lead levels across children living in low, medium, and high-risk housing environments.

It's crucial to use a multiple comparison procedure because when many tests are done, the chance of committing a Type I error (falsely rejecting a true null hypothesis) increases. Procedures such as Tukey's, Bonferroni's, or Dunnett's tests are designed to control the family-wise error rate and give more reliable results when comparing multiple groups. The authors of the study likely used one of these to determine that the differences in mean blood lead levels across different housing risk categories were significant.
Mean Blood Lead Level
The mean blood lead level is a critical measure in environmental health, representing the average amount of lead in the blood of a group of individuals. The units typically used are micrograms per deciliter \( \mathrm{\mu g/dL} \). Lead is a toxic substance, and its presence in the blood can indicate exposure to harmful environmental factors such as lead-based paint, which is often found in older housing.

In this exercise, the researchers differentiated between low, medium, and high-risk housing based on construction periods. The mean blood lead level for children living in each housing category was used to understand their risk of lead exposure. This measure is fundamental in assessing public health concerns and guiding policy to address and mitigate lead exposure in at-risk populations.
Difference in Means
The difference in means is a statistical concept used to compare the average values between two groups. In the exercise, this concept was applied to evaluate whether there was a significant difference in the mean blood lead levels of children from different housing risk categories.

Mean differences are often depicted using confidence intervals. A confidence interval that does not include zero suggests a statistically significant difference between the groups. Set 1 and Set 3 in the exercise do not include zero in any of the intervals, indicating significant differences in all the pairwise comparisons. Conversely, Set 2 includes zero in the interval for \( \mu_{M}-\mu_{H} \), suggesting there may be no significant difference between medium and high-risk housing, which contradicts the author's conclusions. Therefore, only Sets 1 and 3 align with the findings that all pairwise differences were significant.

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

The paper referenced in the previous exercise also gave data for 12 - to 13 -year-old girls. Data consistent with summary values in the paper are shown below. Do the data provide convincing evidence that the mean rating associated with the game description for 12 - to 13 -year-old girls is not the same for all four age restrictive rating labels? Test the appropriate hypotheses using \(\alpha=.05\). $$ \begin{array}{cccc} \text { 7+ label } & \text { 12+ label } & \text { 16+ label } & \text { 18+ label } \\ \hline 4 & 4 & 6 & 8 \\ 7 & 5 & 4 & 6 \\ 6 & 4 & 8 & 6 \\ 5 & 6 & 6 & 5 \\ 3 & 3 & 10 & 7 \\ 6 & 5 & 8 & 4 \\ 4 & 3 & 6 & 10 \\ 5 & 8 & 6 & 6 \\ 10 & 5 & 8 & 8 \\ 5 & 9 & 5 & 7 \\ \hline \end{array} $$

The paper "Women's and Men's Eating Behavior Following Exposure to Ideal-Body Images and Text" (Communication Research [2006]: 507-529) describes an experiment in which 74 men were assigned at random to one of four treatments: 1\. Viewed slides of fit, muscular men 2\. Viewed slides of fit, muscular men accompanied by diet and fitness-related text 3\. Viewed slides of fit, muscular men accompanied by text not related to diet and fitness 4\. Did not view any slides The participants then went to a room to complete a questionnaire. In this room, bowls of pretzels were set out on the tables. A research assistant noted how many pretzels were consumed by each participant while completing the questionnaire. Data consistent with summary quantities given in the paper are given in the accompanying table. Do these data provide convincing evidence that the mean number of pretzels consumed is not the same for all four treatments? Test the relevant hypotheses using a significance level of \(.05\).

Give as much information as you can about the \(P\) -value for an upper-tailed \(F\) test in each of the following situations. a. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=15, F=5.37\) b. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=15, F=1.90\) c. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=15, F=4.89\) d. \(\mathrm{df}_{1}=3, \mathrm{df}_{2}=20, F=14.48\) e. \(\mathrm{df}_{\mathrm{l}}=3, \mathrm{df}_{2}=20, F=2.69\) f. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=50, F=3.24\)

Samples of six different brands of diet or imitation margarine were analyzed to determine the level of physiologically active polyunsaturated fatty acids (PAPUFA, in percent), resulting in the data shown in the accompanying table. (The data are fictitious, but the sample means agree with data reported in Consumer Reports.) $$ \begin{array}{llllll} \text { Imperial } & 14.1 & 13.6 & 14.4 & 14.3 & \\ \text { Parkay } & 12.8 & 12.5 & 13.4 & 13.0 & 12.3 \\ \text { Blue Bonnet } & 13.5 & 13.4 & 14.1 & 14.3 & \\ \text { Chiffon } & 13.2 & 12.7 & 12.6 & 13.9 & \\ \text { Mazola } & 16.8 & 17.2 & 16.4 & 17.3 & 18.0 \\ \text { Fleischmann's } & 18.1 & 17.2 & 18.7 & 18.4 & \end{array} $$ a. Test for differences among the true average PAPUFA percentages for the different brands. Use \(\alpha=.05\). b. Use the \(\mathrm{T}-\mathrm{K}\) procedure to compute \(95 \%\) simultaneous confidence intervals for all differences between means and give the corresponding underscoring pattern.

In the introduction to this chapter, we considered a study comparing three groups of college students (soccer athletes, nonsoccer athletes, and a control group consisting of students who did not participate in intercollegiate sports). The following information on scores from the Hopkins Verbal Learning Test (which measures immediate memory recall) was $$ \begin{array}{lccc} \text { Group } & \begin{array}{c} \text { Soccer } \\ \text { Athletes } \end{array} & \begin{array}{c} \text { Nonsoccer } \\ \text { Athletes } \end{array} & \text { Control } \\ \hline \text { Sample size } & 86 & 95 & 53 \\ \text { Sample mean score } & 29.90 & 30.94 & 29.32 \\ \text { Sample standard } & 3.73 & 5.14 & 3.78 \\ \text { deviation } & & & \\ \hline \end{array} $$ In addition, \(\overline{\bar{x}}=30.19 .\) Suppose that it is reasonable to regard these three samples as random samples from the three student populations of interest. Is there sufficient evidence to conclude that the mean Hopkins score is not the same for the three student populations? Use \(\alpha=.05\).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

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.