/*! 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 56 Do children diagnosed with atten... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Do children diagnosed with attention deficit/ hyperactivity disorder (ADHD) have smaller brains than children without this condition? This question was the topic of a research study described in the paper "Developmental Trajectories of Brain Volume Abnormalities in Children and Adolescents with Attention Deficit/Hyperactivity Disorder" (journal of the American Medical Association [2002]: \(1740-\) 1747). Brain scans were completed for a representative sample of 152 children with ADHD and a representative sample of 139 children without ADHD. Summary values for total cerebral volume (in milliliters) are given in the following table: $$ \begin{array}{lccc} & n & \bar{x} & s \\ \hline \text { Children with ADHD } & 152 & 1,059.4 & 117.5 \\ \text { Children Without ADHD } & 139 & 1,104.5 & 111.3 \end{array} $$ Use a \(95 \%\) confidence interval to estimate the differ- ence in mean brain volume for children with and without ADHD.

Short Answer

Expert verified
The 95% confidence interval calculated from the given data gives the difference in the average brain volume for children with ADHD and those without ADHD with 95% confidence. If the interval contains 0, it suggests that there is no significant difference in mean brain volumes for the two groups.

Step by step solution

01

Identify the given values

First, identify the summary statistics given for each group: \( n_1 = 152, \bar{x_1} = 1059.4, s_1 = 117.5 \) for the ADHD group and \( n_2 = 139, \bar{x_2} = 1104.5, s_2 = 111.3 \) for the non-ADHD group.
02

Calculate the standard error for the difference in means

The formula to compute Standard Error (SE) for difference of two means is \[ SE = \sqrt{\left(\frac{{s_1}^2}{n_1}\right) + \left(\frac{{s_2}^2}{n_2}\right)} \] Substituting the given values, you will obtain the standard error.
03

Calculate the 95% confidence interval for the difference in means

The formula for the 95% confidence interval (\( CI \)) for the difference in two means is \[ CI = (\bar{x_1} - \bar{x_2}) \pm z \cdot SE \] where \( z \) is the z-score corresponding to the desired confidence level (1.96 for 95% confidence level). With the calculated standard error and the given mean values, you will arrive at the confidence interval for the difference in mean brain volume for children with and without ADHD.

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 91Ó°ÊÓ!

Key Concepts

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

ADHD Brain Volume Research
The relevance of research into the brain volume of children with ADHD lies in its potential to enhance our understanding of ADHD's neurological underpinnings. Brain imaging studies, like the one cited in the journal of the American Medical Association, aim to determine if structural differences exist between individuals with and without ADHD. This can offer insights into how such differences may affect behavior and cognitive functions associated with the disorder. High-quality research in this field can lead to more effective diagnostic methods and interventions, tailoring treatments to specific neurological profiles.
Standard Error Calculation
Understanding the concept of standard error (SE) is critical when interpreting data from studies like the ADHD brain volume research. SE measures the amount of variability in sample statistics from one sample to another if repeated samples were taken. To compute the standard error for the difference in means between two groups, you use the formula: \[ SE = \sqrt{\left(\frac{{s_1}^2}{n_1}\right) + \left(\frac{{s_2}^2}{n_2}\right)} \] This allows researchers to estimate the uncertainty surrounding the difference calculated from the sample means. A lower SE indicates that the sample mean is a more precise reflection of the true population mean.
Difference in Means
In research studies that compare two groups, a common statistic of interest is the difference in means. This measures the discrepancy between the average values of the two groups, providing a simple yet powerful means of assessing the contrast in their central tendencies. For example, in the ADHD brain volume study, the difference in mean brain volume between children with ADHD and those without is a focal outcome that could demonstrate a potential impact of the condition on brain development. By comparing the mean cerebral volumes, \( \bar{x_1} - \bar{x_2} \), we capture a snapshot of how the two groups differ on average.
Statistical Significance
Statistical significance is a cornerstone concept in hypothesis testing, helping us determine if our findings are likely to reflect true differences in the population or merely due to random chance. When calculating a 95% confidence interval for the difference in means, as seen in the ADHD research, we're essentially saying there's a 95% chance that the interval contains the true difference in brain volumes. The '95%' is a level of confidence selected to mitigate the likelihood of false-positive results while still remaining sensitive to true effects. If the confidence interval does not include zero, this suggests that the observed difference is statistically significant and unlikely to have occurred because of random variation alone.

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

Descriptions of four studies are given. In each of the studies, the two populations of interest are the students at a particular university who live on campus and the students who live off campus. Which of these studies have samples that are independently selected? Study 1: To determine if there is evidence that the mean amount of money spent on food each month differs for the two populations, a random sample of 45 students who live on campus and a random sample of 50 students who live off campus are selected. Study 2: To determine if the mean number of hours spent studying differs for the two populations, a random sample students who live on campus is selected. Each student in this sample is asked how many hours he or she spend working each week. For each of these students who live on campus, a student who lives off campus and who works the same number of hours per week is identified and included in the sample of students who live off campus. Study 3: To determine if the mean number of hours worked per week differs for the two populations, a random sample of students who live on campus and who have a brother or sister who also attends the university but who lives off campus is selected. The sibling who lives on campus is included in the on campus sample, and the sibling who lives off campus is included in the off- campus sample. Study 4: To determine if the mean amount spent on textbooks differs for the two populations, a random sample of students who live on campus is selected. A separate random sample of the same size is selected from the population of students who live off campus.

The paper referenced in the previous exercise also gave information on calorie content. For the sample of Burger King meal purchases, the mean number of calories was 1,008 , and the standard deviation was \(483 .\) For the sample of McDonald's meal purchases, the mean number of calories was 908 , and the standard deviation was 624 . Based on these samples, is there convincing evidence that the mean number of calories in McDonald's meal purchases is less than the mean number of calories in Burger King meal purchases? Use \(\alpha=0.01\).

Do girls think they don't need to take as many science classes as boys? The article "Intentions of Young Students to Enroll in Science Courses in the Future: An Examination of Gender Differences" (Science Education [1999]: \(55-76\) ) describes a survey of randomly selected children in grades \(4,5,\) and 6 . The 224 girls participating in the survey each indicated the number of science courses they intended to take in the future, and they also indicated the number of science courses they thought boys their age should take in the future. For each girl, the authors calculated the difference between the number of science classes she intends to take and the number she thinks boys should take. a. Explain why these data are paired. b. The mean of the differences was -0.83 (indicating girls intended, on average, to take fewer science classes than they thought boys should take), and the standard deviation was 1.51 . Construct and interpret a \(95 \%\) confidence interval for the mean difference.

The article "More Students Taking AP Tests" (San Luis Obispo Tribune, January 10,2003 ) provided the following information on the percentage of students in grades 11 and 12 taking one or more AP exams and the percentage of exams that earned credit in 1997 and 2002 for seven high schools on the central coast of California. $$ \begin{array}{cccccc} & {\begin{array}{c} \text { Percentage of } \\ \text { Students Taking } \\ \text { One or More } \\ \text { AP Exams } \end{array}} & & {\begin{array}{c} \text { Percentage of } \\ \text { Exams That } \\ \text { Earned College } \end{array}} \\ { 2 - 3 } { 5 - 6 } & & & & {\text { Credit }} \\ { 2 - 3 } { 5 - 6 } \text { School } & 1997 & 2002 & & 1997 & 2002 \\ & 1 & 13.6 & 18.4 & & 61.4 & 52.8 \\ 2 & 20.7 & 25.9 & & 65.3 & 74.5 \\ 3 & 8.9 & 13.7 & & 65.1 & 72.4 \\ 4 & 17.2 & 22.4 & & 65.9 & 61.9 \\ 5 & 18.3 & 43.5 & & 42.3 & 62.7 \\ 6 & 9.8 & 11.4 & & 60.4 & 53.5 \\ 7 & 15.7 & 17.2 & & 42.9 & 62.2 \\ \hline \end{array} $$ a. Assuming that it is reasonable to regard these seven schools as a random sample of high schools located on the central coast of California, carry out an appropriate test to determine if there is convincing evidence that the mean percentage of exams earning college credit at central coast high schools in 1997 and in 2002 were different. b. Do you think it is reasonable to generalize the conclusion of the test in Part (a) to all California high schools? Explain. c. Would it be appropriate to use the paired-samples \(t\) test with the data on percentage of students taking one or more AP exams? Explain.

In a study of the effect of college student employment on academic performance, the following summary statistics for GPA were reported for a sample of students who worked and for a sample of students who did not work (University of Central Florida Undergraduate Research Journal, Spring 2005): $$ \begin{array}{cccc} & \begin{array}{c} \text { Sample } \\ \text { Size } \end{array} & \begin{array}{c} \text { Mean } \\ \text { GPA } \end{array} & \begin{array}{c} \text { Standard } \\ \text { Deviation } \end{array} \\ \begin{array}{c} \text { Students Who Are } \\ \text { Employed } \end{array} & 184 & 3.12 & 0.485 \\ \begin{array}{c} \text { Students Who Are } \\ \text { Not Employed } \end{array} & 114 & 3.23 & 0.524 \\ & & & \end{array} $$ The samples were selected at random from working and nonworking students at the University of Central Florida. Estimate the difference in mean GPA for students at this university who are employed and students who are not employed.

See all solutions

Recommended explanations on Math Textbooks

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.