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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.

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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.

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Most popular questions from this chapter

The authors of the paper "Ultrasound Techniques Applied to Body Fat Measurement in Male and Female Athletes" (Journal of Athletic Training [2009]: \(142-147\) ) compared two different methods for measuring body fat percentage. One method uses ultrasound and the other method uses X-ray technology. The accompanying table gives body fat percentages for 16 athletes using each of these methods (a subset of the data given in a graph that appeared in the paper). For purposes of this exercise, you can assume that the 16 athletes who participated in this study are representative of the population of athletes. Do these data provide convincing evidence that the mean body fat percentage measurement differs for the two methods? Test the appropriate hypotheses using \(\alpha=0.05\). $$ \begin{array}{crr} \text { Athlete } & \text { X-ray } & \text { Ultrasound } \\ \hline 1 & 5.00 & 4.75 \\ 2 & 7.00 & 3.75 \\ 3 & 9.25 & 9.00 \\ 4 & 12.00 & 11.75 \\ 5 & 17.25 & 17.00 \\ 6 & 29.50 & 27.50 \\ 7 & 5.50 & 6.50 \\ 8 & 6.00 & 6.75 \\ 9 & 8.00 & 8.75 \\ 10 & 8.50 & 9.50 \\ 11 & 9.25 & 9.50 \\ 12 & 11.00 & 12.00 \\ 13 & 12.00 & 12.25 \\ 14 & 14.00 & 15.50 \\ 15 & 17.00 & 18.00 \\ 16 & 18.00 & 18.25 \end{array} $$

Some people believe that talking on a cell phone while driving slows reaction time, increasing the risk of accidents. The study described in the paper "A Comparison of the Cell Phone Driver and the Drunk Driver" (Human Factors [2006]: \(381-391\) ) investigated the braking reaction time of people driving in a driving simulator. Drivers followed a pace car in the simulator, and when the pace car's brake lights came on, the drivers were supposed to step on the brake. The time between the pace car brake lights coming on and the driver stepping on the brake was measured. Two samples of 40 drivers participated in the study. The 40 people in one sample used a cell phone while driving. The 40 people in the second sample drank a mixture of orange juice and alcohol in an amount calculated to achieve a blood alcohol level of \(0.08 \%\) (a value considered legally drunk in most states). For the cell phone sample, the mean braking reaction time was 779 milliseconds and the standard deviation was 209 milliseconds. For the alcohol sample, the mean breaking reaction time was 849 milliseconds and the standard deviation was \(228 .\) Is there convincing evidence that the mean braking reaction time is different for the population of drivers talking on a cell phone and the population of drivers who have a blood alcohol level of \(0.08 \%\) ? For purposes of this exercise, you can assume that the two samples are representative of the two populations of interest.

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 article "Plugged In, but Tuned Out" (USA Today, January 20,2010 ) summarizes data from two surveys of kids ages 8 to 18 . One survey was conducted in 1999 and the other was conducted in \(2009 .\) Data on the number of hours per day spent using electronic media, consistent with summary quantities given in the article, are given in the following table (the actual sample sizes for the two surveys were much larger). For purposes of this exercise, assume that the two samples are representative of kids ages 8 to 18 in each of the 2 years the surveys were conducted. Construct and interpret a \(98 \%\) confidence interval estimate of the difference between the mean number of hours per day spent using electronic media in 2009 and \(1999 .\) $$ \begin{array}{llllllllllllllll} 2009 & 5 & 9 & 5 & 8 & 7 & 6 & 7 & 9 & 7 & 9 & 6 & 9 & 10 & 9 & 8 \\ 1999 & 4 & 5 & 7 & 7 & 5 & 7 & 5 & 6 & 5 & 6 & 7 & 8 & 5 & 6 & 6 \end{array} $$

The paper "Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children in a National Household Survey" (Pediatrics [2004]: \(112-118\) ) investigated the effect of fast-food consumption on other dietary variables. For a representative sample of 663 teens who reported that they did not eat fast food during a typical day, the mean daily calorie intake was 2,258 and the sample standard deviation was \(1,519 .\) For a representative sample of 413 teens who reported that they did eat fast food on a typical day, the mean calorie intake was 2,637 and the standard deviation was \(1,138 .\) Use the given information and a \(95 \%\) confidence interval to estimate the difference in mean daily calorie intake for teens who do eat fast food on a typical day and those who do not.

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