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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 cubic milliliters) are given in the following table:$$ \begin{array}{lccc} & n & \bar{x} & s \\ \hline \text { Children with ADHD } & 152 & 1,059.4 & 117.5 \\ \text { Children without ADHD } & 139 & 1,104.5 & 111.3 \\ \hline \end{array} $$ Is there convincing evidence that the mean brain volume for children with ADHD is smaller than the mean for children without ADHD? Test the relevant hypotheses using a 0.05 level of significance.

Short Answer

Expert verified
Whether there is convincing evidence that the mean brain volume for children with ADHD is smaller than the mean for children without ADHD depends on the result of the t-test described in step 4. The specific conclusion comes after comparing the calculated t-score with the critical value.

Step by step solution

01

State the Hypotheses

The null hypothesis \(H_0\) is that there is no difference in the mean brain volume of children with ADHD and those without ADHD. The alternative hypothesis \(H_a\) is that the mean brain volume of children with ADHD is smaller than that of children without ADHD. Mathematically, \(H_0: \mu_{1} = \mu_{2}\) and \(H_a: \mu_{1} < \mu_{2}\), where \(\mu_{1}\) represents the mean brain volume of children with ADHD and \(\mu_{2}\) represents the mean brain volume of children without ADHD.
02

Calculate the Test Statistic

In this case, the test statistic is a t-statistic and can be calculated using the formula for two independent samples: \(t = \frac{{\bar{x}_{1} - \bar{x}_{2}}}{\sqrt{\frac{{s_{1}^{2}}}{n_{1}} + \frac{{s_{2}^{2}}}{n_{2}}}}\), where \(\bar{x}_{1}\) and \(\bar{x}_{2}\) are the sample means, \(s_{1}^{2}\) and \(s_{2}^{2}\) are the sample variances, and \(n_{1}\) and \(n_{2}\) are the sample sizes.
03

Determine the Critical Value

Since this is a one-tailed test with a significance level of 0.05 and the degrees of freedom equals the smaller of \(n_{1}-1\) and \(n_{2}-1\), which is 138, the critical value from t-distribution table approximately equals -1.645.
04

Compare Test Statistic with Critical Value and Make a Decision

If the calculated t-score is less than -1.645, then the null hypothesis is rejected in favor of the alternative hypothesis. If the t-score is greater than -1.645, then there isn't enough evidence to reject the null hypothesis.
05

Conclude the Hypothesis Test

If the null hypothesis was rejected, there is enough evidence to support the claim that the mean brain volume for children with ADHD is smaller than the mean for children without ADHD. If the null hypothesis wasn't rejected, there isn't enough evidence to support the claim.

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Key Concepts

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

Null Hypothesis
In any research involving statistics, it's crucial to start with a null hypothesis (\(H_0\)). This is a default statement asserting that there is no effect or no difference between the groups being studied. It is the assertion that what you expect to prove is not true. For instance, when examining whether children with ADHD have smaller brain volumes compared to those without, the null hypothesis posits there's no difference in brain volume between the two groups, mathematically expressed as \(H_0: \mu_{1} = \mu_{2}\). Establishing a null hypothesis is a fundamental step in hypothesis testing because it provides a basis for comparison and the foundation for statistical analysis.

The null hypothesis is a skeptical perspective that assumes any observed differences are due to chance rather than a real effect. It's not a statement of equality per se, but more a presumption of no association or effect unless proven otherwise with data and statistical methods.
Alternative Hypothesis
Conversely, the alternative hypothesis (\(H_a\) or \(H_1\)), represents what a researcher aims to support. It asserts that there is a significant difference or effect. In this context, where brain volumes of children with and without ADHD are in question, the alternative hypothesis suggests that children with ADHD have a smaller mean brain volume than those without. It can be mathematically defined as \(H_a: \mu_{1} < \mu_{2}\), where \(\mu_{1}\) and \(\mu_{2}\) depict the mean brain volumes for children with and without the condition respectively.

The alternative hypothesis is the statement researchers are trying to verify or find evidence to support. In hypothesis testing, if the evidence suggests that the null hypothesis might be false, then the alternative hypothesis gains credence. The establishment of both null and alternative hypotheses is integral to the research process, providing clear definitions for the direction of the analysis.
T-Statistic
The t-statistic is a ratio that compares the difference between two sample means relative to the variability in the sample data. It takes into account both the sample size and the variance, making it an effective tool in hypothesis testing where we compare two groups.

The t-statistic is calculated using a formula that includes the sample means (\(\bar{x}_{1}\) and \(\bar{x}_{2}\)), the sample standard deviations (\(s_{1}\) and \(s_{2}\)), and the sample sizes (\(n_{1}\) and \(n_{2}\)). The formula for our problem is: \[t = \frac{{\bar{x}_{1} - \bar{x}_{2}}}{{\sqrt{\frac{{s_{1}^{2}}}{n_{1}} + \frac{{s_{2}^{2}}}{n_{2}}}}}\] The t-statistic value can then be used to determine the probability of observing such a result if the null hypothesis was true. The smaller the t-value, the larger the difference between the two groups, which aids in deciding whether to accept or reject the null hypothesis.
Statistical Significance
The concept of statistical significance is used to determine the likelihood that the observed results are not due to chance but to a real effect in the population. This significance is typically tested against a significance level, often denoted as alpha (\(\alpha\)), which is a threshold value that dictates the probability of rejecting a true null hypothesis. Common \(\alpha\) levels include 0.05 or 0.01.

When the p-value associated with a t-statistic is less than this threshold, the results are deemed to be statistically significant. This would mean that the null hypothesis is unlikely to be true and suggests that the alternative hypothesis should be considered. However, 'statistically significant' does not necessarily equate to 'clinically significant'; it simply implies that the finding has statistical support to warrant further study or consideration.
Hypothesis Testing
Lastly, hypothesis testing is a structured method used to determine whether there is enough statistical evidence in favor of a certain belief or hypothesis. It involves several steps, which include formulating the null and alternative hypotheses, deciding on a significance level, calculating a test statistic like the t-statistic, and comparing this value to a critical value.

If the t-statistic is beyond the critical value (considering the chosen significance level), it suggests rejecting the null hypothesis in favor of the alternative. The entire process facilitates informed decision-making based on statistical findings, rather than on speculation or assumptions. Through hypothesis testing, researchers can assess the validity of their predictions about a population, given data sampled from it. In the context of ADHD brain volume research, hypothesis testing allows us to scientifically investigate the proposed differences in brain volumes between affected and unaffected children.

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Most popular questions from this chapter

Two proposed computer mouse designs were compared by recording wrist extension in degrees for 24 people who each used both mouse designs ("Comparative Study of Two Computer Mouse Designs," Cornell Human Factors Laboratory Technical Report RP7992). The difference in wrist extension was calculated by subtracting extension for mouse type \(\mathrm{B}\) from the wrist extension for mouse type \(\mathrm{A}\) for each person. The mean difference was reported to be 8.82 degrees. Assume that this sample of 24 people is representative of the population of computer users. a. Suppose that the standard deviation of the differences was 10 degrees. Is there convincing evidence that the mean wrist extension for mouse type \(A\) is greater than for mouse type \(\mathrm{B}\) ? Use a 0.05 significance level. b. Suppose that the standard deviation of the differences was 26 degrees. Is there convincing evidence that the mean wrist extension for mouse type \(A\) is greater than for mouse type \(\mathrm{B}\) ? Use a 0.05 significance level. c. Briefly explain why different conclusions were reached in the hypothesis tests of Parts (a) and (b).

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 "Ladies First?" A Field Study of Discrimination in Coffee Shops" (Applied Economics [2008]: 1-19) describes a study in which researchers observed wait times in coffee shops in Boston. Both wait time and gender of the customer were observed. The mean wait time for a sample of 145 male customers was 85.2 seconds. The mean wait time for a sample of 141 female customers was 113.7 seconds. The sample standard deviations (estimated from graphs in the paper) were 50 seconds for the sample of males and 75 seconds for the sample of females. Suppose that these two samples are representative of the populations of wait times for female coffee shop customers and for male coffee shop customers. Is there convincing evidence that the mean wait time differs for males and females? Test the relevant hypotheses using a significance level of 0.05

Research has shown that for baseball players, good hip range of motion results in improved performance and decreased body stress. The article "Functional Hip Characteristics of Baseball Pitchers and Position Players" (The American journal of Sports Medicine, \(2010: 383-388\) ) reported on a study involving independent samples of 40 professional pitchers and 40 professional position players. For the sample of pitchers, the mean hip range of motion was 75.6 degrees and the standard deviation was 5.9 degrees, whereas the mean and standard deviation for the sample of position players were 79.6 degrees and 7.6 degrees, respectively. Assuming that these two samples are representative of professional baseball pitchers and position players, estimate the difference in mean hip range of motion for pitchers and position players using a \(90 \%\) confidence interval.

Do female college students spend more time watching TV than male college students? This was one of the questions investigated by the authors of the paper "An Ecological Momentary Assessment of the Physical Activity and Sedentary Behaviour Patterns of University Students" (Health Education Journal [2010]: \(116-125\) ). Each student in a random sample of 46 male students at a university in England and each student in a random sample of 38 female students from the same university kept a diary of how he or she spent time over a 3 -week period. For the sample of males, the mean time spent watching TV per day was 68.2 minutes, and the standard deviation was 67.5 minutes. For the sample of females, the mean time spent watching TV per day was 93.5 minutes, and the standard deviation was 89.1 minutes. Is there convincing evidence that the mean time female students at this university spend watching \(\mathrm{TV}\) is greater than the mean time for male students? Test the appropriate hypotheses using \(\alpha=0.05\).

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