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Head movement evaluations are important because disabled individuals may be able to operate communications aids using head motion. The paper "Constancy of Head Turning Recorded in Healthy Young Humans" (Journal of Biomedical Engineering [2008]\(: 428-436)\) reported the accompanying data on neck rotation (in degrees) both in the clockwise direction (CL) and in the counterclockwise direction (CO) for 14 subjects. For purposes of this exercise, you may assume that the 14 subjects are representative of the population of adult Americans. Based on these data, is it reasonable to conclude that mean neck rotation is greater in the clockwise direction than in the counterclockwise direction? Carry out a hypothesis test using a significance level of 0.01 . $$ \begin{array}{lccccccc} \text { Subject: } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \text { CL: } & 57.9 & 35.7 & 54.5 & 56.8 & 51.1 & 70.8 & 77.3 \\ \text { CO: } & 44.2 & 52.1 & 60.2 & 52.7 & 47.2 & 65.6 & 71.4 \\ \text { Subject: } & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \text { CL: } & 51.6 & 54.7 & 63.6 & 59.2 & 59.2 & 55.8 & 38.5 \\ \text { CO: } & 48.8 & 53.1 & 66.3 & 59.8 & 47.5 & 64.5 & 34.5 \end{array} $$

Short Answer

Expert verified
Based on the t-score and the critical t-value, it can be determined whether to accept the null hypothesis or reject it in favor of the alternative hypothesis. Depending on the answer, the conclusion would be made if it is reasonable or not that the mean neck rotation is greater in the clockwise direction than in the counterclockwise.

Step by step solution

01

Determine the Hypothesis

The null hypothesis \(H_0\) states that the mean neck rotation in the clockwise direction is equal to that in the counterclockwise direction, whereas the alternative hypothesis \(H_1\) posits that the clockwise mean neck rotation is greater than the counterclockwise. So, we have: \(H_0: \mu_{CL} = \mu_{CO}\) \(H_1: \mu_{CL} > \mu_{CO}\) where \(\mu_{CL}\) and \(\mu_{CO}\) are population mean neck rotations in the clockwise and counterclockwise directions, respectively.
02

Calculate Mean and Difference

Calculate the mean rotation in both directions by summing up the rotations and dividing by 14 (total number of subjects). Subsequently, calculate the difference in mean neck rotations.
03

Calculate the standard deviation and Standard Error

Calculate the standard deviation of the differences. Then calculate the Standard Error (SE) by dividing the standard deviation by the square root of the number of samples.
04

Determine the t-score

Using the formula for the t-score in a paired t-test, calculate the t-score. The formula should be \( t = \frac{\bar{d}}{SE} \), where \(\bar{d}\) is the mean difference, and SE is the Standard Error from Step 3.
05

Determine the critical value and compare

Determine the critical t-value for a 0.01 significance level with \( n - 1 \) degrees of freedom. Compare the calculated t-score with the critical t-value. If the t-score is greater than the critical t-value, reject the null hypothesis.

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Key Concepts

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

Significance Level
The significance level, denoted by \( \alpha \), is a threshold that determines when we should reject the null hypothesis in a hypothesis test. If the probability of observing the test statistic under the null hypothesis is less than the significance level, we reject the null hypothesis as it suggests that such an extreme value of the test statistic is unlikely to occur just by random chance. In the context of our exercise, a significance level of 0.01 means we have a maximum tolerance of a 1% chance of committing a Type I error, which is rejecting a true null hypothesis. Establishing such a low significance level reflects the need for a high degree of confidence in the results of the hypothesis test.
Paired t-test
A paired t-test is a statistical method used to compare the means of two related groups. In our exercise, we are comparing neck rotation measurements in two different directions for the same individuals, making it a classic scenario for the paired t-test. This test takes into account that the two samples are not independent and that they are 'paired' because they come from the same individual. The test assesses whether the average difference between the pairs is significantly different from zero. The paired t-test is particularly powerful in 'before and after' studies or studies that measure the effect of a treatment or condition in the same subjects over time.
Population Mean Comparison
Comparing population means is a common objective in statistical analyses, as it allows researchers to determine if there is a significant difference between two or more groups. In our neck rotation exercise, we aim to compare the population mean neck rotations in the clockwise direction \( \mu_{CL} \) with that in the counterclockwise direction \( \mu_{CO} \). The null hypothesis posits that there is no difference between these means, while the alternative hypothesis suggests there is a difference, specifically that the mean in the clockwise direction is greater. The paired t-test will help to determine if the observed difference in sample means reflects a true difference in population means or if it’s likely attributable to random variation.
Standard Error Calculation
The standard error (SE) is a measure of how much sample means are expected to vary from the true population mean. It's calculated by dividing the standard deviation of the sample by the square root of the sample size. In other words, SE gives us an understanding of the precision of our sample mean as an estimate of the population mean. For the paired t-test in our exercise, the SE is calculated using the standard deviation of the differences in neck rotations (CL - CO) between pairs. This SE is then used to calculate the t-score, which will tell us how many standard errors the sample mean difference is from zero. A larger t-score indicates a more significant difference between our paired observations.

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Most popular questions from this chapter

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.

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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 \(\alpha=0.05\). $$ \begin{array}{crr} \text { Athlete } & \text { X-ray } & \text { Ultrasound } \\ \hline 1 & 5.00 & 4.75 \\ 2 & 7.00 & 3.75 \\ 3 & 9.25 & 9.00 \\ 4 & 12.00 & 11.75 \\ 5 & 17.25 & 17.00 \\ 6 & 29.50 & 27.50 \\ 7 & 5.50 & 6.50 \\ 8 & 6.00 & 6.75 \\ 9 & 8.00 & 8.75 \\ 10 & 8.50 & 9.50 \\ 11 & 9.25 & 9.50 \\ 12 & 11.00 & 12.00 \\ 13 & 12.00 & 12.25 \\ 14 & 14.00 & 15.50 \\ 15 & 17.00 & 18.00 \\ 16 & 18.00 & 18.25 \end{array} $$

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