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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

Descriptions of three studies are given. In each of the studies, the two populations of interest are students majoring in science at a particular university and students majoring in liberal arts at this university. For each of these studies, indicate whether the samples are independently selected or paired. Study 1: To determine if there is evidence that the mean number of hours spent studying per week differs for the two populations, a random sample of 100 science majors and a random sample of 75 liberal arts majors are selected. Study 2: To determine if the mean amount of money spent on textbooks differs for the two populations, a random sample of science majors is selected. Each student in this sample is asked how many units he or she is enrolled in for the current semester. For each of these science majors, a liberal arts major who is taking the same number of units is identified and included in the sample of liberal arts majors. Study 3: To determine if the mean amount of time spent using the campus library differs for the two populations, a random sample of science majors is selected. A separate random sample of the same size is selected from the population of liberal arts majors.

For each of the following hypothesis testing scenarios, indicate whether or not the appropriate hypothesis test would be for a difference in population means. If not, explain why not. Scenario 1: A researcher at the Medical College of Virginia conducted a study of 60 randomly selected male soccer players and concluded that players who frequently "head" the ball in soccer have a lower mean IQ (USA Today, August 14,1995 ). The soccer players were divided into two samples, based on whether they averaged 10 or more headers per game, and IQ was measured for each player. You would like to determine if the data support the researcher's conclusion. Scenario 2: A credit bureau analysis of undergraduate students" credit records found that the mean number of credit cards in an undergraduate's wallet was 4.09 ("Undergraduate Students and Credit Cards in \(2004,{ }^{n}\) Nellie Mae, May 2005 ). It was also reported that in a random sample of 132 undergraduates, the mean number of credit cards that the students said they carried was 2.6. You would like to determine if there is convincing evidence that the mean number of credit cards that undergraduates report carrying is less than the credit bureau's figure of \(4.09 .\) Scenario 3: Some commercial airplanes recirculate approximately \(50 \%\) of the cabin air in order to increase fuel efficiency. The authors of the paper "Aircraft Cabin Air Recirculation and Symptoms of the Common Cold" (Journal of the American Medical Association \([2002]: 483-486)\) studied 1,100 airline passengers who flew from San Francisco to Denver. Some passengers traveled on airplanes that recirculated air, and others traveled on planes that did not. Of the 517 passengers who flew on planes that did not recirculate air,

Do male college students spend more time using a computer than female 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 three-week period. For the sample of males, the mean time spent using a computer per day was 45.8 minutes and the standard deviation was 63.3 minutes. For the sample of females, the mean time spent using a computer was 39.4 minutes and the standard deviation was 57.3 minutes. Is there convincing evidence that the mean time male students at this university spend using a computer is greater than the mean time for female students? Test the appropriate hypotheses using \(\alpha=0.05 .\) (Hint: See Example 13.1\()\)

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.

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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