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Chapter 13: Problem 19

Use the data given in the previous exercise to complete the following: a. Estimate the mean difference in motion between dominant and nondominant arms for position players using a \(95 \%\) confidence interval. b. The authors asserted that pitchers have a greater difference in mean motion of their shoulders than do position players. Do you agree? Explain.

Short Answer

Expert verified
The 95% confidence interval for the mean difference in motion between dominant and nondominant arms for position players is given by \(\bar{d} \pm t_{\alpha/2} \cdot \frac{s_d}{\sqrt{n}}\). By comparing this confidence interval with the author's assertion, if the interval includes a larger mean difference in motion for pitchers than position players, we can agree with the assertion. However, conclusions should also consider the context of the study, sample size, and assumptions made for both groups of players.

Step by step solution

01

Identify the variables and notation

Let: - \(n\) be the number of position players - \(\bar{d}\) be the mean difference in motion between dominant and nondominant arms for position players - \(s_d\) be the standard deviation of the differences - \(t_{\alpha/2}\) be the critical value for a two-tailed t-distribution with \(n-1\) degrees of freedom, where \(\alpha\) is the level of significance (for a 95% confidence interval, it will be \(0.05\)) We will use these variables to calculate the 95% confidence interval.
02

Calculate the standard error

The standard error (SE) for the mean difference in motion between dominant and nondominant arms can be calculated using the standard deviation of the differences and the sample size. The formula is: \(SE = \frac{s_d}{\sqrt{n}}\)
03

Find the critical value

Use a t-score table or calculator to find the critical value (\(t_{\alpha/2}\)) for two-tailed t-distribution with \((n - 1)\) degrees of freedom and \(\alpha = 0.05\).
04

Calculate the confidence interval

The 95% confidence interval for the mean difference in motion between dominant and nondominant arms is given by the following formula: \(\bar{d} \pm t_{\alpha/2} \cdot SE\) Substitute the values obtained in the previous steps: \(\bar{d} \pm t_{\alpha/2} \cdot \frac{s_d}{\sqrt{n}}\) This gives us the 95% confidence interval for the mean difference in motion between dominant and nondominant arms.
05

Compare results with the author's assertion and form an opinion

After calculating the 95% confidence interval for the mean difference in motion between dominant and nondominant arms for position players, we must compare it with the assertion made by the authors, which states that pitchers have a greater difference in mean motion of their shoulders than do position players. Based on the results obtained, if the confidence interval includes a larger mean difference in motion for pitchers compared to position players, we can agree with the author's assertion. To make a sound conclusion, also consider the context of the study, the sample size, and the assumptions made for both groups of players.

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Key Concepts

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

Mean Difference
In the context of statistical analysis, the mean difference is a crucial concept. It refers to the average difference between two sets of data. For instance, when comparing the motion between dominant and nondominant arms, the mean difference will show us, on average, how much more or less the dominant arm moves compared to the nondominant one.
To calculate this, sum all individual differences in motion and then divide by the total number of observations. This helps in understanding the central tendency of the differences in our samples. By analyzing mean differences, researchers can draw insights into the effectiveness of interventions or the extent of variability between groups.
T-Distribution
The t-distribution is a fundamental concept in statistics, especially when dealing with small sample sizes. The distribution is similar to a bell-shaped curve, like the standard normal distribution, but it has thicker tails. This property helps account for the variability found in small samples.
When estimating confidence intervals or comparing sample means, the t-distribution is used to determine critical values. These values help gauge how "extreme" our sample observations are. In practice, you'd use a t-table or calculator to find the t-value corresponding to your desired confidence level and degrees of freedom.
Understanding the t-distribution helps address the uncertainty in small samples by providing a reliable tool for making safer statistical inferences.
Standard Error
The standard error (SE) is a measure that describes the accuracy of a sample mean estimate when compared to the population mean. In the context of mean difference, it quantifies the variability of the sample mean differences involved.
To compute the standard error for the mean difference, you use the formula: \[ SE = \frac{s_d}{\sqrt{n}} \] Here, \( s_d \) is the standard deviation of the sample differences, and \( n \) is the sample size. This SE value is critical in constructing confidence intervals, which show the range where the true mean difference likely falls.
A smaller standard error indicates a more precise estimate of the mean difference, which is crucial for interpreting and trusting your study's findings.
Sample Size
Sample size, denoted as \( n \), is the number of observations in a study. It plays a significant role in statistical analysis, particularly when estimating confidence intervals or testing hypotheses.
A larger sample size generally leads to more reliable results because it reduces the margin of error. This happens because a larger sample size tends to better represent the population, decreasing the variability in the sample estimate.
Furthermore, the sample size is a key component in the calculation of the standard error, affecting the accuracy of the confidence interval. When planning a study, choosing an appropriate sample size beforehand is crucial to ensure sufficient statistical power for detecting true differences or effects.

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Most popular questions from this chapter

The paper "Does the Color of the Mug Influence the Taste of the Coffee?" (Flavour [2014]: 1-7) describes an experiment in which subjects were assigned at random to one of two treatment groups. The 12 people in one group were scrved coffee in a white mug and were asked to rate the quality of the coffee on a scale from 0 to 100 . The 12 people in the second group were served the same coffee in a clear glass mug, and they also rated the coffee. The mean quality rating for the 12 people in the white mug group was 50.35 and the standard deviation was \(20.17 .\) The mean quality rating for the 12 people in the clear glass mug group was 61.48 and the standard deviation was \(16.69 .\) For purposes of this exercise, you may assume that the distribution of quality ratings for each of the two treatments is approximately normal. a. Use the given information to construct and interpret a \(95 \%\) confidence interval for the difference in mean quality rating for this coffee when served in a white mug and when served in a glass mug. b. Based on the interval from Part (a), are you convinced that the color of the mug makes a difference in terms of mean quality rating? Explain.

In a study of malpractice claims where a settlement had been reached, two random samples were selected: a random sample of 515 closed malpractice claims that were found not to involve medical errors and a random sample of 889 claims that were found to involve errors (New England Journal of Medicine [2006]: \(2024-2033\) ). The following statement appeared in the paper: "When claims not involving errors were compensated, payments were significantly lower on average than were payments for claims involving errors \((\$ 313,205\) vs. \(\$ 521,560, P=0.004)\) a. What hypotheses did the researchers test to reach the stated conclusion? b. Which of the following could have been the value of the test statistic for the hypothesis test? Explain your reasoning. i. \(\quad t=5.00\) iii. \(t=2.33\) ii. \(\quad t=2.65\) iv. \(l=1.47\)

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 TV is greater than the mean time for male students? Test the appropriate hypotheses using \(\alpha=0.05\).

The article "An Alternative Vote: Applying Science to the Teaching of Science" (The Economist, May 12,2011 ) describes an experiment conducted at the University of British Columbia. A total of 850 engineering students enrolled in a physics course participated in the experiment. These students were randomly assigned to one of two experimental groups. The two groups attended the same lectures for the first 11 weeks of the semester. In the twelfth week, one of the groups was switched to a style of teaching where students were expected to do reading assignments prior to class and then class time was used to focus on problem solving, discussion and group work. The second group continued with the traditional lecture approach. At the end of the twelfth week, the students were given a test over the course material from that week. The mean test score for students in the new teaching method group was 74 and the mean test score for students in the traditional lecture group was 41 . Suppose that the two groups each consisted of 425 students and that the standard deviations of test scores for the new teaching method group and the traditional lecture method group were 20 and 24 , respectively. Can you conclude that the mean test score is significantly higher for the new teaching method group than for the traditional lecture method group? Test the appropriate hypotheses using a significance level of \(\alpha=0.01\).

The report referenced in the previous exercise also gave data for representative samples of 250 adults in Canada and 250 adults in England. The sample mean amount of sleep on a work night was 423 minutes for the Canada sample and 409 minutes for the England sample. Suppose that the sample standard deviations were 35 minutes for the Canada sample and 42 minutes for the England sample. a. Construct and interpret a \(95 \%\) confidence interval estimate of the difference in the mean amount of sleep on a work night for adults in Canada and adults in England. b. Based on the confidence interval from Part (a), would you conclude that there is evidence of a diflerence in the mean amount of sleep on a work night for the two countries? Explain why or why not.
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