91Ó°ÊÓ

Head movement evaluations are important because individuals, especially those who are disabled, may be able to operate communications aids in this manner. The article "Constancy of Head Turning Recorded in Healthy Young Humans" (J. of Biomed. Engr., 2008: 428-436) reported data on ranges in maximum inclination angles of the head in the clockwise anterior, posterior, right, and left directions for 14 randomly selected subjects. Consider the accompanying data on average anterior maximum inclination angle (AMIA) both in the clockwise direction and in the counterclockwise direction. $$ \begin{array}{lccccccc} \text { Subj: } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \text { Cl: } & 57.9 & 35.7 & 54.5 & 56.8 & 51.1 & 70.8 & 77.3 \\ \text { Co: } & 44.2 & 52.1 & 60.2 & 52.7 & 47.2 & 65.6 & 71.4 \\ \text { Subj: } & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \text { Cl: } & 51.6 & 54.7 & 63.6 & 59.2 & 59.2 & 55.8 & 38.5 \\ \text { Co: } & 48.8 & 53.1 & 66.3 & 59.8 & 47.5 & 64.5 & 34.5 \end{array} $$ a. Calculate a point estimate of the population correlation coefficient between Cl AMIA and Co AMIA \(\left(\sum \mathrm{Cl}=\right.\) 786.7, \(\sum \mathrm{Co}=767.9, \sum \mathrm{Cl}^{2}=45,727.31, \sum \mathrm{Co}^{2}=\) \(\left.43,478.07, \sum \mathrm{ClCo}=44,187.87\right)\). b. Assuming bivariate normality (normal probability plots of the \(\mathrm{Cl}\) and Co samples are reasonably straight), carry out a test at significance level .01 to decide whether there is a linear association between the two variables in the population (as do the authors of the cited paper). Would the conclusion have been the same if a significance level of \(.001 \mathrm{had}\) been used?

Step by step solution

01

Calculate Sample Means

First, we calculate the sample means \( \bar{X} \) and \( \bar{Y} \) using the given sums for Cl and Co:\[ \bar{X} = \frac{\sum Cl}{n} = \frac{786.7}{14} = 56.19 \]\[ \bar{Y} = \frac{\sum Co}{n} = \frac{767.9}{14} = 54.85 \]

02

Calculate Standard Deviations

Next, compute the standard deviations for Cl and Co using the given sum of squares:\[ s_X = \sqrt{\frac{\sum \mathrm{Cl}^2 - \frac{(\sum \mathrm{Cl})^2}{n}}{n-1}} = \sqrt{\frac{45727.31 - \frac{786.7^2}{14}}{13}} = \sqrt{\frac{45727.31 - 44217.77}{13}} = 12.68 \]\[ s_Y = \sqrt{\frac{\sum \mathrm{Co}^2 - \frac{(\sum \mathrm{Co})^2}{n}}{n-1}} = \sqrt{\frac{43478.07 - \frac{767.9^2}{14}}{13}} = \sqrt{\frac{43478.07 - 42163.78}{13}} = 11.47 \]

03

Calculate Correlation Coefficient

Use the formula for the sample correlation coefficient \( r \):\[ r = \frac{\sum \mathrm{ClCo} - \frac{\sum \mathrm{Cl} \cdot \sum \mathrm{Co}}{n}}{(n-1) s_X s_Y} = \frac{44187.87 - \frac{786.7 \cdot 767.9}{14}}{13 \cdot 12.68 \cdot 11.47} \]Solving this gives:\[ r = \frac{44187.87 - 43200.31}{189.05} = \frac{987.56}{189.05} \approx 0.198 \]

04

Significance Test for Correlation

We perform a hypothesis test for the correlation using a t-test:- Null hypothesis \( H_0: \rho = 0 \)- Alternative hypothesis \( H_1: \rho eq 0 \)Calculate the t-statistic:\[ t = r \sqrt{\frac{n-2}{1-r^2}} = 0.198 \sqrt{\frac{12}{1-(0.198)^2}} \approx 0.711 \]Compare \( t \) to \( t_{critical} \) from the t-distribution with 12 degrees of freedom at significance level 0.01 (approximately 3.055). Since \( |t| < t_{critical} \), we fail to reject the null hypothesis.

05

Conclusion and Comparison of Significance Levels

Since the t-statistic is not significant at 0.01, it would not be significant at 0.001 either. Therefore, there is no evidence of a linear association between Cl AMIA and Co AMIA at either significance level.

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Key Concepts

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

Bivariate Normal Distribution

When working with two related variables, like the Cl AMIA and Co AMIA in this exercise, the data can be examined through a concept called the bivariate normal distribution. This statistical model assumes that both variables are normally distributed, which simplifies the analysis of their relationship.
The bivariate normal distribution is a crucial assumption for calculating correlation coefficients and testing their significance. It helps us to verify if our data trends align with a bell curve shape, a characteristic of normal distribution.
For our exercise, we should check the normality of each variable separately using normal probability plots. If both plots resemble a straight line, the assumption is validated, and the use of correlation coefficients is justified.

Hypothesis Testing

Hypothesis testing is a method used to determine the strength of the evidence from our data and to decide whether or not there is a significant association between two variables.
In this exercise, we conducted hypothesis testing using the correlation coefficient as a metric to gauge this relationship.
Steps in hypothesis testing include:

• Formulating the null hypothesis (\(H_0\): no correlation exists, \(\rho = 0\))
• Stating the alternative hypothesis (\(H_1\): a correlation does exist, \(\rho eq 0\))
• Calculating a test statistic using the sample correlation coefficient and sample size
• Comparing the test statistic to a critical value from the t-distribution

This process is essential for determining if the observed correlation in our sample occurs by random chance or if it reflects an actual relationship. In this exercise, our test statistic did not exceed the critical value at the 0.01 significance level, so we could not ascertain any significant association.

Statistical Significance

Statistical significance helps us to determine if a result is meaningful in a statistical study. In this context, it lets us know if the correlation we observe is likely to be present in the larger population, rather than just due to sampling error.
For a finding to be statistically significant, the test statistic from our hypothesis testing must exceed a critical value, determined by our chosen significance level (such as 0.01 or 0.001).
Choosing a lower significance level (like 0.001) demands stronger evidence to declare a result significant. Here, neither at 0.01 nor at 0.001 did our test statistic indicate a significant correlation. Thus, we concluded that the correlation is not statistically significant, suggesting any observed association might not hold in the population at large.