/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 Torsion during hip external rota... [FREE SOLUTION] | 91Ó°ÊÓ

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Torsion during hip external rotation and extension may explain why acetabular labral tears occur in professional athletes. The article "Hip Rotational Velocities During the Full Golf Swing" (J. of Sports Science and Med., 2009: 296-299) reported on an investigation in which lead hip internal peak rotational velocity \((x)\) and trailing hip peak external rotational velocity \((y)\) were determined for a sample of 15 golfers. Data provided by the article's authors was used to calculate the following summary quantities: $$ \begin{array}{r} \sum\left(x_{i}-\bar{x}\right)^{2}=64,732.83, \sum\left(y_{i}-\bar{y}\right)^{2}=130,566.96, \\ \sum\left(x_{i}-\bar{x}\right)\left(y_{i}-\bar{y}\right)=44,185.87 \end{array} $$ Separate normal probability plots showed very substantial linear patterns. a. Calculate a point estimate for the population correlation coefficient. b. Carry out a test at significance level \(.01\) to decide whether there is a linear relationship between the two velocities in the sampled population; your conclusion should be based on a \(P\)-value. c. Would the conclusion of (b) have changed if you had tested appropriate hypotheses to decide whether there is a positive linear association in the population? What if a significance level of \(.05\) rather than \(.01\) had been used?

Short Answer

Expert verified
Correlation estimate is 0.4857; no significant relationship at 0.01 level. Significant positive association at 0.05 level.

Step by step solution

01

Understanding the Problem

We need to find the population correlation coefficient, test for a linear relationship between two variables at a 0.01 significance level, and also check if our conclusion would change under different conditions.
02

Calculate the Point Estimate for the Correlation Coefficient

The correlation coefficient, \( r \), is given by the formula \( r = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum (x_i - \bar{x})^2 \sum (y_i - \bar{y})^2}} \). Substituting the provided values, we have: \[ r = \frac{44,185.87}{\sqrt{64,732.83 \times 130,566.96}} \approx 0.4857.\]
03

Hypothesis Testing for Linear Relationship

We need to test the hypothesis \( H_0: \rho = 0 \) against \( H_a: \rho eq 0 \), where \( \rho \) is the population correlation coefficient. The test statistic is calculated as \( t = \frac{r\sqrt{n-2}}{\sqrt{1-r^2}} \). For \( n = 15 \), this is \[ t = \frac{0.4857 \sqrt{15-2}}{\sqrt{1-0.4857^2}} \approx 2.066. \]
04

Determine the Critical Value and P-value

The degrees of freedom is \( n - 2 = 13 \). For a two-tailed test at \( \alpha = 0.01 \), the critical value of \( t \) is approximately \( 3.012 \) (from a t-table or using software). The P-value associated with \( t = 2.066 \) is greater than 0.01, so we do not reject \( H_0 \).
05

Conclusion for Part (b)

Since the P-value is greater than the significance level of \(0.01\), we do not have sufficient evidence to claim a significant linear relationship between the two velocities in the population.
06

Hypothesis Test for Positive Association

If testing for \( H_0: \rho \leq 0 \) against \( H_a: \rho > 0 \), using the same calculation, \( t = 2.066 \) is still not significant at the \( \alpha = 0.01 \) level due to a greater P-value. However, at \( \alpha = 0.05 \), the test would be significant as the critical value for a one-tailed test at \( \alpha = 0.05 \) is smaller (approximately \( 1.771 \), indicating rejection of \( H_0 \) if \( t > 1.771 \)).
07

Conclusion for Part (c)

For a one-tailed test with \( \alpha = 0.01 \), we still would not have rejected \( H_0 \). However, with a significance level of \( \alpha = 0.05 \) and a one-tailed test for \( H_a: \rho > 0 \), we would have concluded a significant positive linear association.

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Key Concepts

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

Linear Regression Analysis
Linear regression analysis is a statistical method used to examine the linear relationship between two variables. In the context of this exercise, we looked at how one rotation velocity influences another. Linear regression helps us understand how changes in one variable might impact another. For example, how lead hip internal peak rotational velocity might impact the trailing hip peak external rotational velocity in a golfer's swing.

To perform linear regression, we calculate several quantities like the correlation coefficient, which indicates the strength and direction of a linear relationship. The correlation coefficient, denoted as \( r \), can range between -1 and 1.
  • If \( r = 1 \), there is a perfect positive linear relationship.
  • If \( r = -1 \), there is a perfect negative linear relationship.
  • If \( r = 0 \), the variables are uncorrelated.
In this exercise, we calculated \( r \approx 0.4857 \), suggesting a moderate positive correlation. This tells us there's a moderate, direct relationship between the two rotational velocities. However, correlation doesn't imply causation. Other factors can influence these variables, and this analysis only captures a linear aspect of their relationship.
Hypothesis Testing in Statistics
Hypothesis testing is a statistical technique which allows us to make decisions or inferences about population parameters. In the context of this exercise, we used hypothesis testing to determine if a linear relationship between the two rotational velocities existed in the sampled population.

We set up our hypotheses as follows:
  • Null Hypothesis \( H_0: \rho = 0 \) - This suggests no linear relationship exists.
  • Alternative Hypothesis \( H_a: \rho eq 0 \) - This suggests a linear relationship exists.
We used the t-distribution to make inferences since we are dealing with correlations. Our calculated test statistic was \( t \approx 2.066 \).

To evaluate this, we compared it against critical values from the t-distribution (based on degrees of freedom, \( n - 2 \)). In this exercise, for the two-tailed test at a 0.01 significance level, the critical value was \( 3.012 \). Since our test statistic didn't exceed this critical value, we failed to reject the null hypothesis, implying insufficient evidence to suggest a linear relationship.
Significance Level in Statistics
The significance level, denoted by \( \alpha \), is a threshold we set to determine whether to reject the null hypothesis in hypothesis testing. It represents the probability of making a Type I error, which is rejecting the null hypothesis when it's true.

In our exercise, we initially used a significance level of \( \alpha = 0.01 \), meaning we are allowing a 1% chance of incorrectly concluding that a linear relationship exists when it does not. However, when using a significance level of \( \alpha = 0.05 \), or a 5% chance, our conclusion can change.
  • At \( \alpha = 0.01 \): The P-value was greater, so we do not reject \( H_0 \).
  • At \( \alpha = 0.05 \): The test for a one-tailed hypothesis \( H_a: \rho > 0 \) becomes significant. The critical value of \( t \approx 1.771 \), indicating we reject \( H_0 \).
Using a higher significance level (\( \alpha = 0.05 \)) increases our sensitivity to detect a true relationship, but also increases the risk of a Type I error. It highlights how the choice of significance level affects our hypothesis test outcomes. The context or domain often determines the appropriate \( \alpha \) level to use.

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Most popular questions from this chapter

The Turbine Oil Oxidation Test (TOST) and the Rotating Bomb Oxidation Test (RBOT) are two different procedures for evaluating the oxidation stability of steam turbine oils. The article "Dependence of Oxidation Stability of Steam Turbine Oil on Base Oil Composition" (J. of the Society of Tribologists and Lubrication Engrs., Oct. 1997: 19-24) reported the accompanying observations on \(x=\) TOST time (hr) and \(y=\) RBOT time (min) for 12 oil specimens. $$ \begin{array}{lrrrrrr} \text { TOST } & 4200 & 3600 & 3750 & 3675 & 4050 & 2770 \\ \text { RBOT } & 370 & 340 & 375 & 310 & 350 & 200 \\ \text { TOST } & 4870 & 4500 & 3450 & 2700 & 3750 & 3300 \\ \text { RBOT } & 400 & 375 & 285 & 225 & 345 & 285 \end{array} $$ a. Calculate and interpret the value of the sample correla tion coefficient (as do the article's authors). b. How would the value of \(r\) be affected if we had le \(x=\) RBOT time and \(y=\) TOST time? c. How would the value of \(r\) be affected if RBOT time were expressed in hours? d. Construct normal probability plots and comment. e. Carry out a test of hypotheses to decide whether RBOT time and TOST time are linearly related.

The article "Some Field Experience in the Use of an Accelerated Method in Estimating 28-Day Strength of Concrete" (J. of Amer: Concrete Institute, 1969: 895) considered regressing \(y=28\)-day standard-cured strength (psi) against \(x=\) accelerated strength (psi). Suppose the equation of the true regression line is \(y=1800+1.3 x\). a. What is the expected value of 28 -day strength when accelerated strength \(=2500\) ? b. By how much can we expect 28-day strength to change when accelerated strength increases by 1 psi? c. Answer part (b) for an increase of 100 psi. d. Answer part (b) for a decrease of 100 psi.

Suppose an investigator has data on the amount of shelf space \(x\) devoted to display of a particular product and sales revenue \(y\) for that product. The investigator may wish to fit a model for which the true regression line passes through \((0,0)\). The appropriate model is \(Y=\beta_{1} x+\epsilon\). Assume that \(\left(x_{1}, y_{1}\right), \ldots,\left(x_{n}, y_{n}\right)\) are observed pairs generated from this model, and derive the least squares estimator of \(\beta_{1}\).

The efficiency ratio for a steel specimen immersed in a phosphating tank is the weight of the phosphate coating divided by the metal loss (both in \(\mathrm{mg} / \mathrm{ft}^{2}\) ). The article "Statistical Process Control of a Phosphate Coating Line" (Wire J. Intl., May 1997: 78-81) gave the accompanying data on tank temperature \((x)\) and efficiency ratio \((y)\). $$ \begin{array}{lrrrrrrr} \text { Temp. } & 170 & 172 & 173 & 174 & 174 & 175 & 176 \\ \text { Ratio } & .84 & 1.31 & 1.42 & 1.03 & 1.07 & 1.08 & 1.04 \\ \text { Temp. } & 177 & 180 & 180 & 180 & 180 & 180 & 181 \\ \text { Ratio } & 1.80 & 1.45 & 1.60 & 1.61 & 2.13 & 2.15 & .84 \\ \text { Temp. } & 181 & 182 & 182 & 182 & 182 & 184 & 184 \\ \text { Ratio } & 1.43 & .90 & 1.81 & 1.94 & 2.68 & 1.49 & 2.52 \\ \text { Temp. } & 185 & 186 & 188 & & & & \\ \text { Ratio } & 3.00 & 1.87 & 3.08 & & & & \end{array} $$ a. Construct stem-and-leaf displays of both temperature and efficiency ratio, and comment on interesting features. b. Is the value of efficiency ratio completely and uniquely determined by tank temperature? Explain your reasoning. c. Construct a scatter plot of the data. Does it appear that efficiency ratio could be very well predicted by the value of temperature? Explain your reasoning.

Verify that if each \(x_{i}\) is multiplied by a positive constant \(c\) and each \(y_{i}\) is multiplied by another positive constant \(d\), the \(t\) statistic for testing \(H_{0}: \beta_{1}=0\) versus \(H_{\mathrm{a}}: \beta_{1} \neq 0\) is unchanged in value (the value of \(\hat{\beta}_{1}\) will change, which shows that the magnitude of \(\hat{\beta}_{1}\) is not by itself indicative of model utility).

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