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Suppose that data are obtained from 20 pairs of \((x, y)\) and the sample correlation coefficient is 0.8 . (a) Test the hypothesis that \(H_{0}: \rho=0\) against \(H_{1}: \rho \neq 0\) with \(\alpha=0.05 .\) Calculate the \(P\) -value. (b) Test the hypothesis that \(H_{1}: \rho=0.5\) against \(H_{1}: \rho \neq 0.5\) with \(\alpha=0.05 .\) Calculate the \(P\) -value. (c) Construct a \(95 \%\) two-sided confidence interval for the correlation coefficient. Explain how the questions in parts (a) and (b) could be answered with a confidence interval.

Short Answer

Expert verified
(a) Reject \( H_0 \) if P-value < 0.05; calculated P-value is very small. (b) Do not reject \( H_0 \) if P-value > 0.05; 0.5 likely within 95% CI. (c) CI conversion shows \( \rho \) much closer to 0.8 than 0.5.

Step by step solution

01

Determine Degrees of Freedom

For both parts (a) and (b) of the problem, we need the degrees of freedom to calculate the test statistic. The degrees of freedom is calculated as the number of data pairs minus 2, i.e., \(n - 2\), where \(n = 20\). So, the degrees of freedom is \(18\).
02

Hypothesis Testing for \(H_0: \rho = 0\)

For part (a), we test the null hypothesis \(H_0: \rho = 0\) using the sample correlation coefficient \( r = 0.8 \). The test statistic \( t \) is given by the formula:\[ t = \frac{r \sqrt{n-2}}{\sqrt{1-r^2}} \]Substituting, we have \( t = \frac{0.8 \sqrt{18}}{\sqrt{1-0.8^2}} \). Calculate \( t \) and compare to the critical \( t \)-value for \(18\) degrees of freedom at \( \alpha = 0.05 \). The test is two-tailed since the alternative \( H_1: \rho eq 0 \).
03

Calculate the P-value for \(H_0: \rho = 0\)

Using the calculated test statistic, find the P-value from the t-distribution table for \(18\) degrees of freedom. Since it is a two-tailed test, double the significance level if you are using half tables.
04

Hypothesis Testing for \(H_0: \rho = 0.5\)

For part (b), test \(H_0: \rho = 0.5\) using Fisher's z-transformation:\[ z = \frac{1}{2} \log \left( \frac{1+r}{1-r} \right) \]Calculate \( z_r \) for \( r = 0.8 \) and \( z_0.5 \) for \( \rho = 0.5 \). The test statistic is:\[ z = \frac{z_r - z_0.5}{\sqrt{\frac{1}{n-3}}} \]Compute z and compare to the critical z-value at \( \,\alpha = 0.05 \) for a two-tailed test.
05

Calculate the P-value for \(H_0: \rho = 0.5\)

Using the standard normal distribution, find the P-value corresponding to the calculated z from Step 4. As the test is two-sided, double the sigma area for the normal distribution.
06

Constructing 95% Confidence Interval

To construct a 95% confidence interval for the correlation coefficient, convert \( r = 0.8 \) to \( z \) using Fisher’s transformation. Calculate the standard error \( \frac{1}{\sqrt{n-3}} \), then:\[ z \pm Z_{\alpha/2} \times \text{SE} \]Convert back to the \( r \) scale for the lower and upper bounds of the confidence interval.
07

Answer Questions Using Confidence Interval

Use the confidence interval to note if zero and/or 0.5 are within it. If zero is not in the interval, the null in part (a) can be rejected; similarly, if 0.5 is not in the interval, the null in part (b) can be rejected.

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Key Concepts

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

Correlation Coefficient
The correlation coefficient, often represented by the symbol \( r \), measures the strength and direction of a linear relationship between two variables. This value ranges from -1 to 1, where an \( r \) of +1 implies a perfect positive relationship, -1 indicates a perfect negative relationship, and 0 indicates no linear relationship.

In the given exercise, the sample correlation coefficient is \( r = 0.8 \). This suggests a strong positive linear relationship between the paired data. A correlation close to 1 means that as one variable increases, the other tends to increase as well, and vice versa. However, it's important to note that correlation does not imply causation.

Understanding the correlation coefficient allows us to execute hypothesis tests to establish the statistical significance of the observed relationship.
P-value Calculation
P-value is a critical concept in hypothesis testing. It helps us determine the significance of our test results. A P-value less than the significance level \( \alpha \) indicates strong evidence against the null hypothesis, prompting its rejection. In this exercise, \( \alpha = 0.05 \).

For part (a), we calculated a test statistic using the formula: \[ t = \frac{r \sqrt{n-2}}{\sqrt{1-r^2}} \]where \( r = 0.8 \) and \( n = 20 \). The resulting test statistic can be compared to critical \( t \)-values.

For part (b), Fisher's z-transformation provides a normal distribution approximation, which helps us compute a z-score for hypothesis testing against \( \rho = 0.5 \). The resulting z-score is then used with a standard normal distribution table to find the P-value. In both tests, if the calculated P-value is less than 0.05, the null hypothesis can be rejected.
Confidence Interval
A confidence interval provides a range of values within which we can say with a certain degree of confidence that the parameter lies. In part (c) of the exercise, we constructed a 95% confidence interval for the correlation coefficient.

Using Fisher's z-transformation, we can transform the sample correlation \( r = 0.8 \) into a z-score. Then, we calculate the standard error by \[ SE = \frac{1}{\sqrt{n-3}} \]to build our confidence interval around the z-score.

This interval is then converted back to the correlation scale to get the actual lower and upper bounds. If a proposed value like zero or 0.5 is outside this interval, it provides evidence to reject the corresponding null hypothesis.
Fisher's z-transformation
Fisher's z-transformation is a mathematical technique used to stabilize the variance of the correlation coefficients. This becomes especially useful when dealing with hypothesis testing involving correlations, as it allows us to take advantage of normal distribution properties for inference.

By transforming the sample correlation coefficient \( r \) to the z-scale, we can apply standard normal distribution methods to calculate z-scores and confidence intervals. The transformation is given by:\[ z = \frac{1}{2} \log \left( \frac{1+r}{1-r} \right) \]

In the exercise, Fisher's z-transformation helps in testing the hypothesis for \( \rho = 0.5 \) and constructing the confidence interval. By allowing us to compute z-scores, it simplifies the process of assessing statistical significance and reliability of the correlation estimate.

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Most popular questions from this chapter

Show that in a simple linear regression model the point \((\bar{x}, \bar{y})\) lies exactly on the least squares regression line.

Suppose that we are interested in fitting a simple linear regression model \(Y=\beta_{0}+\beta_{1} x+\epsilon\) where the intercept, \(\beta_{0},\) is known. (a) Find the least squares estimator of \(\beta_{1}\). (b) What is the variance of the estimator of the slope in part (a)? (c) Find an expression for a \(100(1-\alpha) \%\) confidence interval for the slope \(\beta_{1}\). Is this interval longer than the corresponding interval for the case in which both the intercept and slope are unknown? Justify your answer.

The World Health Organization defines obesity in adults as having a body mass index (BMI) higher than \(30 .\) Of the 250 men in the study mentioned in Exercise \(11-1,23\) are by this definition obese. How good is waist (size in inches) as a predictor of obesity? A logistic regression model was fit to the data: $$\log \left (\frac{p} {1-p}\right)=-41.828+0.9864 \ \mathrm{waist} $$ where \(p\) is the probability of being classified as obese. (a) Does the probability of being classified as obese increase or decrease as a function of waist size? Explain. (b) What is the estimated probability of being classified as obese for a man with a waist size of 36 inches? (c) What is the estimated probability of being classified as obese for a man with a waist size of 42 inches? (d) What is the estimated probability of being classified as obese for a man with a waist size of 48 inches? (e) Make a plot of the estimated probability of being classified as obese as a function of waist size.

\(11-114 .\) Weighted Least Squares. Suppose that we are fitting the line \(Y=\beta_{0}+\beta_{1} x+\epsilon,\) but the variance of \(Y\) depends on the level of \(x\); that is, $$ V\left(Y_{i} \mid x_{i}\right)=\sigma_{i}^{2}=\frac{\sigma^{2}}{w_{i}} \quad i=1,2, \ldots, n $$ where the \(w_{i}\) are constants, often called weights. Show that for an objective function in which each squared residual is multiplied by the reciprocal of the variance of the corresponding observation, the resulting weighted least squares normal equations are $$\begin{array}{c}\hat{\beta}_{0} \sum_{i=1}^{n} w_{i}+\hat{\beta}_{1} \sum_{i=1}^{n} w_{i} x_{i}=\sum_{i=1}^{n} w_{i} y_{i} \\\\\hat{\beta}_{0} \sum_{i=1}^{n} w_{i} x_{i}+\hat{\beta}_{1} \sum_{i=1}^{n} w_{i} x_{i}^{2}=\sum_{i=1}^{n} w_{i} x_{i} y_{i}\end{array}$$ Find the solution to these normal equations. The solutions are weighted least squares estimators of \(\beta_{0}\) and \(\beta_{1}\).

Show that, for the simple linear regression model, the following statements are true: (a) \(\sum_{i=1}^{n}\left(\mathrm{y}_{i}-\hat{\mathrm{y}}_{i}\right)=0\) (b) \(\sum_{i=1}^{n}\left(\mathrm{y}_{i}-\hat{\mathrm{y}}_{i}\right) \mathrm{x}_{i}=0\) (c) \(\frac{1}{n} \sum_{i=1}^{n} \hat{\mathrm{y}}_{i}=\bar{y}\)

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