Chapter 11: Problem 78
A random sample of \(n=25\) observations was made on the time to failure of an electronic component and the temperature in the application environment in which the component was used. (a) Given that \(r=0.83\), test the hypothesis that \(\rho=0\) using \(\alpha=0.05 .\) What is the \(P\) -value for this test? (b) Find a \(95 \%\) confidence interval on \(\rho\). (c) Test the hypothesis \(H_{0}: \rho=0.8\) versus \(H_{1}: \rho \neq 0.8,\) using \(\alpha=0.05 .\) Find the \(P\) -value for this test.
Short Answer
Step by step solution
Understand the context of the problem
Hypothesis test for \(\rho = 0\)
Calculate the P-value for \(\rho = 0\) test
Calculate the 95% confidence interval for \(\rho\)
Hypothesis test for \(\rho = 0.8\)
Conclusion for Hypotheses and Confidence Interval
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Confidence Interval for Correlation
We used Fisher's Z-transformation to convert the sample correlation coefficient \(r\) to a Z-score, then calculated the standard error for Fisher's Z. Here's a breakdown of the steps:
1. **Compute Fisher's Z**: Transform the correlation \(r = 0.83\) using \(z' = \frac{1}{2} \ln \left(\frac{1+r}{1-r}\right) \). This gave us \(z' \approx 1.180\).
2. **Calculate the Standard Error**: The formula \(\sigma_{z'} = \frac{1}{\sqrt{n-3}}\) is used to estimate the standard error. For our sample size \(n = 25\), \(\sigma_{z'} \approx 0.213\).
3. **Determine the Interval**: The 95% confidence interval in the Z domain is \(z' - 1.96\times\sigma_{z'}\) to \(z' + 1.96\times\sigma_{z'}\) which converts back to a correlation using the inverse Fisher transformation. Thus, we concluded the confidence interval for \(\rho\) as approximately \(0.64\) to \(0.92\).
This reflects that, with 95% confidence, the true correlation lies within this range.
Fisher's Z-transformation
In the exercise, Fisher's Z-transformation was applied to \(r = 0.83\) to find the confidence interval and for the hypothesis test against \(\rho = 0.8\). Fisher's transformation formula is \(z' = \frac{1}{2} \ln \left(\frac{1+r}{1-r}\right)\). This transformation results in a Z-value that is easier to handle, particularly for large sample sizes, because it stabilizes variance.
**Why Use Fisher’s Z-transformation?**:
- **Normalizes the Distribution:** By converting the correlation coefficient \(r\), you get a more normally distributed variable, especially useful when \(|r|\) is near 1 or larger samples are involved.
- **Accuracy Enhancement:** It provides more precise confidence intervals and significance tests.
P-value Interpretation
In the hypothesis test where we examined whether \(\rho = 0\), the P-value was calculated from the test statistic \(t \approx 5.7\), resulting in a very small P-value (less than 0.0001). This small P-value strongly indicated that we reject the null hypothesis, affirming that the correlation is indeed significant.
For the hypothesis test \(\rho = 0.8\), the calculated test statistic yielded a P-value of about 0.7. This higher P-value suggested there is not enough evidence to reject the null hypothesis, meaning it is plausible that the actual population correlation could be 0.8.
- **Low P-value (typically < 0.05):** Strong evidence against the null hypothesis, leading to its rejection in favor of the alternative hypothesis.
- **High P-value (typically > 0.05):** Weak evidence against the null hypothesis, often resulting in failure to reject the null.