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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

Expert verified
(a) Reject \(H_0\), \(P < 0.001\). (b) CI: \(0.64\) to \(0.92\). (c) Fail to reject \(H_0\), \(P \approx 0.7\).

Step by step solution

01

Understand the context of the problem

We are given a sample size of \(n=25\) and a sample correlation \(r=0.83\). We're tasked with conducting hypothesis tests on the population correlation \(\rho\) and finding a confidence interval. Each part requires different statistical techniques.
02

Hypothesis test for \(\rho = 0\)

First, test the hypothesis \(H_0: \rho = 0\) against \(H_1: \rho eq 0\) using \(\alpha = 0.05\). The test statistic for correlation is:\[ t = \frac{r \sqrt{n-2}}{\sqrt{1-r^2}} \]Substitute the given values: \(r = 0.83\), \(n = 25\).\[ t = \frac{0.83 \sqrt{25-2}}{\sqrt{1-0.83^2}} = \frac{0.83 \times 4.69}{\sqrt{0.3111}} \approx 5.7 \]The degrees of freedom are \(n - 2 = 23\). Look up or calculate the critical value from a t-distribution table for \(df=23\) at \(\alpha = 0.05\). The critical t-value is approximately \(\pm 2.069\). Since \(t\approx 5.7\) exceeds this, we reject \(H_0\) and conclude there is significant correlation.
03

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

Using a t-distribution table or calculator, find the P-value corresponding to \(t \approx 5.7\) for \(df = 23\). This P-value is very small (typically less than 0.0001), supporting rejection of \(H_0\) strongly.
04

Calculate the 95% confidence interval for \(\rho\)

Use Fisher's Z-transformation:\[ z' = \frac{1}{2} \ln \left(\frac{1+r}{1-r}\right) = \frac{1}{2} \ln \left(\frac{1+0.83}{1-0.83}\right) \approx 1.180 \]Estimate \(\sigma_{z'} = \frac{1}{\sqrt{n-3}} = \frac{1}{\sqrt{22}} \approx 0.213\).Calculate the confidence interval:\[ z' - Z\sigma_{z'} \text{ to } z' + Z\sigma_{z'} \]Where \(Z = 1.96\) (for 95%).\[ 1.180 - 1.96 \times 0.213 \text{ to } 1.180 + 1.96 \times 0.213 \approx 0.761 \text{ to } 1.599 \]Convert back to \(\rho\) using the inverse Fisher transformation:\[ \rho_1 = \frac{e^{2(0.761)}-1}{e^{2(0.761)}+1} \approx 0.64 \]\[ \rho_2 = \frac{e^{2(1.599)}-1}{e^{2(1.599)}+1} \approx 0.92 \]Thus, the 95% CI is approximately \(0.64\) to \(0.92\).
05

Hypothesis test for \(\rho = 0.8\)

Test \(H_0: \rho = 0.8\) versus \(H_1: \rho eq 0.8\) using \(\alpha = 0.05\). Convert \(r=0.83\) and \(\rho=0.8\) to Fisher's Z:\[ z'_r = \frac{1}{2} \ln \left(\frac{1+0.83}{1-0.83}\right) \approx 1.180 \]\[ z'_\rho = \frac{1}{2} \ln \left(\frac{1+0.8}{1-0.8}\right) \approx 1.099 \]The test statistic is:\[ z = \frac{z'_r - z'_\rho}{\sigma_{z'}} = \frac{1.180 - 1.099}{0.213} \approx 0.38 \]Find the corresponding P-value using standard normal distribution (approximately 0.7), which is greater than \(\alpha = 0.05\). Hence, we fail to reject \(H_0\).
06

Conclusion for Hypotheses and Confidence Interval

For part (a), we reject \(H_0\) and conclude \(\rho eq 0\). For part (b), the 95% CI for \(\rho\) is approximately \(0.64\) to \(0.92\). In part (c), we fail to reject \(H_0\) that \(\rho = 0.8\), since the P-value is greater than 0.05.

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Key Concepts

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

Confidence Interval for Correlation
The confidence interval for a correlation coefficient provides a range of values within which the actual population correlation, \(\rho\), is expected to fall, with a given level of confidence. In our example, we constructed a 95% confidence interval for \(\rho\) using Fisher's Z-transformation. This method transforms the correlation coefficient to stabilize the variance and make the confidence interval more accurate.

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
Fisher's Z-transformation is an important technique in statistics used to perform hypothesis testing and construct confidence intervals for correlation coefficients. It transforms the correlation coefficient into a variable that is approximately normal, which simplifies further statistical procedures.

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.
It’s critical in hypothesis testing related to correlation, as we saw with the test statistic \(z = \frac{z'_r - z'_\rho}{\sigma_{z'}}\) computed during the hypothesis test part of the task.
P-value Interpretation
The P-value in the context of hypothesis testing is a critical metric that helps determine the statistical significance of the results. It quantifies the probability of observing a result as extreme as, or more extreme than, the one obtained in our sample data, assuming the null hypothesis is true.

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.
Understanding P-value interpretation is essential in making informed decisions about the hypotheses tested, providing clarity about the results' significance.

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Most popular questions from this chapter

In an article in Statistics and Computing ["An Iterative Monte Carlo Method for Nonconjugate Bayesian Analysis" (1991, pp. \(119-128\) )], Carlin and Gelfand investigated the age \((x)\) and length \((y)\) of 27 captured dugongs (sea cows).$$\begin{aligned}x=& 1.0,1.5,1.5,1.5,2.5,4.0,5.0,5.0,7.0,8.0,8.5,9.0,9.5, \\\& 9.5,10.0,12.0,12.0,13.0,13.0,14.5,15.5,15.5,16.5 \\\& 17.0,22.5,29.0,31.5 \\\y=& 1.80,1.85,1.87,1.77,2.02,2.27,2.15,2.26,2.47,2.19, \\\& 2.26,2.40,2.39,2.41,2.50,2.32,2.32,2.43,2.47,2.56, \\\& 2.65,2.47,2.64,2.56,2.70,2.72,2.57\end{aligned}$$ (a) Find the least squares estimates of the slope and the intercept in the simple linear regression model. Find an estimate of \(\sigma^{2}\) (b) Estimate the mean length of dugongs at age 11 . (c) Obtain the fitted values \(\hat{y}_{i}\) that correspond to each observed value \(y_{i} .\) Plot \(\hat{y}_{i}\) versus \(y_{i},\) and comment on what this plot would look like if the linear relationship between length and age were perfectly deterministic (no error). Does this plot indicate that age is a reasonable choice of regressor variable in this model?

Consider the simple linear regression model \(y=\) \(10+25 x+\varepsilon\) where the random error term is normally and independently distributed with mean zero and standard deviation \(2 .\) Use software to generate a sample of eight observations, one each at the levels \(x=10,12,14,16,18,\) \(20,22,\) and 24 (a) Fit the linear regression model by least squares and find the estimates of the slope and intercept. (b) Find the estimate of \(\sigma^{2}\). (c) Find the standard errors of the slope and intercept. (d) Now use software to generate a sample of 16 observations, two each at the same levels of \(x\) used previously. Fit the model using least squares. (e) Find the estimate of \(\sigma^{2}\) for the new model in part (d). Compare this to the estimate obtained in part (b). What impact has the increase in sample size had on the estimate? (f) Find the standard errors of the slope and intercept using the new model from part (d). Compare these standard errors to the ones that you found in part (c). What impact has the increase in sample size had on the estimated standard errors?

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) \(\frac{1}{n} \sum_{i=1}^{n} \hat{\mathrm{y}}_{i}=\bar{y}\)

Consider the simple linear regression model \(y=10+\) \(30 x+\epsilon\) where the random error term is normally and independently distributed with mean zero and standard deviation \(1 .\) Use software to generate a sample of eight observations, one each at the levels \(x=10,12,14,16,18,20,22,\) and 24. (a) Fit the linear regression model by least squares and find the estimates of the slope and intercept. (b) Find the estimate of \(\sigma^{2}\). (c) Find the value of \(R^{2}\). (d) Now use software to generate a new sample of eight observations, one each at the levels of \(x=10,14,18,22,26,30\) \(34,\) and \(38 .\) Fit the model using least squares. (e) Find \(R^{2}\) for the new model in part (d). Compare this to the value obtained in part (c). What impact has the increase in the spread of the predictor variable \(x\) had on the value?

Determine if the following models are intrinsically linear. If yes, determine the appropriate transformation to generate the linear model. (a) \(Y=\beta_{0} x^{\beta_{1}} \epsilon\) (b) \(Y=\frac{3+5 x}{x}+\epsilon\) (c) \(Y=\beta_{0} \beta_{1}^{x} \epsilon\) (d) \(Y=\frac{x}{\beta_{0} x+\beta_{1}+x \epsilon}\)

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