91Ó°ÊÓ

In statistics, a **bivariate normal distribution** describes the relationship between two variable distributions that are both normally distributed and correlated. For example, in the given exercise, the rates of water flow at two different points, A and B, are assumed to follow a bivariate normal distribution. This assumption is crucial because it allows us to use certain statistical methods, like the Fisher transformation, to make accurate inferences about the correlation between the two variables.

A bivariate normal distribution is characterized by:

• Two means: \(\text{E}(Y_{1})\) and \(\text{E}(Y_{2})\)
• Two variances: \(\text{Var}(Y_{1})\) and \(\text{Var}(Y_{2})\)
• The correlation coefficient \(\rho_{12}\), which measures the linear relationship between the two variables

In this exercise, we are focusing on estimating the correlation coefficient (\(\rho_{12}\)) and its confidence interval, assuming the joint normality of the water flow rates at points A and B. Understanding the bivariate normal distribution is the first step in solving problems involving the correlation of two variables.

The **Fisher transformation** is an important tool when dealing with correlation coefficients. It converts the sample correlation coefficient (\(r\)) into a variable (\(z\)) that follows a normal distribution, making it easier to calculate confidence intervals and perform hypothesis testing.

The formula for the Fisher transformation is:
\[ z = \frac{1}{2} \text{ln} \frac{(1 + r)}{(1 - r)} \]
For instance, in the given exercise, the sample correlation coefficient \(r_{12}\) is 0.83. Applying the Fisher transformation:
\[ z = \frac{1}{2} \text{ln} \frac{(1 + 0.83)}{(1 - 0.83)} = \frac{1}{2} \text{ln} \frac{1.83}{0.17} \]
\[ z \approx 1.188 \]

By transforming \(r\) to \(z\), we can step into the normal distribution world, which simplifies the statistical calculations. This transformation helps in making the mathematics behind correlations much more manageable and reliable, particularly when dealing with large sample sizes like in our exercise.

Calculating the **standard error** helps in understanding the variability of our sample estimate. In simpler terms, it tells us how much the calculated Fisher-transformed value (\(z\)) might vary if we repeated the experiment multiple times.

For the Fisher transformation, the standard error (SE) of \(z\) is:
\[ \text{SE} = \frac{1}{\sqrt{n - 3}} \]
where \(n\) is the sample size. In the exercise, the sample size \(n\) is 147. Plugging in the numbers:
\[ \text{SE} = \frac{1}{\sqrt{147 - 3}} = \frac{1}{12} \approx 0.0833 \]

Knowing the standard error is essential for constructing confidence intervals. It quantifies the range within which we can expect the true correlation coefficient to lie, based on our sample data. This calculation gives us a sense of how precise our sample estimate is.

After finding the confidence interval for the Fisher-transformed value (\(z\)), we need to convert it back to the original correlation coefficient (\(r\)) using the **inverse Fisher transformation**. This step is crucial for interpreting the results in the original context of correlation.

The formula for the inverse Fisher transformation is:
\[ r = \frac{e^{2z} - 1}{e^{2z} + 1} \]

For example, if the lower bound of \(z\) is 0.973, we compute:
\[ r_{\text{lower}} = \frac{e^{1.946} - 1}{e^{1.946} + 1} \approx 0.75 \]

Similarly, if the upper bound of \(z\) is 1.403, we get:
\[ r_{\text{upper}} = \frac{e^{2.806} - 1}{e^{2.806} + 1} \approx 0.886 \]

Thus, the confidence interval for the correlation coefficient \(r\) is [0.75, 0.886]. This process allows us to translate the statistical calculations back into the real-world variable relationships, making the results meaningful and interpretable.