Question

A sample of size 1600 is taken from a population of fathers and sons and the correlation between their heights is found to be 0.80. Then, the correlation limits for the entire population are:

• A. (0.573, 0.750)
• B. (0.773, 0.827)
• C. (0.8, 0.878)
• D. (0.573, 0.80)

Hint:

When dealing with correlation coefficients, remember that their sampling distribution is not normal, especially for values of \(r\) far from 0. Fisher's Z-transformation is the standard method to handle this. If a confidence level isn't given, check if using Z=2 (approx. 95%) or Z=3 (approx. 99.7%) matches one of the options.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To find the confidence limits for a population correlation coefficient (\(\rho\)), we use Fisher's Z-transformation. This method transforms the skewed sampling distribution of \(r\) into an approximately normal distribution, for which a standard confidence interval can be constructed. The limits of this interval are then transformed back to the original correlation scale.

Step 2: Key Formula or Approach:
1. Transform the sample correlation \(r\) to \(z_r\): \( z_r = \frac{1}{2} \ln \left( \frac{1+r}{1-r} \right) \). 2. Calculate the standard error of \(z_r\): \( SE_{z_r} = \frac{1}{\sqrt{n-3}} \). 3. Construct the confidence interval for the transformed correlation: \( z_r \pm Z_{\alpha/2} \times SE_{z_r} \). 4. Transform the interval limits back to the \(r\) scale using the inverse function: \( r = \frac{e^{2z}-1}{e^{2z}+1} \).
Step 3: Detailed Explanation:
Given values: - Sample size, \( n = 1600 \).
- Sample correlation, \( r = 0.80 \).
The confidence level is not specified. The term "correlation limits" often implies a high-confidence range. Let's try constructing a 99.7% confidence interval (which corresponds to \(Z_{\alpha/2} \approx 3\)), as this is a common interpretation for "limits".
1. Fisher's Z-transformation: \[ z_r = \frac{1}{2} \ln \left( \frac{1+0.80}{1-0.80} \right) = \frac{1}{2} \ln \left( \frac{1.8}{0.2} \right) = \frac{1}{2} \ln(9) \approx 0.5 \times 2.1972 = 1.0986 \] 2. Standard Error: \[ SE_{z_r} = \frac{1}{\sqrt{1600-3}} = \frac{1}{\sqrt{1597}} \approx \frac{1}{39.96} \approx 0.0250 \] 3. Confidence Interval for \(z_\rho\) (using Z=3): \[ \text{CI}_z = 1.0986 \pm 3 \times 0.0250 = 1.0986 \pm 0.075 \] - Lower z-limit: \( 1.0986 - 0.075 = 1.0236 \) - Upper z-limit: \( 1.0986 + 0.075 = 1.1736 \) 4. Inverse Transformation: - Lower r-limit: \( r_L = \frac{e^{2 \times 1.0236} - 1}{e^{2 \times 1.0236} + 1} = \frac{e^{2.0472} - 1}{e^{2.0472} + 1} = \frac{7.746 - 1}{7.746 + 1} = \frac{6.746}{8.746} \approx 0.7713 \) - Upper r-limit: \( r_U = \frac{e^{2 \times 1.1736} - 1}{e^{2 \times 1.1736} + 1} = \frac{e^{2.3472} - 1}{e^{2.3472} + 1} = \frac{10.456 - 1}{10.456 + 1} = \frac{9.456}{11.456} \approx 0.8254 \) The resulting limits are approximately (0.771, 0.825), which closely matches option (B).
Step 4: Final Answer:
The correlation limits for the entire population are approximately (0.773, 0.827).