Question

Minimum value of the correlation coefficient 'r' in a sample of 27 pairs from a bivariate normal population, significant at 5% level, is: (Given \(t_{0.05} (25) = 2.06\))

• A. r $>$ 0.25
• B. r $>$ 0.30
• C. r $>$ 0.381
• D. r $>$ 0.19

Hint:

To solve for \(r\) from the t-statistic formula \( t = \frac{r\sqrt{df}}{\sqrt{1-r^2}} \), you can use the rearranged formula: \( |r| = \sqrt{\frac{t^2}{t^2 + df}} \). Using this: \( |r| = \sqrt{\frac{(2.06)^2}{(2.06)^2 + 25}} = \sqrt{\frac{4.2436}{4.2436 + 25}} = \sqrt{\frac{4.2436}{29.2436}} \approx 0.381 \). This can be a faster way to find the critical 'r' value.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The problem asks for the critical value of the Pearson correlation coefficient, 'r', for a given sample size and significance level. To find this, we use the t-test for the significance of the correlation coefficient. The null hypothesis is \(H_0: \rho = 0\) (no correlation in the population), and the alternative is \(H_1: \rho \neq 0\). 'r' is considered significant if the calculated test statistic exceeds the critical t-value. We need to find the minimum 'r' that makes this happen.

Step 2: Key Formula or Approach:
The t-statistic for testing the significance of a correlation coefficient is: \[ t = \frac{r \sqrt{n-2}}{\sqrt{1-r^2}} \] Where: - \( r \) is the sample correlation coefficient. - \( n \) is the number of pairs in the sample. The degrees of freedom (df) for this test is \(n-2\).

Step 3: Detailed Explanation:
We are given the following information: - Sample size, \( n = 27 \). - Significance level, \(\alpha = 0.05\). - Degrees of freedom, \( df = n - 2 = 27 - 2 = 25 \). - The critical t-value is given as \( t_{\text{critical}} = 2.06 \). Note that testing for significance of correlation is a two-tailed test, so this value corresponds to \( t_{\alpha/2, df} = t_{0.025, 25} \). The notation \(t_{0.05}(25)\) in the question is slightly ambiguous but 2.06 is the standard critical value for a two-tailed test at \(\alpha = 0.05\) with 25 df. To find the minimum significant value of \(r\), we set the calculated t-statistic equal to the critical t-value and solve for \(r\). We consider \(|r|\) because the correlation could be positive or negative. \[ \frac{|r| \sqrt{27-2}}{\sqrt{1-r^2}} = 2.06 \] \[ \frac{|r| \sqrt{25}}{\sqrt{1-r^2}} = 2.06 \] \[ \frac{5|r|}{\sqrt{1-r^2}} = 2.06 \] Now, solve for \(r\). Let's square both sides: \[ \frac{25r^2}{1-r^2} = (2.06)^2 \] \[ \frac{25r^2}{1-r^2} = 4.2436 \] \[ 25r^2 = 4.2436 (1-r^2) \] \[ 25r^2 = 4.2436 - 4.2436r^2 \] \[ 25r^2 + 4.2436r^2 = 4.2436 \] \[ 29.2436r^2 = 4.2436 \] \[ r^2 = \frac{4.2436}{29.2436} \approx 0.14511 \] \[ |r| = \sqrt{0.14511} \approx 0.3809 \]
Step 4: Final Answer:
The correlation coefficient 'r' is significant at the 5% level if its absolute value is greater than 0.3809. Therefore, the minimum value for a significant positive correlation is approximately 0.381. The condition is \( r>0.381 \).