Question:medium

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

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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.
Updated On: Feb 18, 2026
  • r $>$ 0.25
  • r $>$ 0.30
  • r $>$ 0.381
  • r $>$ 0.19
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The Correct Option is C

Solution and Explanation

Step 1: Concept Overview:
The problem seeks the critical value of the Pearson correlation coefficient, \(r\), given a sample size and significance level. This involves a t-test to assess the correlation coefficient's significance. The null hypothesis, \(H_0: \rho = 0\), assumes no population correlation, while the alternative hypothesis, \(H_1: \rho eq 0\), suggests otherwise. Significance occurs when the calculated t-statistic exceeds the critical t-value. We aim to determine the smallest \(r\) that satisfies this condition.

Step 2: Core Formula:
The t-statistic for correlation significance is calculated as: \[ t = \frac{r \sqrt{n-2}}{\sqrt{1-r^2}} \] Where: - \( r \) represents the sample correlation coefficient. - \( n \) represents the number of pairs in the sample. The degrees of freedom (df) are \(n-2\).

Step 3: Step-by-Step Solution:
We have: - Sample size: \( n = 27 \). - Significance level: \(\alpha = 0.05\). - Degrees of freedom: \( df = n - 2 = 27 - 2 = 25 \). - Critical t-value: \( t_{\text{critical}} = 2.06 \). Correlation significance testing is two-tailed; this value corresponds to \( t_{\alpha/2, df} = t_{0.025, 25} \). The notation \(t_{0.05}(25)\) might be slightly unclear, but 2.06 is the standard critical value for a two-tailed test with \(\alpha = 0.05\) and 25 df. To find the minimum significant \(r\), equate the calculated t-statistic to the critical t-value and solve for \(r\), considering \(|r|\) for both positive and negative correlations: \[ \frac{|r| \sqrt{27-2}}{\sqrt{1-r^2}} = 2.06 \] \[ \frac{|r| \sqrt{25}}{\sqrt{1-r^2}} = 2.06 \] \[ \frac{5|r|}{\sqrt{1-r^2}} = 2.06 \] Squaring 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 Result:
The correlation coefficient \(r\) is significant at the 5% level if its absolute value exceeds 0.3809. Hence, the minimum significant positive correlation is approximately 0.381. The condition is \( r>0.381 \).
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