Step 1: Understanding the Concept:
This question requires testing the statistical significance of a sample correlation coefficient (\( r \)) using a Student's t-test.
Key Formula or Approach:
The t-test statistic used to test the significance of a correlation coefficient is:
\[ t = \frac{r \sqrt{n-2}}{\sqrt{1-r^2}} \]
This statistic follows a t-distribution with \( n - 2 \) degrees of freedom under the null hypothesis \( H_0: \rho = 0 \).
Step 2: Detailed Explanation:
Let us perform the calculations:
- Given Data:
Sample size, \( n = 625 \implies n-2 = 623 \) degrees of freedom
Sample correlation coefficient, \( r = 0.2 \implies r^2 = 0.04 \)
Critical \( t \)-value at the \( 5% \) level, \( t_{\text{critical}} = 1.96 \)
- Calculate the test statistic (\( t \)):
\[ t = \frac{0.2 \sqrt{623}}{\sqrt{1 - 0.04}} \]
First, find the square roots:
\[ \sqrt{623} \approx 24.96 \]
\[ \sqrt{0.96} \approx 0.98 \]
Now, substitute these back into our equation:
\[ t = \frac{0.2 \times 24.96}{0.98} \approx \frac{4.992}{0.98} \approx 5.09 \]
- Compare with Critical Value:
Our calculated value is \( |t| = 5.09 \).
Comparing this to the critical value:
\[ |t_{\text{calculated}}| = 5.09 > t_{\text{critical}} = 1.96 \]
Since the calculated \( t \)-statistic is significantly larger than the critical value, we reject the null hypothesis.
Therefore, the correlation coefficient \( r = 0.2 \) is statistically significant at the \( 5% \) level.
Step 3: Final Answer:
Yes, the value of \( r \) is significant.
Therefore, the correct choice is Option (A).