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Correlation
Pearson Correlation — Study Hours vs. GPA
Compute Pearson r between study hours and GPA. Includes t-test for significance and 95% CI using Fisher z-transformation.
📊 r(6) = .985, p < .001
Step-by-step solution
🔗 5 · Correlation
5a · Pearson Correlation: Study Hours vs GPA
Correlation
Research Question:
Is there a linear relationship between weekly study hours (X) and GPA (Y) for 8 students?
| i | X (hrs) | Y (GPA) | X−X̄ | Y−Ȳ | (X−X̄)(Y−Ȳ) | (X−X̄)² | (Y−Ȳ)² |
|---|---|---|---|---|---|---|---|
| 1 | 10 | 2.8 | -5 | -0.55 | 2.75 | 25 | 0.30 |
| 2 | 15 | 3.2 | 0 | -0.15 | 0 | 0 | 0.02 |
| 3 | 20 | 3.8 | 5 | 0.45 | 2.25 | 25 | 0.20 |
| 4 | 12 | 3.0 | -3 | -0.35 | 1.05 | 9 | 0.12 |
| 5 | 18 | 3.6 | 3 | 0.25 | 0.75 | 9 | 0.06 |
| 6 | 8 | 2.5 | -7 | -0.85 | 5.95 | 49 | 0.72 |
| 7 | 22 | 4.0 | 7 | 0.65 | 4.55 | 49 | 0.42 |
| 8 | 15 | 3.3 | 0 | -0.05 | 0 | 0 | 0.003 |
| Σ | 120 | 27.2 | 0 | 0 | 17.30 | 166 | 1.843 |
$$r = \frac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sqrt{\sum(x_i-\bar{x})^2 \cdot \sum(y_i-\bar{y})^2}} = \frac{17.30}{\sqrt{166 \times 1.843}} = \frac{17.30}{\sqrt{305.9}} = \frac{17.30}{17.49} = 0.989$$
Significance test
$$t = \frac{r\sqrt{n-2}}{\sqrt{1-r^2}} = \frac{0.989\sqrt{6}}{\sqrt{1-0.978}} = \frac{0.989 \times 2.449}{\sqrt{0.022}} = \frac{2.422}{0.148} = 16.36, \quad df=6$$
95% CI via Fisher-z
$$z_r = \tanh^{-1}(r) = \frac{1}{2}\ln\frac{1+r}{1-r} = \frac{1}{2}\ln\frac{1.989}{0.011} = 2.693$$
$$CI_{z}: 2.693 \pm \frac{1.96}{\sqrt{n-3}} = 2.693 \pm 0.981 \Rightarrow [1.712,\ 3.674] \Rightarrow r \in [.936,\ .998]$$
.989
r
.978
r²
16.36
t(6)
<.001
p
🔴 Strong positive correlation — Study hours are strongly associated with GPA, r=.989, p<.001.
APA-7
Study hours and GPA were strongly positively correlated, r(6) = .99, p < .001, 95% CI [.94, 1.00]. Study hours accounted for 97.8% of the variance in GPA.