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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

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🔗 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, d f = 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

C I_{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

.978

r²

16.36

t(6)

<.001

🔴 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.

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Pearson Correlation — Study Hours vs. GPA

🔗 5 · Correlation

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