Question

814 the table below shows the number of hours ten students spent studying for a test and their scores.
hours spent studying (x) | 0 | 1 | 2 | 4 | 4 | 4 | 6 | 6 | 7 | 8
test scores (y) | 35 | 40 | 46 | 65 | 67 | 70 | 82 | 88 | 82 | 95
write the linear regression equation for this data set. round all values to the nearest hundredth. state the correlation coefficient of this line, to the nearest hundredth. explain what the correlation coefficient suggests in the context of the problem.

Explanation:

Step1: Calculate necessary sums

$n=10$, $\sum x=0+1+2+4+4+4+6+6+7+8=42$, $\sum y=35+40+46+65+67+70+82+88+82+95=670$, $\sum xy=0×35+1×40+2×46+4×65+4×67+4×70+6×82+6×88+7×82+8×95=3204$, $\sum x^2=0^2+1^2+2^2+4^2×3+6^2×2+7^2+8^2=234$, $\sum y^2=35^2+40^2+46^2+65^2+67^2+70^2+82^2×2+88^2+95^2=47388$

Step2: Compute slope $m$

$m=\frac{n\sum xy - \sum x\sum y}{n\sum x^2 - (\sum x)^2}=\frac{10×3204 - 42×670}{10×234 - 42^2}=\frac{32040 - 28140}{2340 - 1764}=\frac{3900}{576}≈6.77$

Step3: Compute y-intercept $b$

$b=\frac{\sum y - m\sum x}{n}=\frac{670 - 6.77×42}{10}=\frac{670 - 284.34}{10}≈38.57$

Step4: Compute correlation coefficient $r$

$r=\frac{n\sum xy - \sum x\sum y}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}}=\frac{3900}{\sqrt{576×(10×47388 - 670^2)}}=\frac{3900}{\sqrt{576×(473880 - 448900)}}=\frac{3900}{\sqrt{576×24980}}≈\frac{3900}{3798.4}≈0.99$

The correlation coefficient $0.99$ suggests a strong positive linear relationship between hours spent studying and test scores—more study hours are strongly associated with higher scores.

Answer:

Linear regression equation: $y≈6.77x + 38.57$
Correlation coefficient: $0.99$