Q.1. The grades of class of 16 students in a class on a midterm (x) and on the final examination (y) are as follows: X 74 59 68 75 40 73 56 76 72 81 94 95 98 67 56 53 Y 81 65 64 64 46 75 65 78 58 67 85 97 96 68 60 67 (a) Find the linear regression line and interpret the coefficients. (b) Find the correlation coefficient and interpret it.

Given $n=16$ data points:
$\sum x = 1187$
$\sum y = 1096$
$\sum x^2 = 89691$
$\sum y^2 = 89444$
$\sum xy = 83547$

Mean of x: $\bar{x} = \frac{\sum x}{n} = \frac{1187}{16} = 74.1875$
Mean of y: $\bar{y} = \frac{\sum y}{n} = \frac{1096}{16} = 68.5$

Sum of squares for x:

$$SS_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 89691 - \frac{1187^2}{16} = 89691 - 88060.5625 = 1630.4375$$

Sum of products of x and y:

$$SS_{xy} = \sum xy - \frac{(\sum x)(\sum y)}{n} = 83547 - \frac{1187 \cdot 1096}{16} = 83547 - 81314.5 = 2232.5$$

The linear regression line is $\hat{y} = b_0 + b_1 x$.

$$\boxed{\hat{y} = -33.0696 + 1.3693x}$$

Interpretation of Coefficients:

• Slope ($b_1 = 1.3693$): For every one-point increase in the midterm grade (x), the final examination grade (y) is predicted to increase by approximately 1.3693 points.
• Y-intercept ($b_0 = -33.0696$): When the midterm grade (x) is 0, the predicted final examination grade (y) is -33.0696. This value is not meaningful in this context as grades cannot be negative and a midterm grade of 0 is outside the observed range of data.

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$$r = \frac{SS_{xy}}{\sqrt{SS_{xx} \cdot SS_{yy}}} = \frac{2232.5}{\sqrt{1630.4375 \cdot 14368}} = \frac{2232.5}{\sqrt{23429300}} = \frac{2232.5}{4840.3822} \approx 0.4612$$ $$\boxed{r = 0.4612}$$

Interpretation of the Correlation Coefficient:
The correlation coefficient $r = 0.4612$ indicates a moderate positive linear relationship between midterm grades and final examination grades. This means that as midterm grades increase, final examination grades tend to increase, but the relationship is not very strong.