Question

nine students obtained the following percentage of marks in the college internal test (x) and in the final university examination (y). Find the correlation coefficient between the marks of the two test. x 51 63 73 46 50 60 47 36 60 y 49 72 74 45 58 66 50 30 35

Answer 1

n = 9

We can determine the covarience using the formula:

$s_{xy} = \frac{\sum x_iy_i - \frac{\sum x_i \times \sum y_i}{n}}{n-1} \\ = \frac{26897 - \frac{486 \times 479}{9}}{9-1} \\ = \frac{26897 - 25866}{8} \\ = 128.87$

Let us next determine the sample variance $s^2$ using the formula:

$s^2 = \frac{\sum x^2_i - \frac{\sum x^2_i}{n}}{n-1} \\ s^2_x = \frac{27220 - \frac{486^2}{9}}{9-1} \\ = \frac{27220 -26244}{8} \\ = 122.0 \\ s^2_y = \frac{27431 - \frac{479^2}{9}}{9-1} \\ = \frac{27431 -25493.44}{8} \\ = 242.19$

The sample standard deviation is the square root of the population sample:

$s_x = \sqrt{122.0} = 11.04 \\ s_y = \sqrt{242.19} = 15.56$

We can determine the correlation coefficient r using the formula:

$r = \frac{s_{xy}}{s_xs_y} \\ = \frac{128.87}{11.04 \times 15.56} \\ = 0.75$