Question

Find the correlation coefficient between age in years (x) and glucose level (y) from the data of 5 people as follows.

Answer 1

0.621

Answer 2

The correlation coefficient (Pearson's r) between two variables x and y for a sample of n data points is given by the formula:

$r = \frac{n \sum (xy) - (\sum x)(\sum y)}{\sqrt{[n \sum x^2 - (\sum x)^2][n \sum y^2 - (\sum y)^2]}}$

Given data:

n = 5

x: 43, 22, 25, 42, 58

y: 99, 65, 79, 75, 87

1. Calculate the sum of x values ($\sum x$), sum of y values ($\sum y$), sum of products of x and y values ($\sum xy$), sum of squares of x values ($\sum x^2$), and sum of squares of y values ($\sum y^2$).

   $\sum x = 43 + 22 + 25 + 42 + 58 = 190$

   $\sum y = 99 + 65 + 79 + 75 + 87 = 405$

   $\sum xy = (43 \times 99) + (22 \times 65) + (25 \times 79) + (42 \times 75) + (58 \times 87) = 4257 + 1430 + 1975 + 3150 + 5046 = 15858$

   $\sum x^2 = 43^2 + 22^2 + 25^2 + 42^2 + 58^2 = 1849 + 484 + 625 + 1764 + 3364 = 8086$

   $\sum y^2 = 99^2 + 65^2 + 79^2 + 75^2 + 87^2 = 9801 + 4225 + 6241 + 5625 + 7569 = 33461$

2. Use the formula for the sample correlation coefficient: $r = \frac{n \sum (xy) - (\sum x)(\sum y)}{\sqrt{[n \sum x^2 - (\sum x)^2][n \sum y^2 - (\sum y)^2]}}$, where n is the number of data points.

3. Substitute the calculated sums into the formula and compute the value of r.

   $r = \frac{5 \times 15858 - (190)(405)}{\sqrt{[5 \times 8086 - (190)^2][5 \times 33461 - (405)^2]}}$

   $r = \frac{79290 - 76950}{\sqrt{[40430 - 36100][167305 - 164025]}}$

   $r = \frac{2340}{\sqrt{4330 \times 3280}} = \frac{2340}{\sqrt{14202400}} \approx \frac{2340}{3768.619} \approx 0.62087$

Rounding to three decimal places, we get 0.621.