HELP PLS!!Given the following data calculate the correlation coefficient. Then describe the correlation based on the correlation

Question

horsena [70]

3 years ago

13

HELP PLS!!Given the following data calculate the correlation coefficient. Then describe the correlation based on the correlation

coefficient.
x y
0 11
3 26
9 17
4 22
6 18

Mathematics

2 answers:

Y_Kistochka [10] · Answer 1 · 2022-03-06T20:23:47+03:00

I'm going to take the easy way, which is step by step, instead of trying to type out the formula for the correlation coefficient and then fill it in. It's way too complex. Step by step will be way simpler. First thing you have to do is find the mean of both x and y. The mean of x is 0+3+9+4+6 all divided by 5. That means that the mean of x is 4.4. For the mean of y: 11+26+17+22+18 all divided by 5. That means that the mean of y is 18.8. That's the first step. Second step is to find the standard deviation of x and y. That formula is $S_{x}=\sqrt{\frac{sum(x-mean(x))^2}{n-1}}$ . We would use the same formula to find the standard deviation for the y using the y coordinates and the mean of y. For the standard deviation for x we have $S_{x}=\sqrt{\frac{(0-4.4)^2+(3-4.4)^2+(9-4.4)^2+(4-4.4)^2+(6-4.4)^2}{5-1}}$ . Simplifying that down we have $S_{x}=\sqrt{\frac{19.36+1.96+21.16+.16+2.56}{4}}$ and $S_{x}=\sqrt{\frac{45.2}{4}}$ which means that the standard deviation for x is 3.36. Now for y: $S_{y}=\sqrt{\frac{(11-18.8)^2+(26-18.8)^2+(17-18.8)^2+(22-18.8)^2+(18-18.8)^2}{4}}$ which simplifies down to $S_{y}=\sqrt{\frac{60.84+51.84+3.24+10.24+.64}{4}}$ which means that the standard deviation for y is 5.63. Keep going...we're getting there. Now we need to take each coordinate pair x and y and multiply (x - mean(x))(y - mean(y)) and add those products together. $(0-4.4)(11-18.8)+(3-4.4)(26-18.8)+(9-4.4)(17-18.8)+(4-4.4)(22-18.8)+(6-4.4)(18-18.8)$ which simplifies down to $34.32-10.08-8.28-1.28-1.28=13.4$ . Now divide that 13.4 by the product of the 2 standard deviation values: (3.36)(5.63): $\frac{13.4}{18.9168}$ . That quotient is .7083650512. Lastly, divide that number by n-1 or 4. That means that the correlation coefficient is .177. A really good correlation coefficient is close to 1. So .998 or even .863 is a good correlation. Our decimal is only .1, so the data is all over the place and the equation you could find on your calculator using the regression program would not fit these points very well at all.