Correlation Coefficient worksheet

1. Pearson correlation (r): 0.72

1. Find how far each x is from the x average, square each gap and add them up. This is Sxx

   $\mathrm{Sxx}$$=83.5$

2. Do the same for y: square each y's gap from the y average and add them up. This is Syy

   $\mathrm{Syy}$$=729.5$

3. For each pair, multiply the x gap by the y gap, then add them up. This is Sxy, and its sign shows the direction

   $\mathrm{Sxy}$$=177.5$

4. Divide Sxy by the square root of Sxx × Syy. This scales it to a number between −1 and 1, which is r

   $r$$=\frac{177.5}{\sqrt{83.5\times 729.5}}$$=0.719188$

5. Square r to get the share of variation the two variables have in common

   $\mathrm{r²}$$={0.719188}^2$$=0.517231$

6. Turn r into a t statistic so it can be tested. With n pairs, it has n − 2 = 6 degrees of freedom

   $t$$=0.719188\times \sqrt{\frac{6}{1-{0.719188}^2}}$$=2.535412$

7. The two-sided p-value is the chance of a |t| at least this large if there were no correlation

   $p$$=\text{2 × P(T >}2.5354\text{)}$$=0.04435454$

2. Pearson correlation (r): 0.67

1. Find how far each x is from the x average, square each gap and add them up. This is Sxx

   $\mathrm{Sxx}$$=98$

2. Do the same for y: square each y's gap from the y average and add them up. This is Syy

   $\mathrm{Syy}$$=308$

3. For each pair, multiply the x gap by the y gap, then add them up. This is Sxy, and its sign shows the direction

   $\mathrm{Sxy}$$=116$

4. Divide Sxy by the square root of Sxx × Syy. This scales it to a number between −1 and 1, which is r

   $r$$=\frac{116}{\sqrt{98\times 308}}$$=0.667682$

5. Square r to get the share of variation the two variables have in common

   $\mathrm{r²}$$={0.667682}^2$$=0.445799$

6. Turn r into a t statistic so it can be tested. With n pairs, it has n − 2 = 6 degrees of freedom

   $t$$=0.667682\times \sqrt{\frac{6}{1-{0.667682}^2}}$$=2.196907$

7. The two-sided p-value is the chance of a |t| at least this large if there were no correlation

   $p$$=\text{2 × P(T >}2.1969\text{)}$$=0.07040158$

3. Pearson correlation (r): 0.84

1. Find how far each x is from the x average, square each gap and add them up. This is Sxx

   $\mathrm{Sxx}$$=129.875$

2. Do the same for y: square each y's gap from the y average and add them up. This is Syy

   $\mathrm{Syy}$$=545.5$

3. For each pair, multiply the x gap by the y gap, then add them up. This is Sxy, and its sign shows the direction

   $\mathrm{Sxy}$$=223.25$

4. Divide Sxy by the square root of Sxx × Syy. This scales it to a number between −1 and 1, which is r

   $r$$=\frac{223.25}{\sqrt{129.875\times 545.5}}$$=0.838748$

5. Square r to get the share of variation the two variables have in common

   $\mathrm{r²}$$={0.838748}^2$$=0.703498$

6. Turn r into a t statistic so it can be tested. With n pairs, it has n − 2 = 6 degrees of freedom

   $t$$=0.838748\times \sqrt{\frac{6}{1-{0.838748}^2}}$$=3.773052$

7. The two-sided p-value is the chance of a |t| at least this large if there were no correlation

   $p$$=\text{2 × P(T >}3.7731\text{)}$$=0.0092555$

4. Pearson correlation (r): 0.92

1. Find how far each x is from the x average, square each gap and add them up. This is Sxx

   $\mathrm{Sxx}$$=59.5$

2. Do the same for y: square each y's gap from the y average and add them up. This is Syy

   $\mathrm{Syy}$$=722.875$

3. For each pair, multiply the x gap by the y gap, then add them up. This is Sxy, and its sign shows the direction

   $\mathrm{Sxy}$$=190.25$

4. Divide Sxy by the square root of Sxx × Syy. This scales it to a number between −1 and 1, which is r

   $r$$=\frac{190.25}{\sqrt{59.5\times 722.875}}$$=0.917349$

5. Square r to get the share of variation the two variables have in common

   $\mathrm{r²}$$={0.917349}^2$$=0.841529$

6. Turn r into a t statistic so it can be tested. With n pairs, it has n − 2 = 6 degrees of freedom

   $t$$=0.917349\times \sqrt{\frac{6}{1-{0.917349}^2}}$$=5.64463$

7. The two-sided p-value is the chance of a |t| at least this large if there were no correlation

   $p$$=\text{2 × P(T >}5.6446\text{)}$$=0.00132546$

5. Pearson correlation (r): 0.86

1. Find how far each x is from the x average, square each gap and add them up. This is Sxx

   $\mathrm{Sxx}$$=183.5$

2. Do the same for y: square each y's gap from the y average and add them up. This is Syy

   $\mathrm{Syy}$$=818$

3. For each pair, multiply the x gap by the y gap, then add them up. This is Sxy, and its sign shows the direction

   $\mathrm{Sxy}$$=332$

4. Divide Sxy by the square root of Sxx × Syy. This scales it to a number between −1 and 1, which is r

   $r$$=\frac{332}{\sqrt{183.5\times 818}}$$=0.856926$

5. Square r to get the share of variation the two variables have in common

   $\mathrm{r²}$$={0.856926}^2$$=0.734322$

6. Turn r into a t statistic so it can be tested. With n pairs, it has n − 2 = 6 degrees of freedom

   $t$$=0.856926\times \sqrt{\frac{6}{1-{0.856926}^2}}$$=4.072317$

7. The two-sided p-value is the chance of a |t| at least this large if there were no correlation

   $p$$=\text{2 × P(T >}4.0723\text{)}$$=0.00655866$

6. Pearson correlation (r): 0.87

1. Find how far each x is from the x average, square each gap and add them up. This is Sxx

   $\mathrm{Sxx}$$=98$

2. Do the same for y: square each y's gap from the y average and add them up. This is Syy

   $\mathrm{Syy}$$=350.875$

3. For each pair, multiply the x gap by the y gap, then add them up. This is Sxy, and its sign shows the direction

   $\mathrm{Sxy}$$=160.5$

4. Divide Sxy by the square root of Sxx × Syy. This scales it to a number between −1 and 1, which is r

   $r$$=\frac{160.5}{\sqrt{98\times 350.875}}$$=0.865537$

5. Square r to get the share of variation the two variables have in common

   $\mathrm{r²}$$={0.865537}^2$$=0.749155$

6. Turn r into a t statistic so it can be tested. With n pairs, it has n − 2 = 6 degrees of freedom

   $t$$=0.865537\times \sqrt{\frac{6}{1-{0.865537}^2}}$$=4.2331$

7. The two-sided p-value is the chance of a |t| at least this large if there were no correlation

   $p$$=\text{2 × P(T >}4.2331\text{)}$$=0.00548134$

7. Pearson correlation (r): 0.98

1. Find how far each x is from the x average, square each gap and add them up. This is Sxx

   $\mathrm{Sxx}$$=120.875$

2. Do the same for y: square each y's gap from the y average and add them up. This is Syy

   $\mathrm{Syy}$$=503.5$

3. For each pair, multiply the x gap by the y gap, then add them up. This is Sxy, and its sign shows the direction

   $\mathrm{Sxy}$$=240.75$

4. Divide Sxy by the square root of Sxx × Syy. This scales it to a number between −1 and 1, which is r

   $r$$=\frac{240.75}{\sqrt{120.875\times 503.5}}$$=0.975884$

5. Square r to get the share of variation the two variables have in common

   $\mathrm{r²}$$={0.975884}^2$$=0.95235$

6. Turn r into a t statistic so it can be tested. With n pairs, it has n − 2 = 6 degrees of freedom

   $t$$=0.975884\times \sqrt{\frac{6}{1-{0.975884}^2}}$$=10.950726$

7. The two-sided p-value is the chance of a |t| at least this large if there were no correlation

   $p$$=\text{2 × P(T >}10.9507\text{)}$$=0.00003443$

8. Pearson correlation (r): 0.72

1. Find how far each x is from the x average, square each gap and add them up. This is Sxx

   $\mathrm{Sxx}$$=106$

2. Do the same for y: square each y's gap from the y average and add them up. This is Syy

   $\mathrm{Syy}$$=390$

3. For each pair, multiply the x gap by the y gap, then add them up. This is Sxy, and its sign shows the direction

   $\mathrm{Sxy}$$=146$

4. Divide Sxy by the square root of Sxx × Syy. This scales it to a number between −1 and 1, which is r

   $r$$=\frac{146}{\sqrt{106\times 390}}$$=0.718071$

5. Square r to get the share of variation the two variables have in common

   $\mathrm{r²}$$={0.718071}^2$$=0.515627$

6. Turn r into a t statistic so it can be tested. With n pairs, it has n − 2 = 6 degrees of freedom

   $t$$=0.718071\times \sqrt{\frac{6}{1-{0.718071}^2}}$$=2.527278$

7. The two-sided p-value is the chance of a |t| at least this large if there were no correlation

   $p$$=\text{2 × P(T >}2.5273\text{)}$$=0.04484416$

9. Pearson correlation (r): 0.91

1. Find how far each x is from the x average, square each gap and add them up. This is Sxx

   $\mathrm{Sxx}$$=109.5$

2. Do the same for y: square each y's gap from the y average and add them up. This is Syy

   $\mathrm{Syy}$$=399.875$

3. For each pair, multiply the x gap by the y gap, then add them up. This is Sxy, and its sign shows the direction

   $\mathrm{Sxy}$$=190.25$

4. Divide Sxy by the square root of Sxx × Syy. This scales it to a number between −1 and 1, which is r

   $r$$=\frac{190.25}{\sqrt{109.5\times 399.875}}$$=0.909192$

5. Square r to get the share of variation the two variables have in common

   $\mathrm{r²}$$={0.909192}^2$$=0.82663$

6. Turn r into a t statistic so it can be tested. With n pairs, it has n − 2 = 6 degrees of freedom

   $t$$=0.909192\times \sqrt{\frac{6}{1-{0.909192}^2}}$$=5.348643$

7. The two-sided p-value is the chance of a |t| at least this large if there were no correlation

   $p$$=\text{2 × P(T >}5.3486\text{)}$$=0.00174686$

10. Pearson correlation (r): 0.61

1. Find how far each x is from the x average, square each gap and add them up. This is Sxx

   $\mathrm{Sxx}$$=98$

2. Do the same for y: square each y's gap from the y average and add them up. This is Syy

   $\mathrm{Syy}$$=839.5$

3. For each pair, multiply the x gap by the y gap, then add them up. This is Sxy, and its sign shows the direction

   $\mathrm{Sxy}$$=174$

4. Divide Sxy by the square root of Sxx × Syy. This scales it to a number between −1 and 1, which is r

   $r$$=\frac{174}{\sqrt{98\times 839.5}}$$=0.606633$

5. Square r to get the share of variation the two variables have in common

   $\mathrm{r²}$$={0.606633}^2$$=0.368003$

6. Turn r into a t statistic so it can be tested. With n pairs, it has n − 2 = 6 degrees of freedom

   $t$$=0.606633\times \sqrt{\frac{6}{1-{0.606633}^2}}$$=1.86915$

7. The two-sided p-value is the chance of a |t| at least this large if there were no correlation

   $p$$=\text{2 × P(T >}1.8691\text{)}$$=0.11080936$