Question

For the accompanying data set, draw a scatter diagram of the data.x 2 6 6 7 9y 3 2 6 9 5B. by hand compute the correlation coefficient. r = _____ (round to three decimals as needed.)C. Fill in the blanks: Because the correlation coefficient is (positive or negative) and the absolute value of the correlation coefficient, ____, is (greater or not greater) than the critical value for this data set,___, (no, a positive, or a negative) linear relation exists between x and y. (round to three decimal places as needed.)Critical values for the correlation coefficient tablen 3 0.9974 0.9505 0.8786 0.8117 0.7548 0.7079 0.66610 0.63211 0.60212 0.57613 0.55314 0.53215 0.51416 0.49717 0.48218 0.46819 0.45620 0.44421 0.43322 0.42323 0.41324 0.40425 0.39626 0.38827 0.38128 0.37429 0.36730 0.361

Answer 1

Answer:

Step-by-step explanation:

Hello!

a.

Given the data corresponding the variables:

x 2 6 6 7 9

y 3 2 6 9 5

The Scatterplot is attached.

b.

To compute the correlation coefficient you need several auxiliary calculations:

∑X= 30

∑X²= 206

∑Y= 25

∑Y²= 155

∑XY= 162

$r= \frac{sumXY-\frac{(sumX)(sumY)}{n} }{\sqrt{[sumX^2-\frac{(sumX)^2}{n} ][sumY^2-\frac{(sumY)^2}{n} ]} }$

$r= \frac{162-\frac{(30)*(25)}{5} }{\sqrt{[206-\frac{(30)^2}{5} ][155-\frac{(25)^2}{5} ]} }$

r= 0.429 ≅ 0.43

c.

The critical value for r has n-2 degrees of freedom, let's say for example you have α:0.05

$r_{n-2;\alpha } = r_{3;0.05 } = 0.878$

For a two-tailed test.

Because the correlation coefficient is positive and the absolute value of the correlation coefficient, _0.43___, is not greater than the critical value for this data set,_0.878__, no linear relationship exists between x and y.

I hope it helps!