Mathematics, 21.02.2020 02:31, evazquez

(a) For the accompanying data set, draw a scatter diagram of the data.
x 2 6 6 7 9
y 3 2 6 9 5

(b) 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.

Answers: 3

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answered: queenz58

a) n=5 $\sum x = 30, \sum y = 25, \sum xy = 162, \sum x^2 =206, \sum y^2 =155$

$r=\frac{5(162)-(30)(25)}{\sqrt{[5(206) -(30)^2][5(155) -(25)^2]}}=0.42967$

So then the correlation coefficient would be r =0.430 rounded

b) Null hypothesis: $\rho =0$

Alternative hypothesis: $\rho \neq 0$

Because the correlation coeffcient is positive and the absolute value of the correlation coefficient 0.430 is not greater than the critical value for this dataset, no linear relation exists between x and y.

And the reason is because we fail to reject the null hypothesis.

Step-by-step explanation:

Part a

We have the following data:

x: 2 6 6 7 9

y: 3 2 6 9 5

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

And in order to calculate the correlation coefficient we can use this formula:

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

For our case we have this:

n=5 $\sum x = 30, \sum y = 25, \sum xy = 162, \sum x^2 =206, \sum y^2 =155$

$r=\frac{5(162)-(30)(25)}{\sqrt{[5(206) -(30)^2][5(155) -(25)^2]}}=0.42967$

So then the correlation coefficient would be r =0.430 rounded

Part b

In order to test the hypothesis if the correlation coefficient it's significant we have the following hypothesis:

Null hypothesis: $\rho =0$

Alternative hypothesis: $\rho \neq 0$

The statistic to check the hypothesis is given by:

$t=\frac{r \sqrt{n-2}}{\sqrt{1-r^2}}$

And is distributed with n-2 degreed of freedom. df=n-2=5-2=3

In our case the value for the statistic would be:

$t=\frac{0.430\sqrt{5-2}}{\sqrt{1-(0.430)^2}}=0.825$

The critical value for n =5 is given by the table attached. We can see that the critical value is $r_{crit}= 0.878$ , and then the final conclusion would be: