Question

Outlier Refer to the accompanying Minitab-generated scatterplot. a. Examine the pattern of all 10 points and subjectively determine whether there appears to be a correlation between x and y. b. After identifying the 10 pairs of coordinates corresponding to the 10 points, find the value of the correlation coefficient r and determine whether there is a linear correlation. c. Now remove the point with coordinates (10, 10) and repeat parts (a) and (b). d. What do you conclude about the possible effect from a single pair of values?

Short answer

a. The pattern indicates an upward trend.Hence, a correlation can be expected for variables x and y.

b. The values are recorded as shown below:

x	y
1	1
2	1
3	1
1	2
1	3
2	2
2	3
3	2
3	3
10	10

The correlation coefficient is 0.906.

Also, there is sufficient evidence to support the existence ofa linear correlation between the two variables.

c. The correlation coefficient is 0, and there is insufficient evidence to support the claim that there is a linear correlation between the two variables.

d. A single pair of values has a substantial effect on the correlation measure.

Step-by-step solution

Given information

The scatterplot generated on Minitab is given.

Analyze the scatterplot

a.

A scatterplot is a two-dimensional graph thatrepresents a pair of values for two variables.

Here,the scatterplot represents an overall upward pattern, which means the values of one variable are expected to increase with the other.

Due to this pattern and moderately close observations, it can be expected that there exists a linear correlation between the two variables.

Compute the measure of the correlation coefficient

b.

The observations obtained from the scatterplot are as follows:

x	y
1	1
2	1
3	1
1	2
1	3
2	2
2	3
3	2
3	3
10	10

The formula for the correlation coefficient is shown below:

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

The valuesare in the table below:

x	y	$x^2$	$y^2$	$xy$
1	1	1	1	1
2	1	4	1	2
3	1	9	1	3
1	2	1	4	2
1	3	1	9	3
2	2	4	4	4
2	3	4	9	6
3	2	9	4	6
3	3	9	9	9
10	10	100	100	100
$\sum x=28$	$\sum y=28$	$\sum x^2=142$	$\sum y^2=142$	$\sum xy=136$

Substitute the values to obtain the correlation coefficient.

$\begin{array}{rl}r& =\frac{10\left(136\right)-\left(28\right)\left(28\right)}{\sqrt{10\left(142\right)-{\left(28\right)}^2}\sqrt{10{\left(142\right)}^2-{\left(28\right)}^2}}\\ & =0.906\end{array}$

Thus, the correlation coefficient is 0.906.

Step 4:Conduct a hypothesis test for correlation

Let$\rho$be the true correlation coefficient.

Form the hypotheses as shown:

$\begin{array}{l}{\mathrm{H}}_{\mathrm{o}}:\rho =0\\ {\mathrm{H}}_{\mathrm{a}}:\rho \ne 0\end{array}$

The samplesize is10(n).

The test statistic is computed as follows:

$\begin{array}{rl}t& =\frac{r}{\sqrt{\frac{1-r^2}{n-2}}}\\ & =\frac{0.906}{\sqrt{\frac{1-{0.906}^2}{10-2}}}\\ & =6.054\end{array}$

Thus, the test statistic is 6.054.

The degree of freedom is computedbelow:

$\begin{array}{rl}df& =n-2\\ & =10-2\\ & =8\end{array}$

The p-value is computedfrom the t-distribution table.

$\begin{array}{rl}p-\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{u}\mathrm{e}& =2P\left(T>t\right)\\ & =2P\left(T>6.054\right)\\ & =2\left(1-P\left(T<6.054\right)\right)\\ & =0.0003\end{array}$

As thep-value is lesser than 0.05, the null hypothesis is rejected.

Therefore, there is sufficient evidence to prove theexistence of a linear correlation between thetwo variables.

Analyze the scatterplot after removing the coordinates (10,10)

c.

The data without coordinate (10,10) is

x	y
1	1
2	1
3	1
1	2
1	3
2	2
2	3
3	2
3	3

The scatterplot hence formed is shown below:

Thus, there appears to be an association between the two variables.

Compute the correlation coefficient

The values are in the table below:

x	y	$x^2$	$y^2$	$xy$
1	1	1	1	1
2	1	4	1	2
3	1	9	1	3
1	2	1	4	2
1	3	1	9	3
2	2	4	4	4
2	3	4	9	6
3	2	9	4	6
3	3	9	9	9
$\sum x=18$	$\sum y=18$	$\sum x^2=42$	$\sum y^2=42$	$\sum xy=36$

Substitute the values to obtain the correlation coefficient:

$\begin{array}{rl}r& =\frac{9\left(36\right)-\left(18\right)\left(18\right)}{\sqrt{9\left(42\right)-{\left(18\right)}^2}\sqrt{9{\left(42\right)}^2-{\left(18\right)}^2}}\\ & =0\end{array}$

Thus, the correlation coefficient is 0.

Conduct a hypothesis test for correlation

Let$\rho$denote the actual correlation coefficient.

The hypotheses areformulatedas shown below

$\begin{array}{l}{\mathrm{H}}_{\mathrm{o}}:\rho =0\\ {\mathrm{H}}_{\mathrm{a}}:\rho \ne 0\end{array}$

The samplesize is9(n).

The test statistic is computed as follows:

$\begin{array}{rl}t& =\frac{r}{\sqrt{\frac{1-r^2}{n-2}}}\\ & =\frac{0}{\sqrt{\frac{1-0^2}{9-2}}}\\ & =0\end{array}$

Thus, the test statistic is 0.

The degree of freedom is computedbelow:

$\begin{array}{rl}df& =n-2\\ & =9-2\\ & =7\end{array}$

The p-value is computed from the t-distribution table.

$\begin{array}{rl}p-value& =2P\left(t>0\right)\\ & =2\left(1-P\left(t<0\right)\right)\\ & =1\end{array}$

As the p-value is greater than 0.05, the null hypothesis fails to be rejected.

Therefore, there is not sufficient evidence to support the claim that the variables have a linear correlation between them.

Discuss the effect of a single pair of values

The result changes to a large extent as one single paired observation is removed from the data. The correlation measure changes from 0.906 to 0 as one pair is removed from the data.