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Explore! Exercises 9 and 10 provide two data sets from 鈥淕raphs in Statistical Analysis,鈥 by F. J. Anscombe, the American Statistician, Vol. 27. For each exercise,

a. Construct a scatterplot.

b. Find the value of the linear correlation coefficient r, then determine whether there is sufficient evidence to support the claim of a linear correlation between the two variables.

c. Identify the feature of the data that would be missed if part (b) was completed without constructing the scatterplot.

x

10

8

13

9

11

14

6

4

12

7

5

y

9.14

8.14

8.74

8.77

9.26

8.10

6.13

3.10

9.13

7.26

4.74

Short Answer

Expert verified

a. The scatterplot is obtainedbelow:

b. The value of the correlation coefficient is 0.8163. There is enough evidence to support the claim that there is a linear correlation between the two variables.

c. The scatterplot shows that the data consists of an outlier thatdeviates the measure of the correlation coefficient toa large extent.

Step by step solution

01

Given information

The samples size is11(n).

The data for two variables is shown.

x

y

10

7.46

8

6.77

13

12.74

9

7.11

11

7.81

14

8.84

6

6.08

4

5.39

12

8.15

7

6.42

5

5.73

02

Sketch a scatterplot

a.

When the data is visualized on a graph in paired form, it is referred to as a scatterplot.Here, one axis represents the values of x, and the other axis represents the values of y.

Steps to sketch a scatterplot:

  1. Make two axes, x and y, for each of the two variables.
  2. Map each paired value corresponding to the scale of the axes.
  3. Thus, a scatter plot for the paired data is obtained.

03

Compute the measure of the correlation coefficient

b.

The formula for the correlation coefficient is given 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 listed in the following table:

x

y

\({x^2}\)

\({y^2}\)

\(xy\)

10

7.46

100

55.6516

74.6

8

6.77

64

45.8329

54.16

13

12.74

169

162.3076

165.62

9

7.11

81

50.5521

63.99

11

7.81

121

60.9961

85.91

14

8.84

196

78.1456

123.76

6

6.08

36

36.9664

36.48

4

5.39

16

29.0521

21.56

12

8.15

144

66.4225

97.8

7

6.42

49

41.2164

44.94

5

5.73

25

32.8329

28.65

\(\sum x = 99\)

\(\sum y = 82.5\)

\(\sum {{x^2}} = 1001\)

\(\sum {{y^2} = } \;659.9762\)

\(\sum {xy\; = \;} 797.47\)

Substitute the values to obtain the value of r.

\(\begin{aligned} r &= \frac{{11\left( {797.47} \right) - \left( {99} \right)\left( {82.50} \right)}}{{\sqrt {11\left( {1001} \right) - {{\left( {99} \right)}^2}} \sqrt {11{{\left( {659.9762} \right)}^2} - {{\left( {82.5} \right)}^2}} }}\\ &= 0.8163\end{aligned}\)

Thus, the correlation coefficient is 0.8163.

04

Step 4:Conduct a hypothesis test for correlation

The statistical hypotheses are formulated below:

\(\begin{array}{l}{{\rm{H}}_{\rm{o}}}:\rho = 0\\{{\rm{{\rm H}}}_{\rm{a}}}:\rho \ne 0\end{array}\)

Here,\(\rho \)isthe actual measure ofthe correlation coefficientfor the variables.

Calculate the test statistic as shown below:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.8163}}{{\sqrt {\frac{{1 - {{0.8163}^2}}}{{11 - 2}}} }}\\ &= 4.239\end{aligned}\)

Thus, the value of the test statistic is 4.239.

The degree of freedom is computedbelow:

\(\begin{aligned} df &= n - 2\\ &= 11 - 2\\ &= 9\end{aligned}\)

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

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T > 4.239} \right)\\ &= 2\left( {1 - P\left( {T < 4.239} \right)} \right)\\ &= 0.002\end{aligned}\)

Since the p-value is lesser than the significance level, the null hypothesis is rejected.

Therefore, there is sufficient evidence to support the existence of a linear correlation between two variables.

05

Analyze the importance of the scatterplot

c.

The scatterplot for the data reveals that one point lies at an extreme,beyond the straight-line pattern established by other points in the data. Resultant ofthis, the correlation measure, which was expected to be close to 1, is lower.

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