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Testing for a Linear Correlation. In Exercises 13鈥28, construct a scatterplot, and find the value of the linear correlation coefficient r. Also find the P-value or the critical values of r from Table A-6. Use a significance level of A = 0.05. Determine whether there is sufficient evidence to support a claim of a linear correlation between the two variables. (Save your work because the same data sets will be used in Section 10-2 exercises.)

Old Faithful Listed below are duration times (seconds) and time intervals (min) to the next eruption for randomly selected eruptions of the Old Faithful geyser in Yellowstone National Park. Is there sufficient evidence to conclude that there is a linear correlation between duration times and interval after times?

Duration

242

255

227

251

262

207

140

Interval After

91

81

91

92

102

94

91

Short Answer

Expert verified

The scatter plot is:

The value of the correlation coefficient is 0.046.

The p-value is 0.921.

There is not enough evidence to support the claim that there is a linear correlation between the two variables.

Step by step solution

01

Given information

The data is recorded for two variables: duration in seconds and time intervals in minutes for the next eruption of a geyser.

Duration

Interval After

242

91

255

81

227

91

251

92

262

102

207

94

140

91

02

Sketch a scatterplot

A scatterplot is a graph oftwo variables that havepaired values. Each variable is scaled on one axis.

Steps to sketch a scatterplot:

  1. Mark two axes, xand y,for duration and interval after, respectively.
  2. Mark the paired data values on the graph corresponding to the axes.

The resultant graph is shown below.

03

Compute the measure of correlation coefficient

The formula for the correlation coefficient is

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

Let the duration be variable x and theinterval after be variable y.

The valuesare listedin the table below:

x

y

\({x^2}\)

\({y^2}\)

\(xy\)

242

91

58564

8281

22022

255

81

65025

6561

20655

227

91

51529

8281

20657

251

92

63001

8464

23092

262

102

68644

10404

26724

207

94

42849

8836

19458

140

91

19600

8281

12740

\(\sum x = 1584\)

\(\sum y = 642\)

\(\sum {{x^2}} = 369212\)

\(\sum {{y^2} = } \;59108\)

\(\sum {xy\; = \;} 145348\)

Substitute the values in the formula:

\(\begin{aligned} r &= \frac{{7\left( {145348} \right) - \left( {1584} \right)\left( {642} \right)}}{{\sqrt {7\left( {369212} \right) - {{\left( {1584} \right)}^2}} \sqrt {7\left( {59108} \right) - {{\left( {642} \right)}^2}} }}\\ &= 0.046\end{aligned}\)

Thus, the correlation coefficient is 0.046.

04

Step 4:Conduct a hypothesis test for correlation

Define\(\rho \)as the true measure ofthe correlation coefficient for the two variables.

For testing the claim, form the hypotheses:

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

The samplesize is7 (n).

The test statistic is computed as follows:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.046}}{{\sqrt {\frac{{1 - {{0.046}^2}}}{{7 - 2}}} }}\\ &= 0.103\end{aligned}\)

Thus, the test statistic is 0.103.

The degree of freedom is

\(\begin{aligned} df &= n - 2\\ &= 7 - 2\\ &= 5.\end{aligned}\)

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

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T > t} \right)\\ &= 2P\left( {T > 0.103} \right)\\ &= 2\left( {1 - P\left( {T < 0.103} \right)} \right)\\ &= 0.921\end{aligned}\)

Thus, the p-value is 0.921.

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

Therefore, there is not enough evidence to conclude that variables x(duration) and y (interval after) have a linear correlation between them.

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Tar

25

27

20

24

20

20

21

24

CO

18

16

16

16

16

16

14

17

Nicotine

1.5

1.7

1.1

1.6

1.1

1.0

1.2

1.4

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Age

2-11

12-17

18-24

25-39

40-49

50-59

60 and older

Percent

7

15

19

19

15

11

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