Question

Testing for a Linear Correlation. In Exercises 13–28, 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

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

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

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.

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.

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.