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.)

22. Crickets and Temperature A classic application of correlation involves the association between the temperature and the number of times a cricket chirps in a minute. Listed below are the numbers of chirps in 1 min and the corresponding temperatures in °F (based on data from The Song of Insects, by George W. Pierce, Harvard University Press). Is there sufficient evidence to conclude that there is a linear correlation between the number of chirps in 1 min and the temperature?

Actress	28	30	29	61	32	33	45	29	62	22	44	54
Actor	43	37	38	45	50	48	60	50	39	55	44	33

Short answer

The scatterplot is shown below:

The value of the correlation coefficient is 0.874.

The p-value is 0.005.

There is enough evidence to support the claim for a linear correlation between chirps in one minute and temperature.

Step-by-step solution

Given information

The data is recorded forchirps of crickets and temperatures in degrees Fahrenheit.

Chirps in 1 min	Temperature
882	69.7
1188	93.3
1104	84.3
864	76.3
1200	88.6
1032	82.6
960	71.6
900	79.6

Sketch a scatterplot

Scatterplot projects a paired set of observationsontwo axes scaled for the two variables.

Steps to sketch a scatterplot:

1. Describe two axes, x and y, for chirps in 1 minute and temperature, respectively.
2. Mark the points on the graph.

The graph is shown below.

Compute the measure of the correlation coefficient

The correlation coefficient formula is

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

Describe variables x and y as chirps in 1 minute and temperature, respectively.

The valuesare listed in the table below:

x	y	\({x^2}\)	\({y^2}\)	\(xy\)
882	69.7	777924	4858.09	61475.4
1188	93.3	1411344	8704.89	110840.4
1104	84.3	1218816	7106.49	93067.2
864	76.3	746496	5821.69	65923.2
1200	88.6	1440000	7849.96	106320
1032	82.6	1065024	6822.76	85243.2
960	71.6	921600	5126.56	68736
900	79.6	810000	6336.16	71640
\(\sum x = 8130\)	\(\sum y = 646\)	\(\sum {{x^2}} = 8391204\)	\(\sum {{y^2} = } \;52626.6\)	\(\sum {xy\; = \;} 663245.4\)

Substitute the values in the formula:

\(\begin{aligned} r &= \frac{{8\left( {663245.4} \right) - \left( {8130} \right)\left( {646} \right)}}{{\sqrt {8\left( {8391204} \right) - {{\left( {8130} \right)}^2}} \sqrt {8\left( {52626.6} \right) - {{\left( {646} \right)}^2}} }}\\ &= 0.874\end{aligned}\)

Thus, the correlation coefficient is 0.874.

Step 4:Conduct a hypothesis test for correlation

Definethe actual measure of the correlation coefficient between chirps and temperature as\(\rho \).

For testing the claim, form the hypotheses:

\(\begin{array}{l}{H_o}:\rho = 0\\{H_a}:\rho \ne 0\end{array}\)

The samplesize is 8(n).

The test statistic is computed as follows:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.874}}{{\sqrt {\frac{{1 - {{\left( {0.874} \right)}^2}}}{{8 - 2}}} }}\\ &= 4.406\end{aligned}\)

Thus, the test statistic is 4.406.

The degree of freedom is

\(\begin{aligned} df &= n - 2\\ &= 8 - 2\\ &= 6.\end{aligned}\)

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

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {t > 4.406} \right)\\ &= 0.0045\\ &\approx 0.005\end{aligned}\)

Thus, the p-value is 0.005.

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

Therefore, there is enough evidence to conclude a linear correlation between chirps in 1 minute and temperature.