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.)

Pizza and the Subway The “pizza connection” is the principle that the price of a slice of pizza in New York City is always about the same as the subway fare. Use the data listed below to determine whether there is a significant linear correlation between the cost of a slice of pizza and the subway fare.

Year	1960	1973	1986	1995	2002	2003	2009	2013	2015
Pizza Cost	0.15	0.35	1	1.25	1.75	2	2.25	2.3	2.75
Subway Fare	0.15	0.35	1	1.35	1.5	2	2.25	2.5	2.75
CPI	30.2	48.3	112.3	162.2	191.9	197.8	214.5	233	237.2

Short answer

The scatter plot is shown below:

The value ofthe correlation coefficient is 0.992.

The p-value is 0.000.

There is enough evidence to support the claim that there is a linear correlation between the two variables(pizza cost and subway fare).

Step-by-step solution

Given information

Association between two variables, pizza cost and subway fare,isbeing studied.

Pizza Cost	Subway Fare
0.15	0.15
0.35	0.35
1	1
1.25	1.35
1.75	1.5
2	2
2.25	2.25
2.3	2.5
2.75	2.75

Sketch a scatterplot

A plot that shows observations from two variablesby scaling them on two axes is referred to as a scatterplot.

Steps to sketch a scatterplot:

1. Mark horizontal axis for price cost and vertical axis for subway fare.
2. Mark points for each paired value with respect to both axes.

The resultant graph is the required scatterplot.

Compute the measure of the 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 pizza cost be defined by variablex and subway fare be defined by variabley.

The valuesare listed in the table below:

x	y	\({x^2}\)	\({y^2}\)	\(xy\)
0.15	0.15	0.0225	0.0225	0.0225
0.35	0.35	0.1225	0.1225	0.1225
1	1	1	1	1
1.25	1.35	1.5625	1.8225	1.6875
1.75	1.5	3.0625	2.25	2.625
2	2	4	4	4
2.25	2.25	5.0625	5.0625	5.0625
2.3	2.5	5.29	6.25	5.75
2.75	2.75	7.5625	7.5625	7.5625
\(\sum x = 13.8\)	\(\sum y = 13.85\)	\(\sum {{x^2}} = 27.685\)	\(\sum {{y^2} = } 28.0925\)	\(\sum {xy\; = \;} 27.8325\)

Substitute the values in the formula:

\(\begin{aligned} r &= \frac{{9\left( {27.8325} \right) - \left( {13.8} \right)\left( {13.85} \right)}}{{\sqrt {9\left( {27.685} \right) - {{\left( {13.8} \right)}^2}} \sqrt {9\left( {28.0925} \right) - {{\left( {13.85} \right)}^2}} }}\\ &= 0.992\end{aligned}\)

Thus, the correlation coefficient is 0.992.

Step 4:Conduct a hypothesis test for correlation

Define\(\rho \)as the actual value of thecorrelation coefficient for pizza cost and subway fare.

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 is 9 (n).

The test statistic is computed as follows:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.992}}{{\sqrt {\frac{{1 - {{0.992}^2}}}{{9 - 2}}} }}\\ &= 20.791\end{aligned}\)

Thus, the test statistic is 20.791.

The degree of freedom is

\(\begin{aligned} df &= n - 2\\ &= 9 - 2\\ &= 7.\end{aligned}\)

Thep-value is computed from the t-distribution table.

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T > t} \right)\\ &= 2P\left( {T > 20.791} \right)\\ &= 2\left( {1 - P\left( {t < 20.791} \right)} \right)\\ &= 0.000\end{aligned}\)

Thus, the p-value is 0.000.

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

Therefore, there is enough evidence to conclude that the variables pizza cost and subway fare have a linear correlation between them.