Q28BSC Testing for a Linear Correlation... [FREE SOLUTION]

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Chapter 10: Q28BSC (page 468)

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.)
Sports Repeat the preceding exercise using diameters and volumes.

Short Answer

Expert verified

The scatterplot is shown below:

The value of the correlation coefficient is 0.978.

The p-value is 0.000.

There is sufficient evidence to support the claim that there exists a linear correlation between diameter and volume.

No, the scatterplot reveals that the trend is non-linear.

Step by step solution

Given information

Refer to Exercise 27 for the data on diameter and volume.

Diameter	Volume
7.4	212.2
23.9	7148.1
4.3	41.6
21.8	5424.6
7	179.6
4	33.5
20.9	4780.1
9.7	477.9

Sketch a scatterplot

A scatterplot is described for a paired set of data obtained from two variables.

Steps to sketch a scatterplot:

Drawx-axisas the diameter and y-axis as the volume data.
Mark allthe paired values.

The resultant scatterplot is given below.

Compute the measure of the correlation coefficient

The correlation coefficient formula is

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

Define x as the diameter and yas the volume.

The valuesare tabulatedbelow:

x	y	\({x^2}\)	\({y^2}\)	\(xy\)
7.4	212.2	54.76	45028.84	1570.28
23.9	7148.1	571.21	51095333.61	170839.59
4.3	41.6	18.49	1730.56	178.88
21.8	5424.6	475.24	29426285.16	118256.28
7	179.6	49	32256.16	1257.2
4	33.5	16	1122.25	134
20.9	4780.1	436.81	22849356.01	99904.09
9.7	477.9	94.09	228388.41	4635.63
\(\sum x = 99\)	\(\sum y = 18297.6\)	\(\sum {{x^2}} = 1715.6\)	\(\sum {{y^2} = } \;103679501\)	\(\sum {xy\; = \;} 396776\)

Substitute the values in the formula:

\(\begin{aligned} r &= \frac{{8\left( {396776} \right) - \left( {99} \right)\left( {18297.6} \right)}}{{\sqrt {8\left( {1715.6} \right) - {{\left( {99} \right)}^2}} \sqrt {8\left( {103679501} \right) - {{\left( {18297.6} \right)}^2}} }}\\ &= 0.978\end{aligned}\)

Thus, the correlation coefficient is 0.978.

Step 4:Conduct a hypothesis test for correlation

Define\(\rho \)asthe correlation measure for diameter and volumes.

For testing the claim, form the hypotheses:

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

The samplesize is8(n).

The test statistic is calculated below:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.978}}{{\sqrt {\frac{{1 - {{\left( {0.978} \right)}^2}}}{{8 - 2}}} }}\\ &= 11.533\end{aligned}\)

Thus, the test statistic is11.533.

The degree of freedom iscalculated below:

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

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

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T > 11.533} \right)\\ &= 0.000\end{aligned}\)

Thus, the p-value is 0.000.

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

Therefore, there issufficient evidence to support the existence of alinear correlation between diameter and volume.

Discuss the scatterplot

The scatterplot does not support the linear associationestablished in the results above. On the contrary, it suggests anon-linear pattern between the two variables.