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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.)

CSI Statistics Use the paired foot length and height data from the preceding exercise. Is there sufficient evidence to conclude that there is a linear correlation between foot lengths and heights of males? Based on these results, does it appear that police can use foot length to estimate the height of a male?

Shoe print(cm)

29.7

29.7

31.4

31.8

27.6

Foot length(cm)

25.7

25.4

27.9

26.7

25.1

Height (cm)

175.3

177.8

185.4

175.3

172.7

Short Answer

Expert verified

The scatter plot is shown below:

The value of the correlation coefficient is 0.826.

The p-value is 0.085.

There is not enough evidence to support the claim of alinear correlation between foot length and height of males.

As two variables are not linearly associated and the scatter plot does not indicate a clear non-linear trend, the foot length cannot be used to predict the height of males.

Step by step solution

01

Given information

Refer to Exercise 17 for the data onshoe print, foot length, and height of males.

Shoe print(cm)

29.7

29.7

31.4

31.8

27.6

Foot length(cm)

25.7

25.4

27.9

26.7

25.1

Height (cm)

175.3

177.8

185.4

175.3

172.7

02

Sketch a scatterplot

Paired values are obtainedon a graph with two axes scaled according to the variables.

Steps to sketch a scatterplot:

  1. Mark horizontal axis for footlengthand vertical axis for the height of males.
  2. Mark each of the points on the graph.
  3. The resultant graph is the scatterplot.

03

Compute the measure of the correlation coefficient

The formula for 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 foot lengthbe defined by variable x and heights of males be defined by variable y.

The valuesare listed in the table below:

x

y

\({x^2}\)

\({y^2}\)

\(xy\)

25.7

175.3

660.49

30730.09

4505.21

25.4

177.8

645.16

31612.84

4516.12

27.9

185.4

778.41

34373.16

5172.66

26.7

175.3

712.89

30730.09

4680.51

25.1

172.7

630.01

29825.29

4334.77

\(\sum x = 130.8\)

\(\sum y = 886.5\)

\(\sum {{x^2}} = 3426.96\)

\(\sum {{y^2} = } \;157271.5\)

\(\sum {xy\; = \;} 23209.27\)

Substitute the values in the formula:

\(\begin{aligned} r &= \frac{{5\left( {23209.27} \right) - \left( {130.8} \right)\left( {886.5} \right)}}{{\sqrt {5\left( {3426.96} \right) - {{\left( {130.8} \right)}^2}} \sqrt {5\left( {157271.5} \right) - {{\left( {886.5} \right)}^2}} }}\\ &= 0.826\end{aligned}\)

Thus, the correlation coefficient is 0.826.

04

Step 4:Conduct a hypothesis test for correlation

Define\(\rho \)as the correlation coefficient for the population of two variables, foot lengths and height of males.

For testing the claim, form the hypotheses:

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

The samplesize is 5 (n).

The test statistic is computed as follows:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.826}}{{\sqrt {\frac{{1 - {{0.826}^2}}}{{5 - 2}}} }}\\ &= 2.538\end{aligned}\)

Thus, the test statistic is 2.538.

The degree of freedom is

\(\begin{aligned} df &= n - 2\\ &= 5 - 2\\ &= 3.\end{aligned}\)

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

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T > t} \right)\\ &= 2P\left( {T > 2.538} \right)\\ &= 2\left( {1 - P\left( {T < 2.538} \right)} \right)\\ &= 0.085\end{aligned}\)

Thus, the p-value is 0.085.

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

Therefore, there is not enough evidence to conclude that foot length and height have a linear correlation between them.

05

Analyze if the foot length can help predict the height of males

From the above result, the variables foot lengths and height of males are not linearly correlated. On the other hand, the scatterplot shows an upward trend but no specific pattern (linear or non-linear).

Thus, the foot lengths cannot be used to predict heights.

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Most popular questions from this chapter

Time and Motion In a physics experiment at Doane College, a soccer ball was thrown upward from the bed of a moving truck. The table below lists the time (sec) that has lapsed from the throw and the height (m) of the soccer ball. What do you conclude about the relationship between time and height? What horrible mistake would be easy to make if the analysis is conducted without a scatterplot?

Time (sec)

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

Height (m)

0.0

1.7

3.1

3.9

4.5

4.7

4.6

4.1

3.3

2.1

Stocks and Sunspots. Listed below are annual high values of the Dow Jones Industrial Average (DJIA) and annual mean sunspot numbers for eight recent years. Use the data for Exercises 1–5. A sunspot number is a measure of sunspots or groups of sunspots on the surface of the sun. The DJIA is a commonly used index that is a weighted mean calculated from different stock values.

DJIA

14,198

13,338

10,606

11,625

12,929

13,589

16,577

18,054

Sunspot

Number

7.5

2.9

3.1

16.5

55.7

57.6

64.7

79.3

Correlation Use a 0.05 significance level to test for a linear correlation between the DJIA values and the sunspot numbers. Is the result as you expected? Should anyone consider investing in stocks based on sunspot numbers?

Explore! Exercises 9 and 10 provide two data sets from “Graphs in Statistical Analysis,” by F. J. Anscombe, the American Statistician, Vol. 27. For each exercise,

a. Construct a scatterplot.

b. Find the value of the linear correlation coefficient r, then determine whether there is sufficient evidence to support the claim of a linear correlation between the two variables.

c. Identify the feature of the data that would be missed if part (b) was completed without constructing the scatterplot.

x

10

8

13

9

11

14

6

4

12

7

5

y

9.14

8.14

8.74

8.77

9.26

8.10

6.13

3.10

9.13

7.26

4.74

Best Multiple Regression Equation For the regression equation given in Exercise 1, the P-value is 0.000 and the adjusted \({R^2}\)value is 0.925. If we were to include an additional predictor variable of neck size (in.), the P-value becomes 0.000 and the adjusted\({R^2}\)becomes 0.933. Given that the adjusted \({R^2}\)value of 0.933 is larger than 0.925, is it better to use the regression equation with the three predictor variables of length, chest size, and neck size? Explain.

Confidence Intervals for a Regression Coefficients A confidence interval for the regression coefficient b1 is expressed

\(\begin{array}{l}{b_1} - E < {\beta _1} < {b_1} + E\\\end{array}\)

Where

\(E = {t_{\frac{\alpha }{2}}}{s_{{b_1}}}\)

The critical t score is found using n –(k+1) degrees of freedom, where k, n, and sb1 are described in Exercise 17. Using the sample data from Example 1, n = 153 and k = 2, so df = 150 and the critical t scores are \( \pm \)1.976 for a 95% confidence level. Use the sample data for Example 1, the Stat diskdisplay in Example 1 on page 513, and the Stat Crunchdisplay in Exercise 17 to construct 95% confidence interval estimates of \({\beta _1}\) (the coefficient for the variable representing height) and\({\beta _2}\) (the coefficient for the variable representing waist circumference). Does either confidence interval include 0, suggesting that the variable be eliminated from the regression equation?

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