Exercises 13–28 use the same data sets as Exercises 13–28 in Section 10-1. In each case, find the regression equation, letting the first variable be the predictor (x) variable. Find the indicated predicted value by following the prediction procedure summarized in Figure 10-5 on page 493.

Use the shoe print lengths and heights to find the best predicted height of a male who has a shoe print length of 31.3 cm. Would the result be helpful to police crime scene investigators in trying to describe the male?

Short Answer

Expert verified

The regression equation is\(\hat y = 125 + 1.73x\).

The best-predicted height with a shoe-print length of 31.3 cm will be 177 cm. As the best prediction is made using the mean value, it will not be helpful for the police to trace a male criminal.

Step by step solution

01

Given information

The given data provides the information of the shoe print (in cm) and the height (in cm), as follows.

02

State the equation of the regression line

The formula for the estimated regression line is

\(y = {b_0} + {b_1}x\).

Here,

\({b_0}\)is the Y-intercept,

\({b_1}\)is the slope,

\(x\)is the explanatory variable, and

\(\hat y\)is the response variable (predicted value).

Let X denote the shoe print (in cm) and Y denote the height (in cm) of the male.

03

Compute the slope and intercept

The calculations required to compute the slope and intercept are as follows.

The sample size is \(\left( n \right) = 5\).

The slope is computed as

\(\begin{array}{c}{b_1} = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{5 \times 26650 - 150.2 \times 886.5}}{{5 \times 4523.1 - {{150.2}^2}}}\\ = 1.73\end{array}\).

The intercept is computed as

\(\begin{array}{c}{b_0} = \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{886.5 \times 4523.1 - 150.2 \times 26650}}{{5 \times 4523.1 - {{150.2}^2}}}\\ = 125.4\end{array}\).

So, the estimated regression equation is

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = 125 + 1.73x\end{array}\).

04

Checking the model

Refer to exercise 17 of section 10-1 for the following result.

1) The scatter plot does not show an approximate linear relationship between the variables.

2) The P-value is 0.294.

As theP-value is greater than the level of significance (0.05), the null hypothesis is failed to be rejected.

Therefore, the correlation is not significant.

Referring to figure 10-5, the criteria for a good regression model are not satisfied.

The best-predicted value of a variable is the sample mean of the response variable.

05

Compute the prediction 

The best-predicted height of a male who has a shoe-print length of 31.3 cm is computed as follows.

The sample mean:

\(\begin{array}{c}\bar y = \frac{{\sum y }}{n}\\ = \frac{{\left( {175.3 + 177.8 + ... + 172.7} \right)}}{5}\\ = 177.3\end{array}\).

Therefore, the best-predicted height of a male who has a shoe-print length of 31.3 cm will be approximately 177 cm.

It will not be helpful to the police in trying to obtain a description of the male because the resultant value is obtained from a bad regression model.

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Most popular questions from this chapter

let the predictor variable x be the first variable given. Use the given data to find the regression equation and the best predicted value of the response variable. Be sure to follow the prediction procedure summarized in Figure 10-5 on page 493. Use a 0.05 significance level.

Head widths (in.) and weights (lb) were measured for 20 randomly selected bears (from Data Set 9 “Bear Measurements” in Appendix B). The 20 pairs of measurements yield\(\bar x = 6.9\)in.,\(\bar y = 214.3\)lb, r= 0.879, P-value = 0.000, and\(\hat y = - 212 + 61.9x\). Find the best predicted value of\(\hat y\)(weight) given a bear with a head width of 6.5 in.

Exercises 13–28 use the same data sets as Exercises 13–28 in Section 10-1. In each case, find the regression equation, letting the first variable be the predictor (x) variable. Find the indicated predicted value by following the prediction procedure summarized in Figure 10-5 on page 493.

Using the listed lemon/crash data, find the best predicted crash fatality rate for a year in which there are 500 metric tons of lemon imports. Is the prediction worthwhile?

In Exercises 9 and 10, use the given data to find the equation of the regression line. Examine the scatterplot and identify a characteristic of the data that is ignored by the regression line.

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 Police sometimes measure shoe prints at crime scenes so that they can learn something about criminals. Listed below are shoe print lengths, foot lengths, and heights of males (from Data Set 2 “Foot and Height” in Appendix B). Is there sufficient evidence to conclude that there is a linear correlation between shoe print lengths and heights of males? Based on these results, does it appear that police can use a shoe print 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

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.)

Sports Repeat the preceding exercise using diameters and volumes.

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