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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 foot lengths and heights to find the best predicted height of a male

who has a foot length of 28 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 value is the mean height of 177 cm. It will not be helpful to the police in trying to obtain a description of the male.

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 for the regression linea

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 foot length (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

\(\begin{array}{c}{b_1} = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{5 \times 23209.27 - 130.8 \times 886.5}}{{5 \times 3426.96 - {{130.8}^2}}}\\ = 3.5226\end{array}\).

The intercept is computed as

\(\begin{array}{c}{b_0} = \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{886.5 \times 3426.96 - 130.8 \times 23209.27}}{{5 \times 3426.96 - {{130.8}^2}}}\\ = 85.15\end{array}\).

The estimated regression equation is

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = 85.15 + 3.5226x\end{array}\).

04

Checking the model

Refer to exercise 18 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 0.085.

As the P-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 simply its sample mean.

05

Compute the prediction 

The best-predicted height of a male who has a foot length of 28 cm is obtained as the mean of the sample responses.

The sample mean is computed as

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

Therefore, the best-predicted height of the male who has a foot length of 28 cm will be 177 cm. As the best-predicted value is the mean height (177 cm), it will not help the police to describe the male.

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Most popular questions from this chapter

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.

\({s_e}\)Notation Using Data Set 1 “Body Data” in Appendix B, if we let the predictor variable x represent heights of males and let the response variable y represent weights of males, the sample of 153 heights and weights results in\({s_e}\)= 16.27555 cm. In your own words, describe what that value of \({s_e}\)represents.

The following exercises are based on the following sample data consisting of numbers of enrolled students (in thousands) and numbers of burglaries for randomly selected large colleges in a recent year (based on data from the New York Times).

The sample data result in a linear correlation coefficient of r= 0.499 and the regression equation\(\hat y = 3.83 + 2.39x\). What is the best predicted number of burglaries, given an enrollment of 50 (thousand), and how was it found?

The following exercises are based on the following sample data consisting of numbers of enrolled students (in thousands) and numbers of burglaries for randomly selected large colleges in a recent year (based on data from the New York Times).

Enrollment (thousands)

53

28

27

36

42

Burglaries

86

57

32

131

157

True or false: If the sample data lead us to the conclusion that there is sufficient evidence to support the claim of a linear correlation between enrollment and number of burglaries, then we could also conclude that higher enrollments cause increases in numbers of burglaries.

Interpreting the Coefficient of Determination. In Exercises 5–8, use the value of the linear correlation coefficient r to find the coefficient of determination and the percentage of the total variation that can be explained by the linear relationship between the two variables.

Bears r = 0.783 (x = head width of a bear, y = weight of a bear)

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