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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 CPI/subway fare data from the preceding exercise and find

the best predicted subway fare for a time when the CPI reaches 500. What is wrong with this prediction?

Short Answer

Expert verified

The regression equation is\(\hat y = - 0.290 + 0.0115x\).

The best-predicted subway fare for a time when the CPI reaches 500 subway fares will be $5.48. The prediction of subway fare at 500 CPI leads to exploitation.

Step by step solution

01

Given information

The given data provides the information of the CPI and subway fare 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.

Let X denote the CPI and Y denote the subway fare (in dollars).

03

Compute the slope and intercept

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

The sample size is \(\left( n \right) = 9\).

The slope is computed as

\(\begin{array}{c}{b_1} = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{9 \times 2753.58 - 1427.4 \times 13.85}}{{9 \times 274678.6 - {{1427.4}^2}}}\\ = 0.01153\end{array}\).

The intercept is computed as

\(\begin{array}{c}{b_0} = \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{13.85 \times 274678.6 - 1427.4 \times 2753.58}}{{9 \times 274678.6 - {{1427.4}^2}}}\\ = - 0.2903\end{array}\).

Thus, the estimated regression equation is

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = - 0.290 + 0.0115x\end{array}\).

04

Check the modela

Refer to exercise 16 of section 10-1 for the following result.

1) The scatter plot shows a strong linear relationship between the variables.

2) The P-value is 0.000.

As theP-value is less than the level of significance (0.05), the null hypothesis is rejected.

Therefore, the correlation is statistically significant.

Referring to figure 10-5, the criteria for a good regression model are satisfied.

The regression equation is used to make predictions.

05

Compute the predicted value

The predicted value of the subway fare when the CPI reaches 500 is

\(\begin{array}{c}\hat y = - 0.290 + 0.0115x\\ = - 0.290 + 0.0115\left( {500} \right)\\ = 5.48\end{array}\).

Thus, the subway fare is predicted as $5.48 when CPI reaches 500.

06

State the error in the predicted valuea

As the CPI value of500 is not close to the range of values in the sample responses for the variable CPI, the prediction of the subway fare when CPI reaches 500 leads to extrapolation.

Thus, the prediction value is not reliable.

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Most popular questions from this chapter

In Exercises 5–8, we want to consider the correlation between heights of fathers and mothers and the heights of their sons. Refer to theStatCrunch display and answer the given questions or identify the indicated items.

The display is based on Data Set 5 “Family Heights” in Appendix B.

Identify the multiple regression equation that expresses the height of a son in terms of the height of his father and mother.

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

Interpreting r. In Exercises 5–8, use a significance level of A = 0.05 and refer to the accompanying displays.

5. Bear Weight and Chest Size Fifty-four wild bears were anesthetized, and then their weights and chest sizes were measured and listed in Data Set 9 “Bear Measurements” in Appendix B; results are shown in the accompanying Statdisk display. Is there sufficient evidence to support the claim that there is a linear correlation between the weights of bears and their chest sizes? When measuring an anesthetized bear, is it easier to measure chest size than weight? If so, does it appear that a measured chest size can be used to predict the weight?

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

z Scores Using only the sunspot numbers, identify the highest number and convert it to a z score. In the context of these sample data, is that highest value “significantly high”? Why or why not?

Effects of an Outlier Refer to the Minitab-generated scatterplot given in Exercise 11 of

Section 10-1 on page 485.

a. Using the pairs of values for all 10 points, find the equation of the regression line.

b. After removing the point with coordinates (10, 10), use the pairs of values for the remaining 9 points and find the equation of the regression line.

c. Compare the results from parts (a) and (b).

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