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Exercises 13鈥28 use the same data sets as Exercises 13鈥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 old/new mpg ratings, find the best predicted new

mpg rating for a car with an old rating of 30 mpg. Is there anything to suggest that the prediction is likely to be quite good?

Short Answer

Expert verified

The regression equation is\(\hat y = 0.863x + 0.808\).

The best-predicted new mpg rating for a car with an old rating of 30 mpg will be approximately 26.7 mpg. The predictions are made from a good regression model and are not extrapolated.

Step by step solution

01

Given information

The given data provides the information of the mpg ratings of 鈥榦ld鈥 and 鈥榥ew鈥 cars, as follows.

02

State the equation for the estimated 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 mpg ratings of the old cars and Y denote the mpg ratings of the new cars.

03

Compute the slope and intercept

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

The sample size is \(\left( n \right) = 11\).

The slope is computed as

\(\begin{array}{c}{b_1} = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( x \right)}^2}}}\\ = \frac{{11 \times 5569 - 255 \times 229}}{{11 \times 6213 - {{255}^2}}}\\ \approx 0.863\end{array}\).

The intercept is computed as

\(\begin{array}{c}{b_0} = \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{229 \times 6213 - 255 \times 5569}}{{11 \times 6213 - {{255}^2}}}\\ \approx 0.808\end{array}\).

So, estimated regression equation is

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = 0.808 + 0.863x\end{array}\).

04

Check the model

Refer to exercise 20 of section 10-1 for the following result.

1) The scatter plot shows an approximate linear relationship between the variables.

2) The P-value is 0.000.

As the P-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.

Thus, the regression equation can be used to make the prediction.

05

Compute the predicted value 

Substitute the 30 mpg rating (an old rating) in the estimated linear regression model for the prediction of the new mpg rating car.

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = 0.808 + 0.863 \times 30\\ = 26.7\end{array}\).

Therefore, the best-predicted new mpg rating for a car with an old rating of 30 mpg will be approximately 26.7 mpg.

06

Express the characteristic that makes the prediction good

The prediction is made using a regression equation of a good regression model. Also, the value that is predicted lies between the range of sampled observations.

Thus, the prediction is not extrapolated. Therefore, it is suggestive that the prediction is likely to be quite good.

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Most popular questions from this chapter

Outlier Refer to the accompanying Minitab-generated scatterplot. a. Examine the pattern of all 10 points and subjectively determine whether there appears to be a correlation between x and y. b. After identifying the 10 pairs of coordinates corresponding to the 10 points, find the value of the correlation coefficient r and determine whether there is a linear correlation. c. Now remove the point with coordinates (10, 10) and repeat parts (a) and (b). d. What do you conclude about the possible effect from a single pair of values?

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 鈥(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?

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.)

Internet and Nobel Laureates Listed below are numbers of Internet users per 100 people and numbers of Nobel Laureates per 10 million people (from Data Set 16 鈥淣obel Laureates and Chocolate鈥 in Appendix B) for different countries. Is there sufficient evidence to conclude that there is a linear correlation between Internet users and Nobel Laureates?

Internet Users

Nobel Laureates

79.5

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9

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In Exercises 5鈥8, use a significance level 0.05 and refer to theaccompanying displays.Cereal Killers The amounts of sugar (grams of sugar per gram of cereal) and calories (per gram of cereal) were recorded for a sample of 16 different cereals. TI-83>84 Plus calculator results are shown here. Is there sufficient evidence to support the claim that there is a linear correlation between sugar and calories in a gram of cereal? Explain.

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