Question

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 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

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

Given information

The given data provides the information of the mpg ratings of ‘old’ and ‘new’ cars, as follows.

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.

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

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

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.

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.

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.