Regression and Predictions. 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 (x) variable. Find the indicated predicted value by following the prediction procedure summarized in Figure 10-5 on page 493.

Using the bill/tip data, find the best predicted tip amount for a dinner bill of $100. What tipping rule does the regression equation suggest?

Bill (dollars)	33.46	50.68	87.92	98.84	63.6	107.34
Tip (dollars)	5.5	5	8.08	17	12	16

Values are given on two variables namely, bill (dollars) and tip (dollars).

Let x represent thebill (dollars).

Let y represent thetip (dollars).

Themean value of xis given as,

\(\begin{array}{c}\bar x = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\ = \frac{{33.46 + 50.68 + .... + 107.34}}{6}\\ = 73.640\end{array}\)

Therefore, the mean value of x is 73.64.

Themean value of yis given as,

\(\begin{array}{c}\bar y = \frac{{\sum\limits_{i = 1}^n {{y_i}} }}{n}\\ = \frac{{5.5 + 5 + .... + 16}}{6}\\ = 10.597\end{array}\)

Therefore, the mean value of y is 10.597.

The standard deviation of xis given as,

\(\begin{array}{c}{s_x} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {33.46 - 73.64} \right)}^2} + {{\left( {50.68 - 73.64} \right)}^2} + ..... + {{\left( {107.34 - 73.64} \right)}^2}}}{{6 - 1}}} \\ = 29.042\end{array}\)

Therefore, the standard deviation of x is 29.042.

The standard deviation of yis given as,

\(\begin{array}{c}{s_y} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({y_i} - \bar y)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {5.5 - 10.597} \right)}^2} + {{\left( {5 - 10.597} \right)}^2} + ..... + {{\left( {16 - 10.597} \right)}^2}}}{{6 - 1}}} \\ = 5.212\end{array}\)

Therefore, the standard deviation of y is 5.211.

The correlation coefficient is given as,

\(r = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right) - {{\left( {\sum y } \right)}^2}} \right)} }}\)

The calculations required to compute the correlation coefficient are as follows:

The correlation coefficient is given as,

\(\begin{array}{c}r = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right) - {{\left( {\sum y } \right)}^2}} \right)} }}\\ = \frac{{6\left( {5308.744} \right) - \left( {441.84} \right)\left( {63.58} \right)}}{{\sqrt {\left( {\left( {6 \times 36754.14} \right) - {{\left( {441.84} \right)}^2}} \right)\left( {\left( {6 \times 809.5364} \right) - {{\left( {63.58} \right)}^2}} \right)} }}\\ = 0.8282\end{array}\)

Therefore, the correlation coefficient is 0.8282.

The slopeof the regression line is given as,

\(\begin{array}{c}{b_1} = r\frac{{{s_Y}}}{{{s_X}}}\\ = 0.8282 \times \frac{{5.212}}{{29.042}}\\ = 0.149\end{array}\)

Therefore, the value of slope is 0.149.

The interceptis computed as,

\(\begin{array}{c}{b_0} = \bar y - {b_1}\bar x\\ = 10.597 - \left( {0.1486 \times 73.640} \right)\\ = - 0.347\end{array}\)

Therefore, the value of intercept is -0.347 which is approximately -0.35.

Theregression equationis given as,

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = - 0.347 + 0.149x\\ \approx - 0.35 + 0.149x\end{array}\)

Thus, the regression equation is \(\hat y = - 0.35 + 0.149x\).

Referring to exercise 25 of section 10-1,

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

2)The P-value is 0.042.

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

Therefore, the correlation is significant.

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

Therefore, the regression equation can be used to predict the value of y.

The best predicted tip amount for a dinner bill of $100is computed as,

\(\begin{array}{c}\hat y = - 0.35 + \left( {0.149 \times 100} \right)\\ = 14.55\end{array}\)

Therefore, thebest predicted tip amount for a dinner bill of $100is $14.55.

The tipping amount is predicted based on the bill by first multiplying the bill amount by 0.149 and then subtracting $0.35 from it. On the other hand, each $100 bill implies approximately $15 bill.