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Solved Problems

Solved Problem 4-1 Judith Thompson runs a florist shop on the Gulf Coast of Texas, specializing in floral arrangements for weddings and other special events. She advertises weekly in the local newspapers and is considering increasing her advertising budget. Before doing so, she decides to evaluate the past effectiveness of these ads. Five weeks are sampled, and the advertising dollars and sales volume for each of these are shown in the following table. Develop a regression equation that would help Judith evaluate her advertising. Find the coefficient of determination for this model.

SALES ($1,000)	ADVERTISING ($100)
11	5
6	3
10	7
6	2
12	8

Solution

SALES Y	ADVERTISING x	${(X - \bar{X})}^{2}$ ${(X - \bar{X})}^{2}$	$(X - \bar{X}) (Y - \bar{Y})$ $(X - \bar{X}) (Y - \bar{Y})$
11	5	${(5 - 5)}^{2} = 0$ ${(5 - 5)}^{2} = 0$	$(5 - 5) (11 - 9) = 0$ $(5 - 5) (11 - 9) = 0$
6	3	${(3 - 5)}^{2} = 4$ ${(3 - 5)}^{2} = 4$	$(3 - 5) (6 - 9) = 6$ $(3 - 5) (6 - 9) = 6$
10	7	${(7 - 5)}^{2} = 4$ ${(7 - 5)}^{2} = 4$	$(7 - 5) (10 - 9) = 2$ $(7 - 5) (10 - 9) = 2$
6	2	${(2 - 5)}^{2} = 9$ ${(2 - 5)}^{2} = 9$	$(2 - 5) (6 - 9) = 9$ $(2 - 5) (6 - 9) = 9$
12	8	${(8 - 5)}^{2} = 9$ ${(8 - 5)}^{2} = 9$	$(8 - 5) (12 - 9) = 9$ $(8 - 5) (12 - 9) = 9$
$\begin{array}{l} Σ Y & = & 45 \\ \bar{Y} & = & 45 / 5 \\ = & 9 \end{array}$ $\begin{array}{l} Σ Y & = & 45 \\ \bar{Y} & = & 45 / 5 \\ = & 9 \end{array}$	$\begin{array}{l} Σ X & = & 25 \\ \bar{X} & = & 25 / 5 \\ = & 5 \end{array}$ $\begin{array}{l} Σ X & = & 25 \\ \bar{X} & = & 25 / 5 \\ = & 5 \end{array}$	$Σ {(X - \bar{X})}^{2} = 26$ $Σ {(X - \bar{X})}^{2} = 26$	$Σ (X - \bar{X}) (Y - \bar{Y}) = 26$ $Σ (X - \bar{X}) (Y - \bar{Y}) = 26$

\begin{array}{l} b_{1} & = & \frac{Σ (X - \bar{X}) (Y - \bar{Y})}{Σ {(X - \bar{X})}^{2}} = \frac{26}{26} = 1 \\ b_{0} & = & \bar{Y} - b_{1} \bar{X} = 9 - (1) (5) = 4 \end{array}

$\begin{array}{l} b_{1} & = & \frac{Σ (X - \bar{X}) (Y - \bar{Y})}{Σ {(X - \bar{X})}^{2}} = \frac{26}{26} = 1 \\ b_{0} & = & \bar{Y} - b_{1} \bar{X} = 9 - (1) (5) = 4 \end{array}$

The regression equation is

\hat{Y} = 4 + 1 X

$\hat{Y} = 4 + 1 X$

To compute $r^{2},$ $r^{2},$ we use the following table:

Y	X	$\hat{Y} = 4 + 1 X$ $\hat{Y} = 4 + 1 X$	${(\hat{Y} - \bar{Y})}^{2}$ ${(\hat{Y} - \bar{Y})}^{2}$	${(Y - \bar{Y})}^{2}$ ${(Y - \bar{Y})}^{2}$
11	5	9	${(11 - 9)}^{2} = 4$ ${(11 - 9)}^{2} = 4$	${(11 - 9)}^{2} = 4$ ${(11 - 9)}^{2} = 4$
6	3	7	${(6 - 7)}^{2} = 1$ ${(6 - 7)}^{2} = 1$	${(6 - 9)}^{2} = 9$ ${(6 - 9)}^{2} = 9$
10	7	11	${(10 - 11)}^{2} = 1$ ${(10 - 11)}^{2} = 1$	${(10 - 9)}^{2} = 1$ ${(10 - 9)}^{2} = 1$
6	2	6	${(6 - 6)}^{2} = 0$ ${(6 - 6)}^{2} = 0$	${(6 - 9)}^{2} = 9$ ${(6 - 9)}^{2} = 9$
12	8	12	${(12 - 12)}^{2} = 0$ ${(12 - 12)}^{2} = 0$	${(12 - 9)}^{2} = 9$ ${(12 - 9)}^{2} = 9$
$\begin{array}{l} Σ Y & = & 45 \\ \bar{Y} & = & 9 \end{array}$ $\begin{array}{l} Σ Y & = & 45 \\ \bar{Y} & = & 9 \end{array}$	$\begin{array}{l} Σ X & = & 25 \\ \bar{X} & = & 5 \end{array}$ $\begin{array}{l} Σ X & = & 25 \\ \bar{X} & = & 5 \end{array}$		$\begin{array}{l} Σ {(Y - \hat{Y})}^{2} & = & 6 \\ SSE \end{array}$ $\begin{array}{l} Σ {(Y - \hat{Y})}^{2} & = & 6 \\ SSE \end{array}$	$\begin{array}{l} Σ {(Y - \bar{Y})}^{2} & = & 32 \\ SST \end{array}$ $\begin{array}{l} Σ {(Y - \bar{Y})}^{2} & = & 32 \\ SST \end{array}$

The slope $(b_{1} = 1)$ $(b_{1} = 1)$ tells us that for each 1-unit increase in X (or $100 in advertising), sales increase by 1 unit (or $1,000). Also, $r^{2} = 0.8125$ $r^{2} = 0.8125$ , indicating that about 81% of the variability in sales can be explained by the regression model with advertising as the independent variable.

Solved Problem 4-2 Use Excel with the data in Solved Problem 4-1 to find the regression model. What does the F test say about this model?

Solution

Program 4.8 provides the Excel output for this problem. We see the equation is

$\hat{Y} = 4 + 1 X$ $\hat{Y} = 4 + 1 X$

The coefficient of determination $(r^{2})$ $(r^{2})$ is shown to be 0.8125. The significance level for the F test is 0.0366, which is less than 0.05. This indicates the model is statistically significant. Thus, there is sufficient evidence in the data to conclude that the model is useful and there is a relationship between X (advertising) and Y (sales).

Program 4.8 Excel 2016 Output for Solved Problem 4-2

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