Chapter 4, Problem 1E

(a)

To determine

Estimated regression line.

Explanation of Solution

The formula for the regression equation is:

  $\begin{array}{l}\text{<mo>Y</mo><mo>^</mo>=a+b<mo>X</mo><mo>^</mo>}\\ \text{Here,}\\ \text{a is estimated intercept coefficient}\\ \text{b is estimated slope coefficient}\end{array}$

Run the ordinary least squares method for the given data in excel. The results drawn are as follows:

  

Use the summary output to find the estimated regression equation as follows:

  $\begin{array}{c}\text{<mo>Y</mo><mo>^</mo>}\text{=}\text{<mo>321</mo>}\text{.24 + 0}\text{.03<mo>A</mo><mo>^</mo><mo> -12</mo>}\text{.44<mo>P</mo><mo>^</mo><mo> + 2</mo>}\text{.08<mo>M</mo><mo>^</mo>}\\ \text{Here,}\\ \text{<mo>Y</mo><mo>^</mo>}\text{= estimated sales in thousands gallons}\\ \text{<mo>A</mo><mo>^</mo>}\text{=}\text{estimated promotional expenditures in \$1,000}\\ \text{<mo>P</mo><mo>^</mo>}\text{=estimated selling price in \$/gallon}\\ \text{<mo>M</mo><mo>^</mo>}\text{=estimated disposable income in \$1,000}\end{array}$

(b)

To determine

The economic interpretation of the estimated intercept (a) and slope (b) coefficients.

Explanation of Solution

Interpretation of estimated intercept (a) coefficient:

When the selling price, promotional expenditure and disposable income are zero, on average the quantity sold of pens is equal to 321.24× $1000 = $321,240.

Interpretation of estimated slope (b) coefficient:

For a given level of selling price and disposable income, an additional $1,000 promotional expenditures lead to a rise in sales by 0.03×1000 = 30 gallons on average.

For a given level of promotional expenditure and selling price, an additional $1,000 disposable income leads to a rise in sales by 2.08×1000 = 2,080 gallons on an average.

For a given level of promotional expenditure and disposable income, an additional $1/gallon lead to falling in sales by 12.44×1000 = 12,440 gallons on an average.

(c)

To determine

The hypothesis that there is no relationship between the variables at 0.05 significance level.

Explanation of Solution

Conduct the t-test to know the statistical significance of the independent variables A, Pand M. The test statistic can be calculated using the following formula:

  $\text{T =}\frac{{\text{b <mo>^</mo>}}_{\text{1}}}{{\text{s}}_{{\text{b}}_{\text{1}}}}$

  $\text{Here,}$

  ${\text{b <mo>^</mo>}}_{\text{1}}\text{ is coefficient of estimated independent variable}$

  ${\text{s}}_{{\text{b}}_{\text{1}}}\text{ is standard error of coefficient of estimated independent variable}$

The t-statistic follows t-distribution with n-1 degrees of freedom.

For variable A, t-test is conducted as follows:

According to the summary output, the t-statistic for A variable is equal to 0.16.

At 5% significance level and 10-1=9 degrees of freedom, the critical value is equal to 2.262.

  

In figure (1), since the calculated t-statistic lies in the acceptance region. Therefore, we accept the null hypothesis. This means that the variable A is not statistically significant.

For variable P, t-test is conducted as follows:

According to the summary output, the t-statistic for P variable is equal to -2.89. At 5% significance level and 10-1=9 degrees of freedom, the critical value is equal to 2.262.

  

In figure (2), since the calculated t-statistic lies in the critical region. Therefore, we reject the null hypothesis. This means that the variable Pis statistically significant.

For variable M, t-test is conducted as follows:

According to the summary output, the t-statistic for M variable is equal to 0.69.

At 5% significance level and 10-1=9 degrees of freedom, the critical value is equal to 2.262.

  

In figure (3), since the calculated t-statistic lies in the acceptance region. Therefore, we accept the null hypothesis. This means that the M variable _is not statistically significant.

(d)

To determine

Coefficient of determination.

Explanation of Solution

The coefficient of determination measures the proportion of variance predicted by the independent variable in the dependent variable. It is denoted as R².

According to the summary output, the value of R² is equal to 0.81. This means that the regression equation predicts 81% of the variance in sales.

(e)

To determine

Analysis of variance on the regression including F-test of the overall significance of the results.

Explanation of Solution

The value of F-statistic is given as 8.40. And the critical value at 0.05 significance level is equal to 0.01.

Since F-statistic is greater than the critical value, thus, the overall model is statistically significant.

(f)

To determine

Best estimate of the product sales when the selling price is $14.50. And an approximate 95 percent prediction interval.

Explanation of Solution

According to the regression statistics of the summary output, for a given level of promotional expenditure and disposable income, $14.50/gallon lead to fall in sales by 12.44×14.50×1000 = 180,380 gallons on an average.

According to the regression statistics in the summary output, a 95% confidence interval for the P variable ranges from -22.99 to -1.89.

(g)

To determine

Price elasticity of demand at a selling price of $14.50.

Explanation of Solution

Formula to calculate elasticity in linear regression model is as follows:

  $\text{Elasticity = b×}\left(\frac{\text{P}}{\text{Y}}\right)$

At given value of P variable equal to 14.50, the estimated value of Y variable is equal to 180,380.

Thus, price elasticity of demand is calculated as follows:

  $\begin{array}{c}\text{Elasticity = -12}\text{.44×}\left(\frac{14.50}{180,380}\right)\\ \text{= -0}\text{.001}\end{array}$

Thus, the price elasticity of demand at a selling price of $14.50 is equal to -0.001.