Question

In: Economics

Refer to this regression equation: (standard errors in parentheses.) Q = 8,400 - 10 P...

Refer to this regression equation:

(standard errors in parentheses.)

Q = 8,400 - 10 P + 5 A + 4 Px + 0.05 I

(1,732) (2.29) (1.36) (1.75) (0.15)

R2 = 0.65
N = 120
F = 35.25
Standard error of estimate = 34.3
Q = Quantity demanded
P = Price = 1,000
A = Advertising expenditures, in thousands = 40
PX = price of competitor's good = 800
I = average monthly income = 4,000

1) Calculate the elasticity for each variable and briefly comment on what information this gives you in each case.

2) Calculate t-statistics for each variable and explain what this tells you.

3) What information does the R2 give you?

4) How would you evaluate the quality of this equation overall? Do you have any concerns? Explain.

5) Should this firm be concerned if macroeconomic forecasters predict a recession? Explain.

Solutions

Expert Solution

It is given that -

$\begin{array}{l}Q = 8400 - \left( {10 \times 1000} \right) + \left( {5 \times 40} \right) + \left( {4 \times 800} \right) + \left( {4000 \times 0.05} \right)\\ \Rightarrow Q = 2000\end{array}$

$\frac{{\Delta Q}}{{\Delta P}} = - 10,\frac{{\Delta Q}}{{\Delta A}} = 5,\frac{{\Delta Q}}{{\Delta Px}} = 4,and\frac{{\Delta Q}}{{\Delta I}} = 0.05$ .

P = 1000, A = 40, Px = 800 and I = 4000.

Question 1. The required elasticities are computed as - $\begin{array}{c}elasticityofquantitywithrespecttofprice = \frac{{\Delta Q}}{{\Delta P}} \times \frac{P}{Q} = ( - )10 \times \frac{{1000}}{{2000}}\\ = \left( - \right)5\end{array}$

$\begin{array}{c}elasticityofquantitywithrespecttofadvertisement = \frac{{\Delta Q}}{{\Delta A}} \times \frac{A}{Q} = 5 \times \frac{{40}}{{2000}}\\ = 0.1\end{array}$

$\begin{array}{c}elasticityofquantitywithrespecttofcompetitor'sprice = \frac{{\Delta Q}}{{\Delta Px}} \times \frac{{Px}}{Q} = 4 \times \frac{{40}}{{2000}}\\ = 0.08\end{array}$

$\begin{array}{c}elasticityofquantitywithrespecttofincome = \frac{{\Delta Q}}{{\Delta I}} \times \frac{I}{Q} = 0.05 \times \frac{{4000}}{{2000}}\\ = 0.1\end{array}$

Question 2.

$Q = \alpha - {\beta _1}P + {\beta _2}A + {\beta _3}Px + {\beta _4}I$
The t-statistics are computed as -

$\begin{array}{c}{H_0}:{\beta _1} = 0\\{H_A}:{\beta _1} \ne 0\\t - statisticsof{\beta _1} = \frac{{{\beta _1} - 10}}{{S\tan darderror{\beta _1}}}\\ = \frac{{0 - 10}}{{2.29}} = - 4.366\end{array}$

$\begin{array}{c}{H_0}:{\beta _2} = 0\\{H_A}:{\beta _2} \ne 0\\t - statisticsof{\beta _2} = \frac{{{\beta _2} - 5}}{{S\tan darderror{\beta _2}}}\\ = \frac{{0 - 5}}{{1.36}} = - 3.676\end{array}$

$\begin{array}{c}{H_0}:{\beta _3} = 0\\{H_A}:{\beta _3} \ne 0\\t - statisticsof{\beta _3} = \frac{{{\beta _3} - 4}}{{S\tan darderror{\beta _3}}}\\ = \frac{{0 - 4}}{{1.75}} = - 2.285\end{array}$

$\begin{array}{c}{H_0}:{\beta _4} = 0\\{H_A}:{\beta _4} \ne 0\\t - statisticsof{\beta _4} = \frac{{{\beta _4} - 0.05}}{{S\tan darderror{\beta _4}}}\\ = \frac{{0 - 0.05}}{{0.15}} = - 0.333\end{array}$
Question 3. The R² explains the total variation in the dependent variable which is jointly explained by all the independent variables together.
Question 4. The quality of the overall equation is evaluated by the R². It is given that the R² is 0.65. That is all the independent variable jointly explains the 65% of the average quantity. It does not matter of concern because r higher than 0.7, that is, r² greater than 0.49 then the regression is not considered as poor in explaining the dependent variable.

Question 5. The recession implies a downturn in economic output and economic income. The given regression equation is -

$Q = \alpha - {\beta _1}P + {\beta _2}A + {\beta _3}Px + {\beta _4}I$

That is any positive change in I positively impacts the average Q. Therefore, fall in 'Income' because of recession as mentioned it will negatively impact the Q. However, the income elasticity is less than 1, that is, 0.1. Thus, if a 1% change in income does not impact much average Q for comparatively less elastic demand with respect to income. Therefore, it is not much a matter of concern for the firm.

Rahul Sunny answered 10 months ago

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