A1. Researcher A advocated using the following econometric model where S stands

Question

A1. Researcher A advocated using the following econometric model where S stands for monthly sales (measured in thousands of dollars) in a given city i, P stands for price (measured in dollars) for the product in the city i and e is a random error for city i After using EViews software, Researcher A obtained the following results Dependent Variable: SALES Method: Least Squares Date: 01/29/13 Time: 12:11 Sample: 1 40 Included observations: 40 Variable Coefficient Std. Error t-Statistic Prob 200.13400 4.000000 PRICE 10.11700 5.000000 R-squared Adjusted R-squared S.E. of regression Sum squared resid Log likelihood Durbin-Watson stat Mean dependent var .000000 11.000000 7.049357 7.142352 29.24786 0.000000 87 S.D. dependent var 10.00000 Akaike info criterion 3800.000 Schwarz criterion 323.8795 F-statistic 2.203037 Prob(F-statistic) QUESTIONS CONTINUE OVER PAGE

Explanation / Answer

Answer to part a)

Hypothesis:

Null hypothesis: there is no significant influence of price on sales

Alternate hypothesis: there is significant influence of price on sales

.

The test statistic T = Slope / standard error

Given: Slope = -10.11700

Standard error = 5

Test Statistic T = -10.11700 / 5

T = -2.0234

.

P value depends on three things:

The P value obtained is: 0.049

[this can be obtained using Excel formula =T.DIST.2T(2.034,39) ]

.

Decision rule:

If P value < significance level , reject the null

If P value > significance level, fail to reject the null

.

Since the P value 0.049 < significance level 0.05 , we reject the null hypothesis
Conclusion: The influence of price is significant on sales

.

Answer to part b)

Goodness of fit measure is R square

The Formula is: R square = 1 – SS residual / SS total

R square = 1 - 3800/3810

R square = 0.0026

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Answer to part c)

A R square value tells us about the percent of variation in the dependent variable that is explained by this model of regression

In this scenario we got Rsquare = 0.0026 ~ 0.26%

Thus only 0.26% of the variation in the sales can be explained by price

.

Answer to part d)

Sales = 200.13400 – 10.117*P

IF P = $5

Sales = 200.134 -10.117 *5

Sales = 149.549 ~ 150

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