The following estimation results are obtained for an equation relating household

Question

The following estimation results are obtained for an equation relating household consumption Y to income X. Both variables are measured in thousands of US dollars. The standard errors of the coefficients are given in parentheses. Question 3 r, = 15 +0.43X, (4.32) (0.15) R. 0.90; n = 372 a. Interpret the coefficient estimate on income X (in an economic sense, this coefficient estimate also measures the marginal propensity to consumel, b. Comment on tooness of fit for the model using the R-squared value C. You believe that income may not be an important determinant of consumption Conduct a relevant hypothesis test using the p-value method and state your conclusion at the 1% significance level. State the hypothesis HO: Test statistic Hi: Pioae or te tes tai Decision Rule Circle the correct response Reject HO in favor of H1/Do not reject HO One of your group members insists that and consumption. Now conduct a and state your conclusion at the 1% significance level. d. there should be a positive link between income relevant hypothesis test using the p-value method State the hypothesis H1: Test statistic P-value for the test statistic

Explanation / Answer

Here dependent variable be household consumption and independent variable be income

We have given that,

The regression equation is,

y = 15 + 0.43*x

R-sq = 0.90

n = 372

SE(b0) = standard error of b0 = 4.32

SE(b1) = Standard error of estimate = 0.15

a) intercept (b0) = 15

slope (b1) = 0.43

Interpretation : For one unit change in x will be 0.43 unit increase in y.

We can see that here there is positive relationship between x and y because the sign of b1 is positive.

b) Goodness of fit test :

Here we have to test the hypothesis that,

H0 : Model fits well to the data

H1 : Model does not fits well to the data

Assume alpha = 0.01

Test statistic = sample correlation coefficient = 0.9487

df = degrees of freedom = 370

By using statistical table of Pearson's correlation coefficient we get critical value = 0.148

Test statistic > critical value

Reject H0 at 1% level of significance.

Conclusion :Model does not fits well to the data

c) Here we have to test the hypothesis that,

H0 : B = 0 Vs H1 : B not= 0

where B is population slope for independent variable

Assume that alpha = level of significance = 1% = 0.01

Now the test statistic is,

t = b / SE(b) = 0.43/0.15 = 2.87

Now we have to find P-value for taking decision.

P-value we can find in EXCEL.

syntax :

=TDIST(x, deg_freedom, tails)

where x is absolute value of test statistic

deg_freedoms = n-2 = 372-2 = 370

tails = 2

P-value = 0.0044

P-value < alpha

Reject H0 at 1% level of significance.

COnclusion : The population slope for income is differ than 0.

We get significant result about t test.

d) Here we have to test,

H0 : Rho = 0 Vs H1 : Rho > 0

where Rho is population correlation between x and y.

Assume that alpha = 0.01

R-sq = 0.90

sample correlation coefficient (r) = sqrt(0.90) = 0.9487

n = 372

The test statistic is,

t = r*sqrt(n-2) / sqrt(1-r^2)

= 0.9487 * sqrt(372-2) / sqrt(1 - 0.90) = 57.71

Now we have to find P-value.

P-value = 2.1841E-187 = 0.000

P-value < alpha

Reject H0 at 1%level of significance

Conclusion : There is positive relationship between x and y.

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