Question

1) requires some form in a consumption function Y = B. + BX: + ui Y consumption X indicates revenue ($ 1000). OLS estimation from 92 observational samples, Y = 8.12 + 0.6X (0.114) A) How would you interpret this OLS result? B) Test the significance of the b# coefficient at the 5% significance level. c) B #=0.8 Test the hypothesis against the bidirectional alternative at the 5% significance level. D) Test the B# = 0.8 hypothesis against its alternative B # <0.8 at 5% significance level. E) Calculate the 95% confidence interval Bi After the interval you found, in option c What can you say about the hypothesis found.

Answer 1

The estimated model is: $\widehat{Y_i}=8.12+0.6X_i$

Standard error: SE=0.114

Observations: n=92

Answer to the question no. a:

It implies that a change in revenue by 1 unit will change the level of consumption by 0.6 units. Or in other words, 100% change in revenue will change the level of consumption by 60%.

Answer to the question no. b:

To test the significance at 5% level we will do the t test. The null hypothesis will be H0: β=0. And the alternative hypothesis will be H1: β≠0. The t statistic can be obtained by:

$t=\frac{\widehat{\beta_1}}{SE_\beta}$

$t=\frac{0.6}{0.114}=5.26$

(It is a two tailed test) The tabulated t statistic at 5% level of significance is 1.984. The calculated t statistic (5.26) is larger than the tabulated t statistic. Thus, the null hypothesis will be rejected at 5% level and can interpret it that the revenue variable has a significant impact on the consumption.

Answer to the question no. c:

Now the null hypothesis is H0: β=0.8. And the alternative hypothesis will be H1: β≠8. So, the t statistic will be:

$t=\frac{\widehat{\beta_1}-\beta_1}{SE_\beta}$

$t=\frac{0.6-0.8}{0.114}=-1.75$

$|t|=|1.75|$

(It is a two tailed test) The tabulated t statistic at 5% level of significance is 1.984. The calculated t statistic (1.75) is lesser than the tabulated t statistic. Thus, the null hypothesis will not be rejected at 5% level and can interpret that the coefficient of revenue variable is not different from 0.8.

Answer to the question no. d:

Now the null hypothesis is H0: β=0.8. And the alternative hypothesis will be H1: β<8. So, the t statistic will be:

$t=\frac{\widehat{\beta_1}-\beta_1}{SE_\beta}$

$t=\frac{0.6-0.8}{0.114}=-1.75$

$|t|=|1.75|$

(It is an one tailed test) The tabulated t statistic at 5% level of significance is 1.66. The calculated t statistic (1.75) is greater than the tabulated t statistic. Thus, the null hypothesis will be rejected at 5% level and can interpret that the coefficient of revenue variable is less than 0.8.

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