Solution to Exercise E6.

Transcription

1 Solution to Exercise E6. The Multiple Regression Model. Inference Exercise E6.1 Beach umbrella rental Part I. Simple Linear Regression Model. a. Regression model: U t = α + β T t + u t t = 1,..., 22 Model 1: OLS, using observations (T = 22) Dependent variable: U const T Mean dependent var S.D. dependent var Sum suared resid S.E. of regression R Adjusted R F (1, 20) P-value(F ) 2.09e 11 Log-likelihood Akaike criterion Schwarz criterion Hannan Quinn ˆρ Durbin Watson SRF: Ŝ t = T t b. Test the statistical significance of the variable temperature. : β = 0 H a : β 0 t = ˆβ 0 ˆσ ˆβ Given that t = > = t (20), the null hypothesis is rejected at the 5% significance level. Therefore, the temperature is a statistically significant variable. c. Two-sided test. : β = 20/2 = 10 H a : β 20/2 t = ˆβ 10 ˆσ ˆβ Since t = = < = t (20), the null hypothesis is not rejected at the 5% significance level. Therefore, if the average temperature increases by 2 o C, the number of rented umbrellas may increase by 20 units. d. Point prediction. Û p = = umbrellas. e. Prediction interval. CI(U p ) 0.95 = [ ; ] The estimated upper limit for the number of rented umbrellas is

2 Part II. General Linear Regression Model. a. Regression model: U t = γ 1 + γ 2 T t + γ 3 P t + γ 4 W W t + v t t = 1,..., 22 Model 3: OLS, using observations (T = 22) Dependent variable: U const T P WW Mean dependent var S.D. dependent var Sum suared resid S.E. of regression R Adjusted R F (3, 18) P-value(F ) 2.11e 09 Log-likelihood Akaike criterion Schwarz criterion Hannan Quinn ˆρ Durbin Watson SRF: Û t = T t P t W W t t = 1,..., 22 b. Test whether the variable temperature is statistically significant. : γ 2 = 0 H a : γ 2 0 t = ˆγ 2 0 ˆσˆγ2 Since t = > = t (18), the null hypothesis is rejected at the 5% significance level. Therefore, the variable temperature is statistically significant. Test whether the variable price is statistically significant. : γ 3 = 0 H a : γ 3 0 t = ˆγ 3 0 ˆσˆγ3 Since t = < = t (18), the null hypothesis is not rejected at the 5% significance level. Therefore, the variable price is not statistically significant. Test whether the variable windy week is statistically significant. : γ 4 = 0 H a : γ 4 0 t = ˆγ 4 0 ˆσˆγ4 Given that t = < = t (18), the null hypothesis is not rejected at the 5% significance level. Therefore, the variable windy week is not statistically significant. Test the overall significance of the explanatory variables. : γ 2 = γ 3 = γ 4 = 0 H a : γ i 0 F = R 2 /(k 1) (1 R 2 )/() F(k 1, ) Since F = > = F 0.05 (3, 18), the null hypothesis is rejected at the 5% significance level. Therefore, the explanatory variables are jointly significant.

3 c. Given that the variables price and windy week are not individually significant, it may be possible that there is a high degree of collinearity between these two variables. Let s test whether the variables price and windy week are jointly significant. : γ 3 = γ 4 = 0 H a : γ i 0 F(, ) Given that F = < = F 0.05 (2, 18), the null hypothesis is not rejected at the 5% significance level. Therefore, variables price and windy week are not jointly significant. The variables price and windy week are neither individually nor jointly significant. Therefore, it may be concluded that there is not a high degree of collinearity in the sample. d. Prediction interval. Windy week: CI(U p ) 0.95 = [ ; ] The estimated upper limit for the number of rented umbrellas would be Non windy week: CI(U p ) 0.95 = [ ; ] The estimated upper limit for the number of rented umbrellas would be e. Model (2). The variables price and windy week are neither individually nor jointly significant. Therefore, the inclusion of the true restrictions β 3 = β 4 = 0 in the estimation of the model reduces the variance of the estimator.

4 Exercise E6.2 Holiday cottages Model A Regression model: RP i = α 1 + α 2 NR i + α 3 BP i + u i i = 1, 2,..., 75 Model 1: OLS, using observations 1 75 Dependent variable: RP const NR BP Mean dependent var S.D. dependent var Sum suared resid S.E. of regression R Adjusted R F (2, 72) P-value(F ) Log-likelihood Akaike criterion Schwarz criterion Hannan Quinn SRF: RP i = NR i BP i a. Test the joint significance of the explanatory variables. : α 2 = α 3 = 0 H a : α i 0 F = R 2 /(k 1) (1 R 2 )/() F(k 1, ) Since F = > = F 0.05 (2, 72), the null hypothesis is rejected at the 5% significance level. Therefore, the explanatory variables are jointly significant. b. Test the statistically significance of the variable number of rooms. : α 2 = 0 H a : α 2 0 t = ˆα 2 0 ˆσˆα2 Since t = < = t (72), the null hypothesis is not rejected at the 5% significance level. Therefore, the variable number of rooms is not statistically significant. c. Confidence interval for the coefficient α 3. CI(α 3 ) 0.95 = [ ; ] If the price of breakfast increases by e1, holding the number of rooms fixed, the variation in the price of the room lies between and euros, with a 95% probability. d. Point prediction. RP p = = euros. e. Prediction interval. CI(RP p ) 0.95 = [ ; ] If a holiday cottage has 10 bedrooms and the price of breakfast is e3, the price of the room lies between and euros, with a 95% probability.

5 Model B a. Regression model: SRF: RP i = λ 1 + λ 2 NR i + λ 3 BP i + λ 4 W IF IF i + λ 5 W IF IP i + λ 6 LOCC i + u i Model 2: OLS, using observations 1 75 Dependent variable: RP const NR BP WIFIF WIFIP LOCC Mean dependent var S.D. dependent var Sum suared resid S.E. of regression R Adjusted R F (5, 69) P-value(F ) Log-likelihood Akaike criterion Schwarz criterion Hannan Quinn RP i = NR i BP i W IF IF i W IF IP i LOCC i b. Test whether the variables access to WiFi and location are jointly significant. : λ 4 = λ 5 = λ 6 = 0 H a : λ i 0 F(, ) Since F = < = F 0.05 (3, 69), the null hypothesis is not rejected at the 5% significance level. Therefore, the variables access to WiFi and location are not jointly significant. c. One-sided test. : λ 6 0 H a : λ 6 > 0 t = ˆα 6 0 ˆσˆα6 Since t = < = t 0.05 (69), the null hypothesis is not rejected at the 5% significance level. Therefore, holiday cottages are not more expensive in the town center. d. Test whether variable access to WiFi is statistically significant. : λ 4 = λ 5 = 0 H a : λ i 0 F(, ) Given that F = < = F 0.05 (2, 69), the null hypothesis is not rejected at the 5% significance level. Therefore, the explanatory variable access to WiFi is not statistically significant.

6 e. Test for the euality of coefficients. : λ 4 = λ 5 H a : λ 4 λ 5 F(, ) Since F = < = F 0.05 (1, 69), the null hypothesis is not rejected at the 5% significance level. Therefore, there is evidence in the sample in favor of the hypothesis that what is really important is to offer access to WiFi; it does not matter whether it is free or not. Model C a. RP i = β 1 + β 2 NR i + β 3 BP i + β 4 W IF IF i + β 5 NP R i + β 6 BER i + β 7 LKR i + u i SRF: Model 3: OLS, using observations 1 75 Dependent variable: RP const NR BP WIFIF NPR BER LKR Mean dependent var S.D. dependent var Sum suared resid S.E. of regression R Adjusted R F (6, 68) P-value(F ) Log-likelihood Akaike criterion Schwarz criterion Hannan Quinn RP i = NR i BP i W IF IF i NP R i BER i LKR i b. Test the statistical significance of the variable having free access to WiFi. : β 4 = 0 H a : β 4 0 t = ˆβ 4 0 ˆσ ˆβ4 Since t = > = t (68), the null hypothesis is rejected at the 5% significance level. Therefore, the variable having free access to WiFi is statistically significant. c. Test the joint significance of the variables proximity to a natural park, proximity to a beach, proximity to a lake or reservoir. : β 5 = β 6 = β 7 = 0 H a : β i 0 F(, )

7 Given that F = > = F 0.05 (3, 68), the null hypothesis is rejected at the 5% significance level. Therefore, the variables proximity to a natural park, proximity to a beach, proximity to a lake or reservoir are jointly significant. d. Test the individual significance of the variables proximity to a natural park, proximity to a beach, proximity to a lake or reservoir. : β i = 0 H a : β i 0 t = ˆβ i 0 ˆσ ˆβi i = 5: Since t = < = t (68), the null hypothesis is not rejected at the 5% significance level. Therefore, the variable proximity to a natural park is not individually significant. i = 6: Since t = > = t (68), the null hypothesis is rejected at the 5% significance level. Therefore, the variable proximity to a beach is individually significant. i = 7: Since t = < = t (68), the null hypothesis is not rejected at the 5% significance level. Therefore, the variable proximity to a lake or reservoir is not individually significant. e. Estimation results for the augmented model (including the variable location). Model 4: OLS, using observations 1 75 Dependent variable: RP const NR BP WIFIF NPR BER LKR LOCC Mean dependent var S.D. dependent var Sum suared resid S.E. of regression R Adjusted R F (7, 67) P-value(F ) Log-likelihood Akaike criterion Schwarz criterion Hannan Quinn Test the statistical significance of the variable location. : β 8 = 0 H a : β 8 0 t = ˆβ 8 0 ˆσ ˆβ8 Since t = < = t (67), the null hypothesis is not rejected at the 5% significance level. Therefore, the variable location is not statistically significant: the price of a room is the same if the holiday cottage is in the town center or far from it.

8 f. Given that the variable location is not statistically significant, the properties of the OLS estimator used in the previous model are not affected: it is linear, unbiased and it has the smallest variance in the class of linear and unbiased estimators. g. The variables number of rooms, price of breakfast, proximity to a natural park, proximity to a lake or reservoir and location are not individually significant. Let s test whether they are jointly significant. : β 2 = β 3 = β 5 = β 7 = β 8 = 0 H a : β i 0 F(, ) Since F = < = F 0.05 (5, 67), the null hypothesis is not rejected at the 5% significance level. Therefore, the variables number of rooms, price of breakfast, proximity to a natural park, proximity to a lake or reservoir and location are not jointly significant. The restricted model is: The estimation results are: RP i = β 1 + β 4 W IF IF i + β 6 BER i + u i Model 5: OLS, using observations 1 75 Dependent variable: RP const WIFIF BER Mean dependent var S.D. dependent var Sum suared resid S.E. of regression R Adjusted R F (2, 72) P-value(F ) Log-likelihood Akaike criterion Schwarz criterion Hannan Quinn The estimator used is linear, unbiased (because the restrictions are true) and its variance is smaller than the variance of the estimator used in the unrestricted model.

9 Exercise E6.3 Soy milk Part I. General Linear Regression Model. a. Regression model: S t = β 1 + β 2 P t + β 3 AE t + β 4 AE 2 t + u t Model 2: OLS, using observations 1990: :06 (T = 270) Dependent variable: S const P AE s AE Mean dependent var S.D. dependent var Sum suared resid S.E. of regression R Adjusted R F (3, 266) P-value(F ) 1.89e 56 Log-likelihood Akaike criterion Schwarz criterion Hannan Quinn ˆρ Durbin Watson SRF: Ŝ t = P t AE t AE 2 t b. Overall significance test. : β 2 = β 3 = β 4 = 0 H a : β i 0 F = R 2 /(k 1) (1 R 2 )/() F(k 1, ) Since F = > = F 0.05 (3, 266), the null hypothesis is rejected at the 5% significance level. Therefore, the explanatory variables are jointly significant. Test the individual significance of the variable price. : β 2 = 0 H a : β 2 0 t = ˆβ 2 0 ˆσ ˆβ2 Since t = > = t (266), the null hypothesis is rejected at the 5% significance level. Therefore, the variable price is individually significant. Test the individual significance of the variable advertising expenditures. : β 3 = β 4 = 0 H a : β i 0 F(, ) Since F = > = F 0.05 (2, 266), the null hypothesis is rejected at the 5% significance level. Therefore, the variable advertising expenditures is individually significant.

10 c. Two-sided test. : β 4 = 0 H a : β 4 0 t = ˆβ 4 0 ˆσ ˆβ4 Since t = > = t (266), the null hypothesis is rejected at the 5% significance level. Therefore, the relationship between sales and advertising expenditures is not linear but uadratic. d. Test a linear restriction. : β 2 = = H a : β F(, ) Since F = > = F 0.05 (1, 266), the null hypothesis is rejected at the 5% significance level. Therefore, a 50 cents increase in the price does not generate a decrease in sales of e750. e. Prediction interval. CI(S p ) 0.95 = [ ; ] If the price is 75 cents and the advertising expenditures are e20000, sales would lie between and thousands of euros. Therefore, a sale of 25 thousands of euros would be feasible. Part II. Trend and seasonality. a. Regression model: S t = β 1 +β 2 P t +β 3 AE t +β 4 AE 2 t +β 5 time+β 6 dm1t+β 7 dm2 t +...+β 16 dm11 t +u t Model 4: OLS, using observations 1990: :06 (T = 270) Dependent variable: S const P AE s AE time dm dm dm dm dm dm dm dm dm dm dm

11 Mean dependent var S.D. dependent var Sum suared resid S.E. of regression R Adjusted R F (15, 254) P-value(F ) 2.6e 137 Log-likelihood Akaike criterion Schwarz criterion Hannan Quinn ˆρ Durbin Watson b. Estimated sales for December: Ŝ December = P t AE t AE 2 t time Estimated sales for August: Ŝ August = P t AE t AE 2 t time c. Test the statistical significance of the trend variable. : β 5 = 0 H a : β 5 0 t = ˆβ 5 0 ˆσ ˆβ5 Since t = > = t (254), the null hypothesis is rejected at the 5% significance level. Therefore, the trend variable is statistically significant. d. Test the statistical significance of seasonality. : β 6 =... β 16 = 0 H a : β i 0 F(, ) Given that F = < = F 0.05 (11, 254), the null hypothesis is not rejected at the 5% significance level. Therefore, seasonality is not a statistically significant variable. e. Estimation results of the augmented model. Model 5: OLS, using observations 1990: :06 (T = 270) Dependent variable: S const P AE s AE time dm dm dm dm dm dm dm dm dm dm dm s P

12 Two-sided test: Mean dependent var S.D. dependent var Sum suared resid S.E. of regression R Adjusted R F (16, 253) P-value(F ) 5.1e 144 Log-likelihood Akaike criterion Schwarz criterion Hannan Quinn ˆρ Durbin Watson : β 17 = 0 H a : β 17 0 t = ˆβ 17 0 ˆσ ˆβ17 Since t = > = t (253), the null hypothesis is rejected at the 5% significance level. Therefore, the relationship between sales and prices is not linear but uadratic. f. First, test whether seasonality is statistically significant in the augmented model. Since it is not statistically significant, specify the regression model: S t = β 1 + β 2 P t + β 3 P 2 t + β 4 AE t + β 5 AE 2 t + β 6 time + u t t = 1990 : 1,..., 2012 : 6 The estimation results are: Model 6: OLS, using observations 1990: :06 (T = 270) Dependent variable: S const P s P AE s AE time Mean dependent var S.D. dependent var Sum suared resid S.E. of regression R Adjusted R F (5, 264) P-value(F ) 1.9e 157 Log-likelihood Akaike criterion Schwarz criterion Hannan Quinn ˆρ Durbin Watson