We would like to test the accuracy of a no-change benchmark for output growth relative to a VAR model for the past 40 years. Below is the result of the test of equal predictive accuracy measured in terms of mean squared forecast error. What do you conclude, which model is better? Comment. Here the 'fe_diff' is defined as squared forecast errors implied by the no-change forecast minus the squared forecast errors implied by the VAR. reg = 1m(fe_diff ~ 1) summary(reg) Call: 1m formula = fe_diff ~ 1 Residuals: Min 1Q Median 3Q Max -5.7154 -0.2042 -0.0621 0.1989 5.0282 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 0.10293 0.08623 1.194 0.234 Residual standard error: 1.091 on 159 degrees of freedom
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The regression results are attached. Residuals: Min 1Q Median 3Q Max -0.48179 -0.24662 -0.00726 0.22012 0.51987 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -4.504e-01 1.019e+01 -0.044 0.965202 AGST 6.012e-01 1.030e-01 5.836 1.27e-05 *** HarvestRain -3.958e-03 8.751e-04 -4.523 0.000233 *** WinterRain 1.043e-03 5.310e-04 1.963 0.064416 . Age 5.847e-04 7.900e-02 0.007 0.994172 FrancePop -4.953e-05 1.667e-04 -0.297 0.769578 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 0.3019 on 19 degrees of freedom Multiple R-squared: 0.8294, Adjusted R-squared: 0.7845 F-statistic: 18.47 on 5 and 19 DF, p-value: 1.044e-06 Interpret the results.
Madhur L.
OLS Regression Results Dep. Variable: mpg R-squared: 0.827 Model: OLS Adj. R-squared: 0.814 Method: Least Squares F-statistic: 64.36 Date: Wed, 12 Feb 2020 Prob (F-statistic): 5.33e-11 Time: 16:51:29 Log-Likelihood: -70.390 No. Observations: 30 AIC: 146.8 Df Residuals: 27 BIC: 151.0 Df Model: 2 Covariance Type: nonrobust coef std err t P>|t| [0.025 0.975] Intercept 37.3942 1.661 22.518 0.000 33.987 40.802 wt -3.9512 0.682 -5.796 0.000 -5.350 -2.552 hp -0.0307 0.010 -2.999 0.006 -0.052 -0.010 Omnibus: 4.039 Durbin-Watson: 1.819 Prob(Omnibus): 0.133 Jarque-Bera (JB): 2.994 Skew: 0.771 Prob(JB): 0.224 Kurtosis: 3.140 Cond. No. 565. Warnings: [1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
Sri K.
The model output is shown below: Call: lm(formula = df$oral_fluency ~ df$reading_time) Residuals: Min 1Q Median 3Q Max -19.2676 -11.3875 0.6343 10.6530 22.1715 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -21.6094 18.9685 -1.139 0.2752 df$reading_time 2.0408 0.9238 2.209 0.0457 * --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 14.49 on 13 degrees of freedom Multiple R-squared: 0.2729, Adjusted R-squared: 0.217 F-statistic: 4.88 on 1 and 13 DF, p-value: 0.04573 Use the information provided in the model output to calculate by hand the 95% confidence interval of 'b' the estimate of the predictor 'reading_time'. What is the lower bound of the 95% confidence interval of 'b'? Does this suggest that 'b' is significantly different from 0?
Adi S.
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