7) The data used above was split into 2 samples and the following regression res

Question

7) The data used above was split into 2 samples and the following regression results were obtained from the split data SUMMARY OUTPUT Multiple R R Square Adjusted R Square Standard Error Observations 0.950 0.902 0.818 48.5 ANOVA MS 25347 108 F Significance F 0.0031 152083 16478 Coe 1 Stat 0.2 P-value Standard Error 184.7 Lower 95% 467.7 ents 95% 405.6 491.6 Land (acres) House Size(sq ft) 278.9 0.089 1.52 31.6 20.1 0.0 0.035 0.95 20.1 27.7 2.537 0.039 1.6 0.15 0.006 3.76 16.0 45.4 0.71 79.1 85.6 0.2 SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error 0.706 0485 32.27 ANOVA MS ance F 19998 8331 28330 3.201 1041 Tota Coefficients Standard Error Stat P-value Lower 95% U 95% 155.0 28.3 0.060 0.291 8.8 291.8 0.035 1.489 276.2 348.4 0.155 0.907 Land (acres) House Size(sq ft) 138.8 0.041 0.520 6.5 0.2 1456 0.8 0.183 0.591 0.2 19.7 17.6 0.2 0.9 0.8 46.3 b) Why is heteroscedasticity a problem? c) Based on a comparison of the two sets of output, does it appear that there is heteroscedasticity in the data set? Explain. Be sure to write down your null and alternative hypothesis, calculate the test statistic, and find your critical value (use a 10% level of significance).

Explanation / Answer

a) Heteroscedasticity : It refers to the data with unequal variability across a set of second predicter variable. While running a Regression technique, its mandatory to check if the data is Heteroscedasticity in nature or not(It may ruin the results)

b) It is a Problem because:

OLS will not give you the estimator with the smallest variance (i.e. your estimators will not be useful).
    Significance tests will run either too high or too low.
    Standard errors will be biased, along with their corresponding test statistics and confidence intervals.

c)

Null Hypothesis : Variable's of residual is constant

Alternate Hypothesis : Residual are not constatn i.e Heteroscedasticity is present

The Breush-Pagan test and the NCV test can be used to check the p value and decide weather to accept or reject the null hypothesis

Note : Since the data is not present further calculation cannot be performed

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