Econometrics for Calculation of Selling Price of Home - Math Problem Example

Econometrics [Name] [Course] [Lecturer] [Unit Code] [Date of Submission] Q1. The figure below indicates a scatter graph of the selling price of home, dollars (sprice) versus age of home at time of sale, years (age). The scatter plot indicates a downward slope from left to right. The downward slope implies that the age of a home at time of sale and the selling price of a home are negatively related; as the age of a home at time of sale increases, the selling price decreases. Q2. Below is model 1: The above model (1) indicates that 1154.3 is the constant selling price of a home in dollars, the selling price of a home goes up by 10680 dollars if the living area increases by a unit square feet, a unit increase in the age of a home decreases the selling price of the home by 11.33 dollars, an extra single bedroom decreases the selling price of a home by 15552.4 dollars whereas an extra single bathroom decreases the selling price of a home by 7019.3 dollars. Livarea has a positive (+) sign which is in actuality in line with my expectations, Age has a negative (-) sign which is also in line with my expectations, Beds has a negative (-) sign which is not in line with my expectations, and, Baths has a negative (-) sign which is not in line with my expectations. Q3. Below is the hypothesis testing for each individual coefficient; : : Test statistic used is: : Computed value = = Using a 5% significance level and 1495 degrees of freedom, the critical value, 1.96. 1.96, hence, we reject the null hypothesis and conclude that at 5% significance level; the price of a home depends on living area. = Using a 5% significance level and 1495 degrees of freedom, the critical value, 1.96. 1.96, hence, we reject the null hypothesis and conclude that at 5% significance level; the price of a home depends on its age at time of sale. = Using a 5% significance level and 1495 degrees of freedom, the critical value, 1.96. -7.90 < -1.96, hence, we fail to reject the null hypothesis and conclude that at 5% significance level; the price of a home does not depend on the number of bedrooms. = Using a 5% significance level and 1495 degrees of freedom, the critical value, 1.96. - < -1.96, hence, we fail to reject the null hypothesis and conclude that at 5% significance level; the price of a home does not depend on the number of bathrooms. Q4. The difference in prices is given by - ( = 1131.64 – 1041 = $90.64. The 95% Confidence interval is given by; 90.64 ± = 90.64 ± = 90.64 ± 0.66 = (89.98, 91.3). The 95% Confidence interval implies that the difference in price of a home 2 years old and a home which is 10 years old lies between 89.98 and 91.3. Q5. Testing the overall significance of the model; : , :,, , , or all are nonzero Test statistic: Critical value = = = 3.37 Computed value: = 6.8795e+010 6.8795e+010 < 3.371, we fail to reject the null hypothesis at 1% significance level and conclude that all and/or each of the variables are not statistically significant. Q6. Hypothesis testing for increase in living area The difference in price is given by ( - = $21,360. Testing hypothesis; : : Test statistic: Critical value: = = 1.645 Computed value: = = 9647 Given that 9647 > 1.645, we reject the null hypothesis and conclude that at 5% significance level, the increase in price is more than $20,000. Q7. Below is model 2; F-test; : : Test statistic: Critical value: = = 3.85 For livareasquared the computed value = = 1.5e+13/1.3e+15 = 115.38 Given that 115.38 > 3.85, we reject the null hypothesis and conclude that at 5% significance level, there is enough evidence that living area squared is a significant predictor of the selling price of a home. For agesquared computed value = = 3.0e+14/1.3e+15 = 23.08 Given that 23.08 > 3.85, we reject the null hypothesis and conclude that at 5% significance level, there is enough evidence that age squared is a significant predictor of the selling price of a home. Q8. Below is model 3; The above model 3 indicates that 10.75 is the constant log of the selling price of a home. The log of the selling price of a home increases by 0.08 as a result of a unit increase in living area square feet. The log of the selling price of a home decreases by 0.001 due to a unit increase in living area square feet squared. The log of the selling price of a home decreases by 0.01 as a result of a unit increase in age of a house. The log of the selling price of a home increases by 0.0001 due to a unit increase in age squared. The log of the selling price of a home decreases by 0.08 as a result of a unit increase in the number of bedrooms. Q9. Using model 3; The log of the selling price of a home = 11.62. The antilog of 11.62 = 111301.72, hence, the price of a 10 year old house with a living area of 2,000 square feet, and three bedrooms is $111301.72. Question 10 Choosing between model 2 and model 3, model 3 is preferable. A log-linear model 3 is more sensitive and it is much easier to interpret. Also, generally, in statistics transformation of variables in a model produces more meaningful and accurate results and this is true for model 3. Logarithms make variations between large values less meaningful and add meaning to small differences. Also, logarithms help to reduce errors. Therefore, the use of the natural logarithm of the dependent variable in model 3 played a big role in ensuring that the errors are minimised. Read More