Statistical Data Analysis - Assignment Example

1) Present histograms of price and dist variables in levels and in natural logarithms with normal density plots fitted. Comment on the distribution of the variables in levels and logs. Price variable shows a normal distribution with mean, $22511.51. However dist variable is not normally distributed. It is skewed to the left. Accordingly the weighted distance to 5 employment centers in most of the houses are less than 6. Mean of dist variable is 3.795. Distributions of log price and log dist variables are normal. Conduct a one-sample Kolmogorov-Smirnov test for normality at 5% significance. Compare and contrast the findings from the histograms and from the tests with level and logarithmic specifications. H0: sample is not distributed normally H1: sample is distributed normally Decision: Probability value is less than 0.05significance level. Therefore we reject the null hypothesis and accept the alternative hypothesis. i.e. price variable is normally distributed Probability value is less than 0.05significance level. Therefore reject the null hypothesis and accept the alternative hypothesis. i.e. log price variable is also normally distributed Thus the results of histogram and Kolmogorov-Smirnov test are consistent. Variable Description Combined K-S Value 1. dist Weighted distance to 5 employment centers 0.000 2. ldist Logarithm of weighted distance to 5 employment centers 0.016 H0: sample is not distributed normally H1: sample is distributed normally Decision: Probability value is less than 0.05significance level. Therefore reject the null hypothesis and accept the alternative hypothesis. i.e. dist variable is normally distributed Probability value is less than 0.05significance level. Therefore reject the null hypothesis and accept the alternative hypothesis. i.e. log-dist variable is also normally distributed Histogram showed the distribution of dist variable as skewed to the left. However Kolmogorov-Smirnov test yields a normal distribution. Thus the results of two tests are not consistent. 2) Conduct paired-sample tests at 5% significance for equality of mean value of the house price under the following scenarios: (i) houses below and above the average value for rooms; To compare the mean house price values under the above scenario unpaired, two-sample t-test with equal variances can be used. Mean value of rooms in the given sample is 6.28. Accordingly there are 278 houses with number of rooms bellow the average (sample A) and 228 houses with number of rooms above the average (sample B). Hence the total observations in two samples are different. Therefore we have to conduct unpaired two sample t-test. Mean Difference -8918.208 t-statistics -12.3611 P-value H0: diff=0 0.0000 H1: diff < 0 0.0000 H2: diff > 0 1.0000 Decision: Probability value of H0 is less than 0.05 significant level. Therefore reject H0 which states there is no statistically significant difference between the mean price of houses having rooms less than average and more than average. (ii) houses below and above the average value for nox; Average nox value of the given data set is 5.549. There are 292 observations in the above average category while there are 214 observations in the below average category. Therefore unpaired, two-sample t-test with equal variances can be used. Mean Difference 6199.578 t-statistics 7.9261 P-value H0: diff=0 0.0000 H1: diff < 0 1.0000 H2: diff > 0 0.0000 Decision: Probability value of H0 is less than 0.05 significant level. Therefore we reject the null hypothesis which states mean price of houses which are situated in lower nitrous oxide levels are not statistically different from those houses situated in higher nitrous oxide areas. (iii) Houses below and above the average value for crime. Average number of crimes committed per capita is 3.611in the given data set. Accordingly there are 128 and 378 numbers of observations in the above and below average categories respectively. Therefore, unpaired, two-sample t-test with equal variances can be used. Mean Difference 8471.173 t-statistics 9.8062 P-value H0: diff=0 0.0000 H1: diff < 0 1.0000 H2: diff > 0 0.0000 Decision: Probability value of H0 is less than 0.05 significant level. Therefore we reject the null hypothesis which states, mean price of houses which are situated in low crime areas are not statistically different compared to the high crimes areas. Do you think the scenarios and the tests results are useful for testing meaningful hypotheses about house prices? Yes. The above analysis shows the significance of individual variables in determining the house prices. Accordingly house specific characteristics such as the number of rooms available, as well as location specific characteristics such as nitrous oxide level in atmosphere and per capita crimes recording from the area affect the housing market. Therefore, demand and supply curve of housing markets needs to incorporate all the above factors. 3) Specify and estimate 3 different linear models of house prices. Variable Model (1) Model (2) Model (3) Dependent Explanatory price price price crime -122.5*** -109.0** -104.5** (33.85) (34.57) (36.26) nox -1853.2*** -930.6* (365.3) (375.5) rooms 4062.7*** 4232.1*** 4930.8*** (416.9) (425.8) (434.7) dist -1231.4*** -698.5*** -1017.2*** (168.0) (134.3) (178.4) radial 293.2*** 268.5*** 178.7* (65.95) (67.38) (69.38) proptax -122.4*** -160.9*** -146.9*** (34.33) (34.30) (36.72) stratio -1102.7*** -918.6*** (125.9) (123.5) lowstat -519.6*** -572.3*** -558.3*** (47.62) (47.62) (50.89) _cons 41534.0*** 27192.5*** 12311.1** (5065.2) (4306.0) (4090.9) N 506 506 506 adj. R-sq 0.711 0.696 0.667 Standard errors in parentheses * p Read More