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# Analyse using stat - Assignment Example

Summary
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…

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Download file to see previous pages 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. ...
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: ...Download file to see next pagesRead More
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