Implement raw datatables and computations, using both graphical and tabular methods of displaying data and results

The US economy is in crisis brought on by the financial markets, the deep slump of the housing, high unemployment, and the slowdown of consumer spending. This has caused problems for buyers and sellers in the real estate industry. Even in this downturn, there are still people looking for homes and this is a buyers’ market. Because of the slow US economy, there is an abundance of homes available for purchase brought on by high unemployment, and foreclosures.

Because of the abundance of homes on the market, buyers are able to purchase more amenities, such as homes with more than one bedroom, bathrooms, a pool, and with more square footage. Even in this stalled economy, those additional amenities all have an impact on the selling price of a home. The objective of this research using the Real Estate Data sets will determine if there is a difference in the meanprice per square foot between three and four bedroom homes. The sample will consist of 26 three bedroom and 26 four bedroom homes. This research will determine if the hypotheses should be accepted or rejected. Given the data mentioned above:

The verbal hypothesis the research will determine if there is a difference in the mean price per square foot between three and four bedroom homes.. The mean for the three bedroom homes is 2,165 with a standard deviationof 210. The mean for the four bedroom homes is 2,308 with a standard deviation of 251. The hypothesis will determine if the price square foot is equal or not equal between the two samples.

The numerical hypothesis is:

H0: µ1 = µ2

H1: µ1 ≠ µ2

Five-Step Hypothesis Test

As stated, the information given gives us

3 Bedrooms 4 Bedrooms

x1 = 2165 x2 = 2308

s1 = 210 s2 = 251

n1 = 26 n2 = 26

The hypothesis tests are:

H0: µ1 = µ2

H1: µ1 ≠ µ2

Therefore, the following represents data used and steps taken to verify the hypothesis using the five step process.

Descriptive statistics4 bedroom

Size

count 26

mean 2,307.69

sample variance 63,138.46

sample standard deviation 251.27

minimum 1900

maximum 2900

range 1000

population variance 60,710.06

population standard deviation 246.39

standard error of the mean 49.28

confidence interval 95.% lower 2,206.20

confidence interval 95.% upper 2,409.18

half-width 101.49

empirical rule

mean – 1s 2,056.42

mean + 1s 2,558.97

percent in interval (68.26%) 73.1%

mean – 2s 1,805.14

mean + 2s 2,810.24

percent in interval (95.44%) 92.3%

mean – 3s 1,553.87

mean + 3s 3,061.51

percent in interval (99.73%) 100.0%

Descriptive statistics 3 bedroom

Size

count 26

mean 2,165.38

sample variance 43,953.85

sample standard deviation 209.65

minimum 1600

maximum 2500

range 900

population variance 42,263.31

population standard deviation 205.58

standard error of the mean 41.12

confidence interval 95.% lower 2,080.70

confidence interval 95.% upper 2,250.06

half-width 84.68

empirical rule

mean – 1s 1,955.73

mean + 1s 2,375.04

percent in interval (68.26%) 65.4%

mean – 2s 1,746.08

mean + 2s 2,584.69

percent in interval (95.44%) 96.2%

mean – 3s 1,536.43

mean + 3s 2,794.34

percent in interval (99.73%) 100.0%

Calculations

Degrees of freedom = v = 50 n1 + n2 – 2

Two-tail critical value = t = (+-) 2.009 Appedix D table in text

Significance value = a = 0.05 95% Confidnece Level

Pooled varience sp2 is (n1 – 1)s12+(n2 – 1)s22 53550.5 2677525

sp2 = 53550.50 n1 + n2 – 2 50

Using sp2 the test statistic is x1-x2

(assuming equal variance) t = -2.2281 SQRT(sp2/n1 +sp2/n2) -2.228055212

Pooled standard deviation is sp

sp = 231.4098 SQRT(sp2) 231.4098096

(sp lies between s1 and s2 therefore the arithmetic is correct.)

The test statistic t = -2.2281 falls within the rejection region so we can reject the hypothesis of equal means.

P-value p = 0.0304 0.030401527

Using sp2 the test statistic is x1-x2

(assuming unequal variance) t = -2.2281 SQRT(s12/n1 +s22/n2) -143

64.18153341 44100 63001

Welch-Satterthwaite test [(s12/n1)+(s22/n2)]2 16968379 4119.269231

Degrees of freedom = v’ = 48 ((s12/n1)2/n1-1)+((s22/n2)2/n2-1) 349937.0415 1696.153846 2876937.87 115077.5148

Two-tail critical value = t = (+-) 2.011 2423.115385 5871488.167 234859.5267

Assumption Test Statistic d.f. Critical Value Decision

Case 1 (equal variances) t = -2.2281 50 t = (+-) 2.009 Reject

Case 2 (unequal variances) t = -2.2281 48 t = (+-) 2.011 Reject

Implement raw data tables and computations, using both graphical and tabular methods of displaying data and results

The solution provides step by step method for the calculation of ANOVA for real estate data set . Formula for the calculation and Interpretations of the results are also included. Interactive excel sheet is included. The user can edit the inputs and obtain the complete results for a new set of data.