Hypothesis Testing in Real Estate: Analyzing Sales Claims

Hypothesis Testing for Regional Real Estate Company 1 Hypothesis Testing for Regional Real Estate Company Tiarre Crawford Southern New Hampshire University

Hypothesis Testing for Regional Real Estate Company 2 Introduction Hired by the Reginal Real estate company I have been asked to analyze one of the company’s Pacific regions newly designed advertisement. Based off the advertisement the average square foot of home sales in the pacific region is $280. However, the salesperson makes claims that the average cost per square foot in the Pacific region is lower than $280. To help make sure that the salesperson claims are true I will randomly sample 750 homes in the Pacific regions. Using that data, I will then test the salesperson hypothesis. Introduction To generate my sample size of 750 I first started off by using the =Rand() function in Excel to randomize the total Pacific regions data. From there I choose 750 homes from the regions list then simply analyze the data I collected. Setup The population parameter being tested is the average cost per square footage of homes in the Pacific region, which is formulated as u=$280 per the salesperson advertisement .The null hypothesis ( H0) represents the average cost per square foot area in the Pacific region that is either greater than or equal to $280. The alternative hypothesis (HA) represents the mean cost per square foot area in the Pacific region that would be less than $280. I will test this hypothesis by using a left-tailed test which can be use that to tell if the alternative hypothesis is correct. Data Analysis Preparations Sample Mean $262 Target $280 Standard Error $5.79 Test Statistics -3.09625687 Degrees of Freedom 749 HO=280 $280

Hypothesis Testing for Regional Real Estate Company 3 Level of Significance A= 0.05 Ha=<280 -$280 P-Value 0.001016432 Test Statistics -3.096 The level of significance was 0.05. My test statistics of the sample population of 750 came out to be -3.096. After running my Left-Tailed test I calculated a p-value of 0.001 which is much smaller than my significance 0.05. By looking the Histogram, you can tell that the graph is skewed to the left representing smaller values of the data. So, the assumptions made by the salesperson that the actual average cost per square footage in the Pacific region is less than the average $280 was indeed true and meets the standards of the alternative hypothesis. Calculations [Calculate the p value using one of the following tests:

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Hypothesis Testing for Regional Real Estate Company 4 =T.DIST.RT([test statistic], [degree of freedom]) =T.DIST([-], [-3.096,749,1]= 0.001016432 =T.DIST.2T([test statistic], [degree of freedom]) I was able to create a normal curve graph using excel so I used my histogram data. Seeing that my histogram is being skewed to the left the p value will be at the very tailed end on the left. The test statistic being -3.096 would also fall to the left of the chart. Test Decision My p-value being that of 0.001 is less than (<) the significance level of 0.05, determines that there is supportive evidence that goes against the salesperson null hypothesis. Therefore, I will make the decision to reject the null hypothesis. Conclusion In conclusion, using left tailed test turned out to be the best test for finding out if the salespersons null hypothesis was indeed true or not. The initial alternative hypothesis stated that the average cost per square footage would be less than(-$280) the average cost of the null hypothesis($280). P-value 0.001 < significant level of 0.05 provided this to be true and of statistically significant therefore I rejected the null hypothesis. For that reason, the salesperson advertisement would be inaccurate and needs to be updated with the new data findings.

MAT 240 Module Five Assignment 1

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