Hypothesis Testing for Regional Real Estate Company
Introduction
The Regional Real Estate Company needs an analysis of real estate data to inform critical decisions. One of the Company’s salespersons within the Pacific region has recently returned to the office with a newly designed advertisement. The advertisement claims that the average cost per square foot of his home sales is above the average cost per square foot in the Pacific Region. The company has hired me as a statistical analyst to test the claim before he can approve the advertisement for use. The claim indicates that the average cost per square foot of his home sales is $275. Therefore, the current analysis tests the hypothesis to offer a basis for the decision.
Setup
The population parameter under consideration includes the average cost of the rooms per square foot. The national mean listing price is 342,365. The population comprises the listing and average costs of houses per square foot data for the regions, including New England, Mid-Atlantic, South Atlantic, East Northcentral, Mountain and Pacific, and West North Central.
Taking the average cost per square foot of the Pacific region as x= 264.02, µ=275, the null and alternative hypothesis becomes;
Null hypotheses (H0): The salesperson’s average cost per square foot price is equal to the Pacific region average.
Alternative hypotheses (H1): The salesperson’s average cost per square foot price is greater than the Pacific region average.
That is: H0 : µ = 264.02 vs. H1 : µ > 264.02
Given the above hypothesis, the test is right-tailed since the aim is to test whether the person’s average cost per square foot is greater than the average in the Pacific region.
Data Analysis Preparations
Describing the sample
The sample contains 1001 data points within the Pacific. This is a large sample, with a mean of 264.01 and the Standard deviation of 161.756. The sample variance is 26165.16, while the median is 202.96. Degrees of freedom = n-1 = 1001-1= 1000.
Descriptive statistics of the sample
| House listing price | Cost per square foot | Square footage | |
| Mean | 489754.8133 | 264.0163927 | 1894.866847 |
| Standard Error | 8809.604291 | 5.112634168 | 7.565482218 |
| Median | 394547.5 | 202.9658424 | 1872 |
| Mode | 699050 | 206.1653338 | 1840 |
| Standard Deviation | 278723.4057 | 161.7565058 | 239.361145 |
| Sample Variance | 77686736889 | 26165.16716 | 57293.75771 |
| Kurtosis | 3.92881148 | 4.502129716 | 0.32374841 |
| Skewness | 1.931879442 | 2.086402642 | 0.081794213 |
| Range | 1470750 | 967.4515958 | 1486 |
| Minimum | 179300 | 103.8323782 | 1185 |
| Maximum | 1650050 | 1071.283974 | 2671 |
| Sum | 490244568.1 | 264280.4091 | 1896761.714 |
| Count | 1001 | 1001 | 1001 |
| Confidence Level(95.0%) | 17287.43079 | 10.03272183 | 14.84604142 |
Provide a histogram of the sample.
Specify whether the assumptions or conditions to perform your identified test have been met.
The assumption that should be met to ensure that the test is applicable is that the data points in the sample are independent (Salkind, 2015). This assumption is met since the data points are independent. The distribution is also normal.
Identify the appropriate test statistic, then calculate the test statistic and identify your significance level.
The test statistic that forms the basis of making the decision regarding the claim is a one-sample t-test obtained as;
Standard error = = = 5.112
= = –2.1483
The significance level is taken as 5%, 0.05, which forms the basis for making the decision.
Calculations
The appropriate test statistic;
;Standard error = = = 5.112
= = –2.1483
Calculate the p-value using the following test:
=T.DIST.RT([test statistic], [degree of freedom])
= T.DIST.RT(-2.1483, 1000)
= 0.9840
Using the normal curve graph, the distribution, the p-value obtained is larger than the critical value and would appear in the rejection region. The test statistic value is less than 0.05, which implies that the value falls in the acceptance region (Turner, 2020).
Test Decision
Discussion on how the p-value relates to the significance level, comparing the p-value and significance level, and making a decision to reject or fail to reject the null hypothesis.
The significance level forms the basis for the decision about rejecting or accepting the null hypothesis. The decision rule is upper-tailed, where the null hypothesis will be rejected if the obtained p-value is greater than the critical value. In this case, the p-value obtained is smaller than the critical value/significance level of (0.05), implying that we do not reject the null hypothesis and conclude that the salesperson’s average cost per square foot price is equal to the Pacific region average.
Conclusion
How does the test decision relate to the hypothesis, and are your conclusions statistically significant?
The analysis aimed to test whether the salesperson’s average cost per house is above the region’s average in order to validate or refute the claim in the research. The analysis indicates the p-value, which is the test statistic, is less than the level of significance selected for the analysis (p=0.05). Since the decision rule was an upper-tail, we accept the null hypothesis and conclude that the salesperson’s average cost per square foot price is equal to the Pacific region average. This implies that the average cost per square foot of his home sales is not $275, as the advertisement claims. Therefore, it is recommended to review the model used to arrive at the value of $275 and correct the biasedness before publishing the advertisement.
References
Salkind, N. (2015). Excel Statistics: A Quick Guide. Third Edition. SAGE Publications.
Turner, D. P. (2020). Sampling Methods in Research Design. Headache: The Journal of Head and Face Pain, 60(1), 8–12
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Question 
You have been hired by the Regional Real Estate Company to help them analyze real estate data. One of the company’s Pacific region salespeople just returned to the office with a newly designed advertisement. It states that the average cost per square foot of his home sales is above the average cost per square foot in the Pacific region. He wants you to make sure he can make that statement before approving the use of the advertisement. The average cost per square foot of his home sales is $275. In order to test his claim, you collect a sample of 1,001 home sales for the Pacific region.
Prompt
Design a hypothesis test and interpret the results using significance level α = .05.
Use the House Listing Price by Region document to help support your work on this assignment. You may also use the Descriptive Statistics in Excel and Creating Histograms in Excel tutorials for support.
Hypothesis Testing for Regional Real Estate Company
Specifically, you must address the following rubric criteria using the Module Five Assignment Template.
Setup: Define your population parameter, including hypothesis statements, and specify the appropriate test.
Define your population parameter.
Write the null and alternative hypotheses. Note: Remember, the salesperson believes that his sales are higher.
Specify the name of the test you will use.
Identify whether it is a left-tailed, right-tailed, or two-tailed test.
Identify your significance level.
Data Analysis Preparations: Describe sample summary statistics, provide a histogram and summary, check assumptions, and find the test statistic and significance level.
Provide the descriptive statistics (sample size, mean, median, and standard deviation).
Provide a histogram of your sample.
Describe your sample by writing a sentence describing the shape, center, and spread of your sample.
Determine whether the conditions to perform your identified test have been met.
Calculations: Calculate the p-value, describe the p-value and test statistic in regard to the normal curve graph, discuss how the p-value relates to the significance level, and compare the p-value to the significance level to reject or fail to reject the null hypothesis.
Determine the appropriate test statistic, then calculate the test statistic.
Note: This calculation is (mean – target)/standard error. In this case, the mean is your regional mean (Pacific), and the target is 275.
Calculate the p-value.
Note: For right-tailed, use the T.DIST.RT function in Excel, left-tailed is the T.DIST function, and two-tailed is the T.DIST.2T function. The degree of freedom is calculated by subtracting 1 from your sample size.
Choose your test from the following:
=T.DIST.RT([test statistic], [degree of freedom])
=T.DIST([test statistic], [degree of freedom], 1)
=T.DIST.2T([test statistic], [degree of freedom])
Using the normal curve graph as a reference, describe where the p value and test statistic would be placed.
Test Decision: Discuss the relationship between the p-value and the significance level, including a comparison between the two, and decide whether to reject or fail to reject the null hypothesis.
Discuss how the p-value relates to the significance level.
Compare the p value and significance level, and make a decision to reject or fail to reject the null hypothesis.
Conclusion: Discuss how your test relates to the hypothesis and discuss the statistical significance.
Explain in one paragraph how your test decision relates to your hypothesis and whether your conclusions are statistically significant.
Guidelines for Submission
Submit the completed Module Five Assignment Template as a Word document that includes your response and supportive charts.