The purpose of this analysis is to assist in confirming or denying the advertisement put forth...

 

  
 
 
 

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The purpose of this analysis is to assist in confirming or denying the advertisement put forth by the salesperson of the Region Real Estate Company that the average cost per square foot of his home sales is $275 above the average cost per square foot in the Pacific region. Setup The population parameter for this hypothesis is based on 1,001 homes in the Pacific region, using the price per square foot. A hypothesis is a statement that makes a claim about the parameters of two competing hypothesis about a population. A null hypothesis Ho: #2275. An alternative hypothesis Ha is a statement that contradicts the null hypothesis, proving that true value of the population parameter is not the same as the hypothesized value or the parameters for the two populations are different, for this would be Ha: <275. The salesperson is to prove that (population mean) is less than $275. Due to the alternative hypothesis, our test will be left- tailed test. Data Analysis Preparations The sample used is 1,001 homes from the Pacific region contains data collected containing the homes listing price, cost per square foot, and square footage. Cost per square foot Mean 264.016393 Standard Erro 5.11263417 Median Mode 202.965842 206.165334 Standard Devi 161.756506 Sample Variar 26165.1672 Kurtosis 4.50212972 Skewness 2.08640264 967.451596 103.832378 1071.28397 264280.409 Range Minimum Maximum Sum Count 1001 Frequency 400 350 300 250 200 150 100 50 (3104,5161) (3161, 521) (25275 (5275,537) Cost post square foot in the Pacific Region 53325109 (53,5446) (5446,5101) (3501140 (55405417) bos 20 Cost S per square foot (5474,5731) TEL (578,5845) (545) NGA ZON (HEATS WAS BLOK, SLAT The distribution for the Pacific region cost per square foot is skewed to the right, with the center about $203 (median) with the majority of the listing between $161 to $218 square foot with several outliers with the largest being about $1,071. Normality is met because the sample size is 1,001 which allows us to say the CLT has been met and we can use the normal distribution. The random condition has also been met, due to, we were told the 1,001 listing were based on a random sample. It is reasonable to assume that the pacific region has more than 10,010. A test statistic is a value calculated from sample data during hypothesis testing that measures the degree of agreement between the sample statistic and the parameter in the null hypothesis. The t-statistic is computed by t= (mean-target) / standard error, 1= (264.02-275)/5.113= -2.148. Using the "/" value, a "p" value can be found to identify statistical significance. The significance level for this scenario is a = 0.05. The significance value of 0.05 is the probability of rejecting the null hypothesis when it is true, giving a 5% risk of concluding that a difference exists when there is no actual difference. Calculating a "p" for this scenario requires the use of a left tailed "7" distribution. The "p" value, - TDIST(-2.148,1001,1) because -2.148 is the "f" statistic, 1,000 is the degree of freedom = (1,001-1) and 1 is always the cumulative number at the end of the excel function for the left tailed value. The "p" value is calculated at 0.016. Since our "p" value (0.016) is less than 0.05, we reject the null hypothesis and having convincing evidence that average square foot for the Pacific region is less than $275 square foot. Below is an example of how this would be on a 0.05, the "p" value would fall left of the 0.05 mark. Probability Density 0.40 0.35 0.30 0.25 0.20 0.15 0.10- 0.05 0.00 H-30 a = 0.05 Calculations H-20 68-95-99.7 Rule HIO 68.27% 95.45% 99.73% H+0 μ+20 H+30 Test Decision The conclusion is that the null hypothesis is rejected because p=0.016<0.05 (the significance level). Because the "p" value is less than the specified significance level, two possibilities exist; one: the null hypothesis is true, and the observed data is relatively unusual with a sample statistic that is extremely due to chance or, two: the null hypothesis is false, and the alternative hypothesis provides a more reasonable explanation for the population parameter. The "p" value (0.016) means there is a 1.6% probability that a result with an extreme deviation from the null hypothesis is due to chance. This low "p" value shows a statistically significant decision has been made to reject the null hypothesis. Conclusion The hypothesis testing completed for this scenario would allow the salesperson to advertise the average cost per square foot of his home sale is above average. The conclusion is statistically significant because of a very low "p" value (0.016). The salesperson claim that his home sales sell at a higher-than-average cost per square foot should be accepted and he should continue with his advertisement, there is sufficient evidence and statistical significance to endorse his claim. [Explain in one paragraph how your test decision relates to your hypothesis and whether your conclusions are statistically significant.] Introduction [Include in this section a brief overview, including the purpose of this analysis.] Setup [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 and identify whether it is a left-tailed, right- tailed, or two-tailed test.] Data Analysis Preparations [Describe the sample.] I [Provide the descriptive statistics of the sample.] [Provide a histogram of the sample.] [Specify whether the assumptions or conditions to perform your identified test have been met.] [Identify the appropriate test statistic, then calculate the test statistic and identify your significance level.] Calculations [Calculate the p value using one of the following tests: -T.DIST RT([test statistic], [degree of freedom]) =T.DIST([test statistic], [degree of freedom], 1) =T.DIST.21([test statistic], [degree of freedom]) 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.] [Use the normal curve graph as a reference to describe where the p value and test statistic would be placed.] Test Decision [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.] I [Explain in one paragraph how your test decision relates to your hypothesis and whether your conclusions are statistically significant.] Conclusion The purpose of this analysis is to assist in confirming or denying the advertisement put forth by the salesperson of the Region Real Estate Company that the average cost per square foot of his home sales is $275 above the average cost per square foot in the Pacific region. Setup The population parameter for this hypothesis is based on 1,001 homes in the Pacific region, using the price per square foot. A hypothesis is a statement that makes a claim about the parameters of two competing hypothesis about a population. A null hypothesis Ho: #2275. An alternative hypothesis Ha is a statement that contradicts the null hypothesis, proving that true value of the population parameter is not the same as the hypothesized value or the parameters for the two populations are different, for this would be Ha: <275. The salesperson is to prove that (population mean) is less than $275. Due to the alternative hypothesis, our test will be left- tailed test. Data Analysis Preparations The sample used is 1,001 homes from the Pacific region contains data collected containing the homes listing price, cost per square foot, and square footage. Cost per square foot Mean 264.016393 Standard Erro 5.11263417 Median Mode 202.965842 206.165334 Standard Devi 161.756506 Sample Variar 26165.1672 Kurtosis 4.50212972 Skewness 2.08640264 967.451596 103.832378 1071.28397 264280.409 Range Minimum Maximum Sum Count 1001 Frequency 400 350 300 250 200 150 100 50 (3104,5161) (3161, 521) (25275 (5275,537) Cost post square foot in the Pacific Region 53325109 (53,5446) (5446,5101) (3501140 (55405417) bos 20 Cost S per square foot (5474,5731) TEL (578,5845) (545) NGA ZON (HEATS WAS BLOK, SLAT The distribution for the Pacific region cost per square foot is skewed to the right, with the center about $203 (median) with the majority of the listing between $161 to $218 square foot with several outliers with the largest being about $1,071. Normality is met because the sample size is 1,001 which allows us to say the CLT has been met and we can use the normal distribution. The random condition has also been met, due to, we were told the 1,001 listing were based on a random sample. It is reasonable to assume that the pacific region has more than 10,010. A test statistic is a value calculated from sample data during hypothesis testing that measures the degree of agreement between the sample statistic and the parameter in the null hypothesis. The t-statistic is computed by t= (mean-target) / standard error, 1= (264.02-275)/5.113= -2.148. Using the "/" value, a "p" value can be found to identify statistical significance. The significance level for this scenario is a = 0.05. The significance value of 0.05 is the probability of rejecting the null hypothesis when it is true, giving a 5% risk of concluding that a difference exists when there is no actual difference. Calculating a "p" for this scenario requires the use of a left tailed "7" distribution. The "p" value, - TDIST(-2.148,1001,1) because -2.148 is the "f" statistic, 1,000 is the degree of freedom = (1,001-1) and 1 is always the cumulative number at the end of the excel function for the left tailed value. The "p" value is calculated at 0.016. Since our "p" value (0.016) is less than 0.05, we reject the null hypothesis and having convincing evidence that average square foot for the Pacific region is less than $275 square foot. Below is an example of how this would be on a 0.05, the "p" value would fall left of the 0.05 mark. Probability Density 0.40 0.35 0.30 0.25 0.20 0.15 0.10- 0.05 0.00 H-30 a = 0.05 Calculations H-20 68-95-99.7 Rule HIO 68.27% 95.45% 99.73% H+0 μ+20 H+30 Test Decision The conclusion is that the null hypothesis is rejected because p=0.016<0.05 (the significance level). Because the "p" value is less than the specified significance level, two possibilities exist; one: the null hypothesis is true, and the observed data is relatively unusual with a sample statistic that is extremely due to chance or, two: the null hypothesis is false, and the alternative hypothesis provides a more reasonable explanation for the population parameter. The "p" value (0.016) means there is a 1.6% probability that a result with an extreme deviation from the null hypothesis is due to chance. This low "p" value shows a statistically significant decision has been made to reject the null hypothesis. Conclusion The hypothesis testing completed for this scenario would allow the salesperson to advertise the average cost per square foot of his home sale is above average. The conclusion is statistically significant because of a very low "p" value (0.016). The salesperson claim that his home sales sell at a higher-than-average cost per square foot should be accepted and he should continue with his advertisement, there is sufficient evidence and statistical significance to endorse his claim. [Explain in one paragraph how your test decision relates to your hypothesis and whether your conclusions are statistically significant.] Introduction [Include in this section a brief overview, including the purpose of this analysis.] Setup [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 and identify whether it is a left-tailed, right- tailed, or two-tailed test.] Data Analysis Preparations [Describe the sample.] I [Provide the descriptive statistics of the sample.] [Provide a histogram of the sample.] [Specify whether the assumptions or conditions to perform your identified test have been met.] [Identify the appropriate test statistic, then calculate the test statistic and identify your significance level.] Calculations [Calculate the p value using one of the following tests: -T.DIST RT([test statistic], [degree of freedom]) =T.DIST([test statistic], [degree of freedom], 1) =T.DIST.21([test statistic], [degree of freedom]) 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.] [Use the normal curve graph as a reference to describe where the p value and test statistic would be placed.] Test Decision [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.] I [Explain in one paragraph how your test decision relates to your hypothesis and whether your conclusions are statistically significant.] Conclusion