Problem Set #2

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Boston University *

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QM222

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Economics

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Feb 20, 2024

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Problem Set # 2: Due October 11 @ 11PM Please use this document to type in your answers (in bold typeface) and then save and upload on your section site). Note : There are 2 Parts to this Pset, each using a different data set provided to you . Part 1 : A large mortgage company, House4U, has been tasked by the CEO to reduce the amount of time it takes to approve new applications for mortgage loans. The approval process must go through formal underwriting, which includes a review of the application and obtaining and reviewing the following: credit bureau report , pay stubs, copies of tax returns, additional work history information, bank statements and other asset information, title search, and the property appraisal. You are hired as a consultant and begin by collecting the data needed to begin your analysis. You select a random sample of 36 mortgage loans which were approved during the last month. You calculate the sample mean and standard deviation in Excel ( Use the dataset called “Approval_time.xlsx in the Problem Set folder ). 1. You want to know about the average time to approval for ALL House4U loans (not just the sample), but you don’t have time to collect data on all loans. With 95% confidence, what is the average time it takes House4U to approve mortgage loan applications? Show you work. We found the sample mean to be 54.1 and the standard deviation to be 24. Then we calculated the standard error to be 24/√36= 4. With 95%, we can conclude that the average time it takes House4U to approve mortgage loan applications is between 46.1 and 62.1, since they are 2 standard errors away from the average. 2. What is the approximate probability that House4U loans have an average time to approval of 46 days or less? Explain your reasoning using distance from mean. There is about a 2.5% chance that the average time of approval is 46 days or less. Since 46.1 is the start of the 95% confidence interval, everything less than 46.1 falls out of one side of the interval, thus half of 5% is 2.5%. 3. Assume that a sample of 49 approved mortgages was used, instead of 36, and the same average # days and standard deviation for days until approval was found. In this case, do you think the probability of an average mortgage loan approval time of 46 days or less would be greater or lesser than your answer to c above? Explain your answer in 1-2 sentences. I think the probability of the average being 46 days or less would be less likely than before. The new standard error would be 2.67 and within 2 intervals, the average would be 48.76 to 59.44. Therefore, 46 falls even more outside the 95% confidence interval. 4. Go back to the original data (36 loans in sample data). Do you think that the probability that House4U has a mean approval time of 46 days or less would be greater or lesser if the
standard deviation was only 18 days rather than 24 days (using n=36)? Explain why or why not in 1-2 sentences. The probability of the mean approval time being 46 days or less if the standard deviation was 18 would still be less than if the original standard deviation was 24. With a standard deviation of 18, the new standard error would be 3, and within 2 confidence intervals the average would be from 48.1 to 60.1. The probability of the average being 46 or less would be less than 2.5%. 5. One might believe that the time to approve a loan follows a normal distribution. However, when interest rates are low there are many more applicants. As a result the distribution of time to close each application is bimodal and looks like the diagram below. Probability Days until approval If the average remains 54 days until approval and the standard deviation remains 24 days, would your answer to question 2 change? Explain in 1-2 sentences. Yes, my answer in question 2 would change. Since we can no longer say that the sample is normally distributed, we cannot use the central limit theorem. Instead, we could approximate the possibility by using different percentiles. 6. You decided to look at older sample data (before interest rates dropped so low) and found that the approval time for individual mortgage loans followed a normal distribution. Using the same mean of 54 days and standard deviation of 24 days what is the approximate probability that an individual application for a mortgage will take 78 days or more? Because the sample is normally distributed, we can use the standard deviation to measure the variability of the data. Within one confidence interval of 68%, the mean is from 30 to 78. Therefore, on the greater side of the distribution, there would be a 16% chance that an individual application for a mortgage will take 78 or more days.
7. Suppose that the National Association of Mortgage Brokers published a report stating that the mean time for approval of all mortgage loans in the US is µ = 50 days . The CEO of House4U is concerned that his firm’s average approval time is different from the national average a. State the Null and Alternative Hypothesis The null hypothesis is that the mean time for approval of all mortgage loans in the US is 50 days and the alternative hypothesis is that the mean is not 50 days. b. With 95% confidence, does the data from the random sample of House4U approval times support the CEO’s concern? Show your calculations and conclude with your finding. We found the sample mean of House4U approval times to be 54.1 days and within 2 confidence intervals, the mean was from 46.1 to 62.1. Since 50 days falls in between this interval, we can accept the null hypothesis with 95% confidence. Part 2 : Use the Brookline Condos (2009-2010) data in Problem Set folder. Run three different regressions to predict price. Type each regression equation, with the standard errors in parentheses underneath each estimated coefficient. Then interpret the coefficient on the explanatory variable for each equation below. 8. Regression A: Number of Bedrooms a. Regression equation: y=118492.68+191862.13(# of bedrooms) standard errors= 15042.79, 6656.48 b. Interpretation of coefficient: The base price of the home without any bedrooms is $118492.68. For every bedroom, the price increases by $191862.13. 9. Regression B: Number of Rooms a. Regression equation: y=-40960.09+116625.29(# of rooms) standard errors= 19922.68, 3980.35 b. Interpretation of coefficient: Since the base price is negative, there wouldn’t be any costs until you have rooms. For every room, the price increases by $116625.29. 10. Regression C: Highrise. Note: You need to create an indicator, “dummy”, variable that equals 1 if the condo is in a High Rise building, from Building Style in Column F a. Regression equation: y=541875.45-129145.65(highrise)
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standard errors= 8754.12, 19565.78 b. Interpretation of coefficient: The price starts at $541875.45, but purchasing in a highrise building means you are paying $129145.65 less. 11. Which of the three explanatory variables’ coefficients (from Regressions A, B and C, above) are statistically significant at the 5% level? Explain. I think all three of the coefficients are statistically significant because they all have a P- value that is less than 5%. 12. Can you reject the hypothesis that there is no price difference between condos in high rise buildings and in other types of structures (at the 5% level of significance)? State both the null and alternative hypothesis and explain your answer. The null hypothesis would be that there is no price difference between condos in high rise buildings and other structures. The alternative hypothesis would be that there is indeed a price difference between high rise buildings and other types of structures. Regression C focuses on the high rise building type, and since the P-value is less than 5%, there is less than a 5% probability that the null is correct. Thus, we can reject the null hypothesis and accept the alternative hypothesis. 13. Using each regression, make a prediction of the sale price for a 5 room condo, with 3 bedrooms, in a high rise building. Show your calculations . a. Prediction Regression 1: y=118492.68+191862.13(3) y= $694078.39 b. Prediction Regression 2: y=-40960.09+116625.29(5) y= $542166.36 c. Prediction Regression 3: y=541875.45-129145.65 y= $412729.8 14. Your intern collects more data indicating whether each house featured hardwood floors. She created a new indicator variable that equals 1 if the house has hardwood floors and 0 otherwise. She runs a regression and gets the following result: Sale Price = 425,000 + 200,000*HardwoodFloors a. What is the average sale price of a home with hardwood floors? Show your calculations 425000+200000(1)= $625000 b. What is the average sale price of a home without hardwood floors? Show your calculations.
425000+200000(0)= $425000 15. Imagine that your intern had instead defined the variable to be NoHardwoodFloors and set it equal 1 if the house does not have hardwood floors. She runs a new regression that looks like this: Sale Price = b 0 + b 1 *NoHardwoodFloors a. What is the value of b 0 in the new regression? ___ 625000 ___________ b. What is the value of b 1 in the new regression? _____ -200000 _________ 16. An additional room (of any type) adds an average of $125,000 to the sale price in Back Bay (Boston). A Brookline politician claims that real estate prices in Brookline do not respond the same way to changes in the number of rooms as in Back Bay (Boston). a. State both the Null and Alternative Hypothesis Null: Prices in Brookline do not respond the same way to changes in the number of rooms as in Back Bay Alternative: Prices in Brookline respond the same way to changes in the numbers of rooms as in Back Bay. b. Does your Regression B model (# of rooms) support his claim with 95% confidence? Show your work. 116625.29+2(3980.35)= $124585.99 116625.29-2(3980.35)= $108664.59 Within two standard errors, the average price for an additional room in Brookline is from 108664.59 to 124585.99. Since $125000 is not within that interval, we can reject the null hypothesis with 95% confidence.