1. In real estate, it is common knowledge that where the house is located is ext
ID: 3258154 • Letter: 1
Question
1.
In real estate, it is common knowledge that where the house is located is extremely important (many people summarize this by saying the three most important factors in home sales are location, location, and location). Many characteristics of a house are tied to its location, one of which is age. Some neighborhoods are considered "old" while others are considered "new," with new homes generally selling at a higher price.
One way to compare "old" and "new" neighborhoods is to examine the difference in the mean age of houses in those neighborhoods. Ben, a real estate agent, is interested in determining whether the mean age of houses located in the Peaceful Pines neighborhood (Group 1) differs from the mean age of houses in the Whispering Willows neighborhood (Group 2). To examine this question, he will perform a hypothesis test using a sample of six houses from each neighborhood. The data (representing the ages in years of each home) are as shown below:
Peaceful Pines
41, 49, 43, 46, 48, 43
Whispering Willows
54, 43, 39, 46, 42, 43
Assuming Case 1 (equal population standard deviations), calculate the test statistic for the hypothesis test that Ben would be interested in conducting, to two decimal places.
2.
The manager of a chain of coffee shops is interested in determining if there has been a decrease in the mean drive-through service time at his store on weekdays between 6 and 10 a.m. since the introduction of a new ordering technology that allows customers to see as well as hear the drive through attendant on a screen next to the menu. Here, "service time" is the total time, in seconds,between when the customer finishes ordering and when he or she receives the order from the window.
The manager goes through the point-of-sale data and randomly selects a sample of 46 orders from the weekday 6-10 a.m. shift before the time before the technology was implemented (Group 1)and 46 service times for this time period after it was implemented (Group 2). The mean service time before implementation was 286.7 with a standard deviation of 28.6, while the mean service time after implementation was 212 with a standard deviation of 28. Calculate the test statistic for testing the hypothesis described above, to two decimal places. Take all calculations toward the answer to four (4) decimal places. Assume the population standard deviations are equal (Case 1).
Explanation / Answer
Q.1 Null Hypothesis : H0 : Age of peaceful pines houses are equal to age of WHispering willows.PP = WW
Alternative Hypothesis : Ha : Age of peaceful pines are not equal to age of whispering willows. PP WW
Mean of Age of peaceful pines house xPP= 45
Standard deviation of peaceful pines house sPP= 3.16
Mean AGe of WHispering Willows xWW= 44.5
Standard deviation of Whispering willows sWW= 5.17
Pooled standard deviation sp = sqrt (spp2 + sww2 /2 ) = sqrt (3.162 + 5.172 /2) =4.2845
Test Statistic
t = (xpp - xww) /sp sqrt (2/n) = (45 - 44.5)/ 4.2845 * sqrt(2/6) = 0.5/ 2.4737 = 0.2021
Q.2 Here difference between service time is d = 30 sec
H0 : d = 30 sec
Ha : d > 30 sec
As number of observations are same for both samples
Pooled standard deviation sp = sqrt [(s12 + s22 )/2] = sqrt [ (28.62 + 282)/2 ] = 28.3016
Test statistic:
t = (xbefore impl - xafter imple) -d / sp sqrt (2/n) = (286.7 - 212) - 30 / 28.3016 * sqrt (2/46) = 44.7/ 5.9013 = 7.5746
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.