. A Home Owners' Association has determined that the average number of days a house was in the market for sale was 90 days, before it was sold. A real estate agency believes that in certain section of Long Island, the average number of days the houses remained in the market before sales was less than 90. It selected a random sample of 10 homes that were sold in this section in order to justify what it believes. The following data represents the number of days that each of these 10 homes stayed in the market before sale? 87, 95, 78, 83, 110, 75, 82, 92, 90, 80 At 0.01 level of confidence and assuming that population is approximately normal, is the real estate agency justified in its belief?
To determine if the real estate agency is justified in its belief that the average number of days a house stays in the market before sale in this section of Long Island is less than 90, we can conduct a one-sample t-test.
Step 1: State the hypotheses:
Null hypothesis (H0): The average number of days a house stays in the market before sale in this section is 90.
Alternate hypothesis (Ha): The average number of days a house stays in the market before sale in this section is less than 90.
Step 2: Set the significance level:
The significance level is given as 0.01.
Step 3: Compute the test statistic:
First, we need to calculate the sample mean and sample standard deviation for the 10 homes:
Sample mean (x̄): (87 + 95 + 78 + 83 + 110 + 75 + 82 + 92 + 90 + 80) / 10 = 88.2
Sample standard deviation (s): √[(87 - 88.2)^2 + (95 - 88.2)^2 + (78 - 88.2)^2 + (83 - 88.2)^2 + (110 - 88.2)^2 + (75 - 88.2)^2 + (82 - 88.2)^2 + (92 - 88.2)^2 + (90 - 88.2)^2 + (80 - 88.2)^2] / (10 - 1) = 9.62
Next, we can calculate the t-score using the formula: t = (x̄ - μ) / (s / √n), where μ is the hypothesized population mean (90), s is the sample standard deviation, and n is the sample size (10).
t = (88.2 - 90) / (9.62 / √10) = -1.66
Step 4: Determine the critical value:
Since the alternative hypothesis is one-tailed (less than), we need to find the critical value for a one-tailed t-test at a 0.01 significance level and 9 degrees of freedom. Looking up in a t-distribution table, the critical value is approximately -2.821.
Step 5: Compare the test statistic with the critical value:
Since the test statistic (-1.66) is greater than the critical value (-2.821), we fail to reject the null hypothesis.
Thus, there is not enough evidence to support the claim that the average number of days a house stays in the market before sale in this section of Long Island is less than 90.
Therefore, the real estate agency is not justified in its belief.
To test whether the real estate agency's belief that the average number of days houses remained in the market before sales is less than 90, we can conduct a one-sample t-test.
Step 1: Set up the hypotheses:
- Null hypothesis (H0): The average number of days houses remained in the market before sales is equal to 90.
- Alternative hypothesis (Ha): The average number of days houses remained in the market before sales is less than 90.
Step 2: Determine the significance level:
The significance level, also known as alpha (α), is given as 0.01.
Step 3: Calculate the test statistic:
The test statistic for a one-sample t-test is given by:
t = (x̄ - μ) / (s / √n)
where x̄ is the sample mean, μ is the population mean, s is the sample standard deviation, and n is the sample size.
Let's calculate the test statistic using the given data:
Sample mean (x̄) = (87 + 95 + 78 + 83 + 110 + 75 + 82 + 92 + 90 + 80) / 10 = 89.2
Population mean (μ) = 90 (given)
Sample standard deviation (s) = √[(87-89.2)^2 + (95-89.2)^2 + (78-89.2)^2 + (83-89.2)^2 + (110-89.2)^2 + (75-89.2)^2 + (82-89.2)^2 + (92-89.2)^2 + (90-89.2)^2 + (80-89.2)^2] / 9 = 8.064
Calculate the standard deviation of the sample mean (s/√n):
s/√n = 8.064 / √10 ≈ 2.55
Plug the values into the test statistic formula:
t = (89.2 - 90) / 2.55 ≈ -0.27
Step 4: Determine the critical value:
The critical value is obtained from a t-distribution table or calculator using the significance level and degrees of freedom (df = n - 1). For a one-tailed test at a 0.01 level of significance with 9 degrees of freedom, the critical value is approximately -2.821.
Step 5: Make a decision:
If the test statistic is less than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the test statistic (-0.27) is greater than the critical value (-2.821), so we fail to reject the null hypothesis.
Therefore, at a 0.01 level of confidence, we do not have enough evidence to support the real estate agency's belief that the average number of days houses remained in the market before sales is less than 90.