David wants to rent an unfurnished apartment for next semester. He took a random sample of 9 apartments adversised in the local Halifax paper, and recorded the rental rates. The rents (in $ per month) were:
500 650 600 505 450 550 515 495 640
a) Find a 90% confidence interval for the mean monthly rent for unfurnished apartments in this community.
b) You wish to test the null hypothesis that the true mean monthly rent for an unfurnished apartment is $500. What is the value of the test statistic?
c) A student at the University of Calgary carried out a similar study, but using a sample of 16 apartments. The value of the test statistic for that study was 2.71. What would be the p-value to test against the alternative that the mean rent is not equal to $500.
1.93
b) Your sample mean = 545
Your sd = 68.9553
Your standard error = 68.9553 / √ 9
SE = 68.9553/3
New standard error = 22.9851
=
t = (545-500) / 22.9851
t = 1.9578
c) n-1=15 = P( |t| > 2.71) =0.0161
a) Well, David certainly did his research. Now, let's calculate that confidence interval. From David's sample, the mean rent is $560 with a standard deviation of $71.14. Using a t-distribution with 8 degrees of freedom (n-1), we can calculate the margin of error. For a 90% confidence interval, the critical value is 1.86. So, the margin of error is 1.86 x (71.14/sqrt(9)) = $44.90. Therefore, the 90% confidence interval for the mean monthly rent is $515.10 to $604.90.
b) Ah, the null hypothesis trying to test its value. To find the test statistic, we'll use the formula:
test statistic = (sample mean - hypothesized mean) / (standard deviation / sqrt(sample size)).
Plugging in the values, we get (560 - 500) / (71.14 / sqrt(9)) = 2.83. So, the value of the test statistic is 2.83.
c) Now, we have a student from the University of Calgary joining the fun. With a test statistic of 2.71 and the alternative hypothesis that the mean rent is not equal to $500, we need to find the p-value. Unfortunately, I can't compute that for you right now. But remember, if the p-value is less than the significance level (usually 0.05), we reject the null hypothesis. Keep up the good work, stats enthusiasts!
a) To find a 90% confidence interval for the mean monthly rent for unfurnished apartments in this community, you can use the t-distribution since the sample size is small.
Step 1: Calculate the sample mean (x̄) and the sample standard deviation (s) of the rental rates.
Sample mean (x̄) = (500 + 650 + 600 + 505 + 450 + 550 + 515 + 495 + 640) / 9 = 551.11
Sample standard deviation (s) = √[ (Σ(xi - x̄)²) / (n-1) ]
= √[ (500 - x̄)² + (650 - x̄)² + ... + (640 - x̄)² / (9-1) ]
= √[ (500 - 551.11)² + (650 - 551.11)² + ... + (640 - 551.11)² / 8 ]
≈ 69.765
Step 2: Calculate the standard error (SE) of the mean, which is the standard deviation divided by the square root of the sample size.
SE = s / √n = 69.765 / √9 = 23.255
Step 3: Find the t-score corresponding to the desired confidence level. For a 90% confidence interval with 8 degrees of freedom (n-1 = 9-1 = 8), the t-score is approximately 1.860 (obtained from a t-distribution table or calculator).
Step 4: Calculate the margin of error (ME), which is the product of the standard error and the t-score.
ME = t-score * SE = 1.860 * 23.255 ≈ 43.191
Step 5: Calculate the lower and upper bounds of the confidence interval.
Lower bound = x̄ - ME = 551.11 - 43.191 ≈ 507.919
Upper bound = x̄ + ME = 551.11 + 43.191 ≈ 594.301
Therefore, the 90% confidence interval for the mean monthly rent for unfurnished apartments in this community is approximately $507.92 to $594.30.
b) To test the null hypothesis that the true mean monthly rent for an unfurnished apartment is $500, you can use a one-sample t-test.
Step 1: Calculate the t-score using the formula:
t-score = (x̄ - μ) / (s / √n)
where x̄ is the sample mean, μ is the hypothesized mean, s is the sample standard deviation, and n is the sample size.
t-score = (551.11 - 500) / (69.765 / √9) = 2.641
Therefore, the value of the test statistic is 2.641.
c) To find the p-value to test against the alternative that the mean rent is not equal to $500, you can use the t-distribution.
Step 1: Determine the degrees of freedom (df) for the t-distribution.
df = n - 1 = 16 - 1 = 15
Step 2: Use a t-distribution table or calculator to find the p-value corresponding to the t-score. For a t-score of 2.71 and df = 15, the p-value is approximately 0.014.
Therefore, the p-value to test against the alternative that the mean rent is not equal to $500 is approximately 0.014.
a) To find a 90% confidence interval for the mean monthly rent for unfurnished apartments in this community, we can start by calculating the sample mean and sample standard deviation.
Sample size (n) = 9
Rental rates: 500, 650, 600, 505, 450, 550, 515, 495, 640
Sample mean (x̄) = (500 + 650 + 600 + 505 + 450 + 550 + 515 + 495 + 640) / 9 = 552.78
Sample standard deviation (s) can be calculated using the formula:
s = √ [(Σ(x - x̄)^2) / (n - 1)]
where Σ represents the summation and (x - x̄) represents the difference between each observation and the sample mean.
Substituting the given values:
s = √ [((500-552.78)^2 + (650-552.78)^2 + (600-552.78)^2 + (505-552.78)^2 + (450-552.78)^2 +
(550-552.78)^2 + (515-552.78)^2 + (495-552.78)^2 + (640-552.78)^2) / (9 - 1)]
s = √ [(3,850,053.11) / (8)]
s = √ [481,256.64]
s ≈ 693.42
To find the confidence interval, we can use the formula:
CI = (x̄ - (Z * (s / √n)), x̄ + (Z * (s / √n)))
where CI represents the confidence interval, Z represents the standard score corresponding to the desired confidence level, and √n represents the square root of the sample size.
For a 90% confidence level, the Z score is approximately 1.645 (obtained from a standard normal distribution table).
CI = (552.78 - (1.645 * (693.42 / √9)), 552.78 + (1.645 * (693.42 / √9)))
CI = (552.78 - (1.645 * 231.14), 552.78 + (1.645 * 231.14))
CI ≈ (512.69, 592.87)
Therefore, the 90% confidence interval for the mean monthly rent for unfurnished apartments in this community is approximately $512.69 to $592.87.
b) To test the null hypothesis that the true mean monthly rent for an unfurnished apartment is $500, we can use the t-test statistic.
The t-test statistic is given by the formula:
t = (x̄ - μ) / (s / √n)
where x̄ is the sample mean, μ is the hypothesized population mean (in this case, $500), s is the sample standard deviation, and √n is the square root of the sample size.
Plugging in the given values:
t = (552.78 - 500) / (693.42 / √9)
t = (52.78) / (231.14 / 3)
t = 52.78 / 77.05
t ≈ 0.684
Therefore, the value of the test statistic is approximately 0.684.
c) To find the p-value for the alternative hypothesis that the mean rent is not equal to $500, we can use the t-distribution table and the given test statistic.
Since the test statistic (2.71) is positive, we need to find the probability of obtaining a more extreme result in either tail of the distribution.
The degrees of freedom (df) for this study is (n - 1), which is 16 - 1 = 15.
Looking up the critical value in the t-distribution table with 15 degrees of freedom, we find that for a two-tailed test at the significance level of 0.05 (p-value of 0.05), the critical value is approximately 2.131.
Since the test statistic (2.71) is greater than the critical value (2.131), we can conclude that the p-value is less than 0.05.
Therefore, the p-value to test against the alternative that the mean rent is not equal to $500 is less than 0.05 (significant at the 0.05 level).