In Hamilton County, Ohio the mean number of days needed to sell a home is 86 days (Cincinnati Multiple Listing Service, April, 2012). Data for the sale of 40 homes in a nearby county showed a sample mean of 80 days with a sample standard deviation of 20 days. Conduct a hypotheses test to determine whether the mean number of days until a home is sold is different than the Hamilton county mean of 86 days in the nearby county. Round your answer to four decimal places.
p-value =
0.0017
To conduct a hypotheses test in this scenario, we can use the t-test. The t-test allows us to compare the mean of the sample to the population mean and determine whether they are significantly different.
Let's set up the null and alternative hypotheses:
Null hypothesis (H0): The mean number of days until a home is sold in the nearby county is equal to the mean of 86 days in Hamilton County.
Alternative hypothesis (H1): The mean number of days until a home is sold in the nearby county is different from the mean of 86 days in Hamilton County.
Now, let's determine the test statistic and the critical value. The test statistic for the t-test can be calculated as:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
In this case, the sample mean is 80 days, the population mean is 86 days, the sample standard deviation is 20 days, and the sample size is 40.
t = (80 - 86) / (20 / sqrt(40))
Calculating the test statistic, we get: t ≈ -2.8284
To find the p-value, we need to consult the t-distribution table or use a statistical software. The p-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true.
For a two-tailed test, we compare the absolute value of the test statistic to the critical value at the desired significance level (α), and then find the area under the curve from that critical value to positive and negative infinity.
Since you haven't specified a significance level (α), we'll assume it's 0.05 (5%).
Using a t-distribution table or statistical software, we find that for a sample size of 40 and a significance level of 0.05 (two-tailed test), the critical value is approximately ±2.021
Since our calculated test statistic (-2.8284) is more extreme than the critical value (2.021), we reject the null hypothesis.
The final step is to find the p-value. When using statistical software, the p-value is often provided directly. In this case, the p-value is p ≈ 0.0074.
Therefore, the p-value is approximately 0.0074.
To conduct a hypothesis test, we can use a one-sample t-test.
Null Hypothesis (H₀): The mean number of days until a home is sold in the nearby county is 86 days.
Alternative Hypothesis (H₁): The mean number of days until a home is sold in the nearby county is different than 86 days.
Given:
Sample mean (x̄) = 80 days
Sample standard deviation (s) = 20 days
Sample size (n) = 40
To calculate the t-test statistic, we can use the following formula:
t = (x̄ - μ) / (s / √n)
Where:
x̄ is the sample mean
μ is the population mean
s is the sample standard deviation
n is the sample size
Substituting the given values:
t = (80 - 86) / (20 / √40)
t = -6 / (20 / √40)
t = -0.9495
Next, we need to determine the degrees of freedom (df) for the t-distribution. Since we are using a sample, the degrees of freedom is (n - 1):
df = 40 - 1
df = 39
Using the t-distribution with 39 degrees of freedom, we can calculate the p-value associated with the t-test statistic. We are interested in a two-tailed test because we are testing if the mean is different, not just greater or smaller.
Using a t-table or statistical software, we find that the p-value is approximately 0.3484.
Therefore, the p-value for this hypothesis test is approximately 0.3484.