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Sagot :
Certainly! Let's conduct a hypothesis test to determine whether the mean number of days until a home is sold in the nearby county is different from the mean in Hamilton County, which is 86 days.
### Step-by-Step Solution:
1. State the Hypotheses:
- Null Hypothesis (H₀): [tex]\(\mu = 86\)[/tex]
- The mean number of days to sell a home in the nearby county is 86 days (same as in Hamilton County).
- Alternative Hypothesis (H₁): [tex]\(\mu \ne 86\)[/tex]
- The mean number of days to sell a home in the nearby county is not equal to 86 days (different than in Hamilton County).
2. Gather the Sample Statistics:
- Sample mean ([tex]\(\bar{x}\)[/tex]): 80 days
- Sample standard deviation (s): 20 days
- Sample size (n): 40
3. Calculate the Standard Error of the Mean (SEM):
The formula for the standard error of the mean is:
[tex]\[ \text{SEM} = \frac{s}{\sqrt{n}} \][/tex]
Using the given data:
[tex]\[ \text{SEM} = \frac{20}{\sqrt{40}} \approx 3.1623 \][/tex]
4. Calculate the Z-Score:
The Z-score formula for the sample mean is:
[tex]\[ z = \frac{\bar{x} - \mu}{\text{SEM}} \][/tex]
Using the given data:
[tex]\[ z = \frac{80 - 86}{3.1623} \approx -1.8974 \][/tex]
5. Determine the p-value:
The p-value is found by looking at the standard normal distribution for the calculated Z-score. For a two-tailed test, the p-value is:
[tex]\[ p\text{-value} = 2 \times P(Z \leq -1.8974) \][/tex]
Using Z-tables or statistical software, we get:
[tex]\[ p\text{-value} \approx 0.0578 \][/tex]
6. Compare the p-value to the Significance Level (α):
We are given [tex]\(\alpha = 0.05\)[/tex]. To determine whether to reject the null hypothesis, we compare the p-value to [tex]\(\alpha\)[/tex]:
[tex]\[ \text{p-value} = 0.0578 \][/tex]
Since [tex]\(0.0578 > 0.05\)[/tex], we fail to reject the null hypothesis.
### Conclusion:
The p-value is [tex]\(0.0578\)[/tex]. Since the p-value is greater than the significance level [tex]\(\alpha = 0.05\)[/tex], we do not have enough evidence to reject the null hypothesis. Therefore, we conclude that there is not enough evidence to state that the mean number of days to sell a home in the nearby county is different from the mean in Hamilton County.
So, the p-value is [tex]\(0.0578\)[/tex] and we do not reject the null hypothesis.
### Step-by-Step Solution:
1. State the Hypotheses:
- Null Hypothesis (H₀): [tex]\(\mu = 86\)[/tex]
- The mean number of days to sell a home in the nearby county is 86 days (same as in Hamilton County).
- Alternative Hypothesis (H₁): [tex]\(\mu \ne 86\)[/tex]
- The mean number of days to sell a home in the nearby county is not equal to 86 days (different than in Hamilton County).
2. Gather the Sample Statistics:
- Sample mean ([tex]\(\bar{x}\)[/tex]): 80 days
- Sample standard deviation (s): 20 days
- Sample size (n): 40
3. Calculate the Standard Error of the Mean (SEM):
The formula for the standard error of the mean is:
[tex]\[ \text{SEM} = \frac{s}{\sqrt{n}} \][/tex]
Using the given data:
[tex]\[ \text{SEM} = \frac{20}{\sqrt{40}} \approx 3.1623 \][/tex]
4. Calculate the Z-Score:
The Z-score formula for the sample mean is:
[tex]\[ z = \frac{\bar{x} - \mu}{\text{SEM}} \][/tex]
Using the given data:
[tex]\[ z = \frac{80 - 86}{3.1623} \approx -1.8974 \][/tex]
5. Determine the p-value:
The p-value is found by looking at the standard normal distribution for the calculated Z-score. For a two-tailed test, the p-value is:
[tex]\[ p\text{-value} = 2 \times P(Z \leq -1.8974) \][/tex]
Using Z-tables or statistical software, we get:
[tex]\[ p\text{-value} \approx 0.0578 \][/tex]
6. Compare the p-value to the Significance Level (α):
We are given [tex]\(\alpha = 0.05\)[/tex]. To determine whether to reject the null hypothesis, we compare the p-value to [tex]\(\alpha\)[/tex]:
[tex]\[ \text{p-value} = 0.0578 \][/tex]
Since [tex]\(0.0578 > 0.05\)[/tex], we fail to reject the null hypothesis.
### Conclusion:
The p-value is [tex]\(0.0578\)[/tex]. Since the p-value is greater than the significance level [tex]\(\alpha = 0.05\)[/tex], we do not have enough evidence to reject the null hypothesis. Therefore, we conclude that there is not enough evidence to state that the mean number of days to sell a home in the nearby county is different from the mean in Hamilton County.
So, the p-value is [tex]\(0.0578\)[/tex] and we do not reject the null hypothesis.
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