Question
Data from the Denver Metro Association of Realtors indicates that the mean price of a home in Denver, Colorado, in 2018 was 260.7 thousand dollars. A random sample of 98 homes sold in 2019 had a mean price of 286 thousand dollars. Can you conclude that the mean price in 2019 is greater than the mean price in 2018? Assume the population standard deviation is o=148. Use the a=0.05 level of significance and the P-value method.
A) state the appropriate null and alternate hypothesis.
Hv0 : [?]
Hv1 : [?]
This hypothesis test is a [blank] test.
A) state the appropriate null and alternate hypothesis.
Hv0 : [?]
Hv1 : [?]
This hypothesis test is a [blank] test.
Answers
Answer
is it right, left or two tailed
Answer
(b) Compute the value of the test statistic. Round the answer to two decimal places.
Answer
(c) Compute the P-value. Round the answer to four decimal places.
Answer
(d) Determine whether to reject Hv0. Use the a=0.05 level of significance.
Answered by
GPT-4o mini
To determine whether the mean price of homes in 2019 is greater than the mean price in 2018, we can set up the hypotheses as follows:
**Null Hypothesis (H0):** The mean price in 2019 is less than or equal to the mean price in 2018.
\[
H0: \mu \leq 260.7
\]
**Alternative Hypothesis (H1):** The mean price in 2019 is greater than the mean price in 2018.
\[
H1: \mu > 260.7
\]
This is a **one-tailed test** because we are specifically testing whether the mean price in 2019 is greater than the mean price in 2018.
---
Now, let's summarize:
A) State the appropriate null and alternate hypothesis.
- \( H0: \mu \leq 260.7 \)
- \( H1: \mu > 260.7 \)
This hypothesis test is a **one-tailed test**.
**Null Hypothesis (H0):** The mean price in 2019 is less than or equal to the mean price in 2018.
\[
H0: \mu \leq 260.7
\]
**Alternative Hypothesis (H1):** The mean price in 2019 is greater than the mean price in 2018.
\[
H1: \mu > 260.7
\]
This is a **one-tailed test** because we are specifically testing whether the mean price in 2019 is greater than the mean price in 2018.
---
Now, let's summarize:
A) State the appropriate null and alternate hypothesis.
- \( H0: \mu \leq 260.7 \)
- \( H1: \mu > 260.7 \)
This hypothesis test is a **one-tailed test**.
Answered by
GPT-4o mini
The hypothesis test is a **right-tailed test**. This is because we are testing if the mean price in 2019 is **greater than** the mean price in 2018. A right-tailed test is used when the alternative hypothesis indicates an increase or a greater value.
To summarize:
- The alternative hypothesis (\( H1: \mu > 260.7 \)) indicates we are looking for evidence of an increase in the mean price, thus making it a **right-tailed test**.
To summarize:
- The alternative hypothesis (\( H1: \mu > 260.7 \)) indicates we are looking for evidence of an increase in the mean price, thus making it a **right-tailed test**.
Answered by
GPT-4o mini
To compute the test statistic for the hypothesis test, we can use the formula for the z-test statistic in this case since we know the population standard deviation.
The formula for the z-test statistic is:
\[
z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}
\]
Where:
- \(\bar{x}\) = sample mean
- \(\mu_0\) = population mean (from 2018)
- \(\sigma\) = population standard deviation
- \(n\) = sample size
Given:
- \(\bar{x} = 286\) (mean price in 2019)
- \(\mu_0 = 260.7\) (mean price in 2018)
- \(\sigma = 148\)
- \(n = 98\)
Now, let's plug in the values:
\[
z = \frac{286 - 260.7}{148/\sqrt{98}}
\]
First, calculate the standard error (\(\sigma/\sqrt{n}\)):
\[
\sigma/\sqrt{n} = \frac{148}{\sqrt{98}} \approx \frac{148}{9.899} \approx 14.94
\]
Now, we can calculate the z-value:
\[
z = \frac{286 - 260.7}{14.94} \approx \frac{25.3}{14.94} \approx 1.69
\]
Rounding to two decimal places, the value of the test statistic is:
\[
z \approx 1.69
\]
The formula for the z-test statistic is:
\[
z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}
\]
Where:
- \(\bar{x}\) = sample mean
- \(\mu_0\) = population mean (from 2018)
- \(\sigma\) = population standard deviation
- \(n\) = sample size
Given:
- \(\bar{x} = 286\) (mean price in 2019)
- \(\mu_0 = 260.7\) (mean price in 2018)
- \(\sigma = 148\)
- \(n = 98\)
Now, let's plug in the values:
\[
z = \frac{286 - 260.7}{148/\sqrt{98}}
\]
First, calculate the standard error (\(\sigma/\sqrt{n}\)):
\[
\sigma/\sqrt{n} = \frac{148}{\sqrt{98}} \approx \frac{148}{9.899} \approx 14.94
\]
Now, we can calculate the z-value:
\[
z = \frac{286 - 260.7}{14.94} \approx \frac{25.3}{14.94} \approx 1.69
\]
Rounding to two decimal places, the value of the test statistic is:
\[
z \approx 1.69
\]
Answered by
GPT-4o mini
To compute the P-value for the test statistic \( z = 1.69 \) in a right-tailed test, we can use the standard normal distribution.
1. **Find the P-value for \( z = 1.69 \)**:
The P-value corresponds to the probability that the standard normal variable is greater than 1.69.
2. **Using standard normal distribution tables** or a calculator, we can find the area to the right of \( z = 1.69 \).
- First, find the area to the left of \( z = 1.69 \):
\[
P(Z < 1.69) \approx 0.9545
\]
- Therefore, the area to the right (the P-value) is:
\[
P(Z > 1.69) = 1 - P(Z < 1.69) = 1 - 0.9545 = 0.0455
\]
Thus, the P-value is approximately:
\[
\text{P-value} \approx 0.0455
\]
Rounded to four decimal places, the P-value is:
\[
\text{P-value} \approx 0.0455
\]
1. **Find the P-value for \( z = 1.69 \)**:
The P-value corresponds to the probability that the standard normal variable is greater than 1.69.
2. **Using standard normal distribution tables** or a calculator, we can find the area to the right of \( z = 1.69 \).
- First, find the area to the left of \( z = 1.69 \):
\[
P(Z < 1.69) \approx 0.9545
\]
- Therefore, the area to the right (the P-value) is:
\[
P(Z > 1.69) = 1 - P(Z < 1.69) = 1 - 0.9545 = 0.0455
\]
Thus, the P-value is approximately:
\[
\text{P-value} \approx 0.0455
\]
Rounded to four decimal places, the P-value is:
\[
\text{P-value} \approx 0.0455
\]
Answered by
GPT-4o mini
To determine whether to reject the null hypothesis \( H_0 \) at the \( \alpha = 0.05 \) level of significance, we compare the P-value to the significance level.
**Given:**
- \( \alpha = 0.05 \)
- P-value = 0.0455
**Decision Rule:**
- If the P-value is less than or equal to the significance level (\( \alpha \)), we reject the null hypothesis \( H_0 \).
- If the P-value is greater than the significance level, we fail to reject the null hypothesis.
**Comparison:**
- Since \( 0.0455 < 0.05 \), we reject the null hypothesis \( H_0 \).
**Conclusion:**
At the \( \alpha = 0.05 \) level of significance, we have sufficient evidence to conclude that the mean price of homes in 2019 is greater than the mean price in 2018.
**Given:**
- \( \alpha = 0.05 \)
- P-value = 0.0455
**Decision Rule:**
- If the P-value is less than or equal to the significance level (\( \alpha \)), we reject the null hypothesis \( H_0 \).
- If the P-value is greater than the significance level, we fail to reject the null hypothesis.
**Comparison:**
- Since \( 0.0455 < 0.05 \), we reject the null hypothesis \( H_0 \).
**Conclusion:**
At the \( \alpha = 0.05 \) level of significance, we have sufficient evidence to conclude that the mean price of homes in 2019 is greater than the mean price in 2018.
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