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.

Question

A) state the appropriate null and alternate hypothesis.
Hv0 : [?]
Hv1 : [?]
This hypothesis test is a [blank] test.

Answer 1

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.

Answer 2

is it right, left or two tailed

Answer 3

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.

Answer 4

(b) Compute the value of the test statistic. Round the answer to two decimal places.

Answer 5

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:

\[ z \approx 1.69 \]

Answer 6

(c) Compute the P-value. Round the answer to four decimal places.

Answer 7

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:

\[ \text{P-value} \approx 0.0455 \]

Answer 8

(d) Determine whether to reject Hv0. Use the a=0.05 level of significance.

Answer 9

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.