Based on Orlando Housing Market, the average Orlando price is 260K in February 2021.  A random sample of 36 homes sold in the month of March 2021 had a mean of 270K and a standard deviation of 24K. Can you conclude that the mean price is the same as the mean from the month of February? use

Before you start the 5 steps below, can you identify the claim in symbolic form and write it down on scratch paper?

1.)   Identify the null and alternative hypotheses.  Enter =, <, >, or "not equal" for

:

:

2.)  Find the test statistic:  Test Statistic =
(Round to two decimal places)

3.)  Find the p-value:  p-value =
(Round to four decimal places)

4.)   State the decision about the null hypothesis.   (Type Reject or Fail to Reject in the answer box.)

5.)  Conclusion:  There
sufficient evidence to conclude that the mean price  is the same as the mean from the month of February  (Type is or isn't into the answer box)

Let's go through the hypothesis testing step by step.

### Step 1: Identify the claim in symbolic form

We are testing whether the mean price in March 2021 is the same as the mean price in February 2021, which is 260K. The claim in symbolic form is:

\[
H_0: \mu = 260 \text{ (Null Hypothesis)}
\]
\[
H_a: \mu \neq 260 \text{ (Alternative Hypothesis)}
\]

### Step 2: Find the test statistic

Since the population standard deviation is unknown and the sample size is 36 (which is greater than 30), we can use the t-test for the test statistic.

The test statistic is given by:

\[
t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}
\]

Where:
- \(\bar{x} = 270\) K (sample mean)
- \(\mu = 260\) K (population mean under the null hypothesis)
- \(s = 24\) K (sample standard deviation)
- \(n = 36\) (sample size)

Substituting the values:

\[
t = \frac{270 - 260}{\frac{24}{\sqrt{36}}}
\]

Let's calculate this.The test statistic is \( t = 2.50 \).

### Step 3: Find the p-value

To find the p-value, we need to use the t-distribution with \( n - 1 \) degrees of freedom. In this case, the degrees of freedom \( df = 36 - 1 = 35 \). Since this is a two-tailed test (because \( H_a: \mu \neq 260 \)), we will find the p-value corresponding to \( t = 2.5 \) and double it.

Let's calculate the p-value.The p-value is approximately \( p = 0.0173 \).

### Step 4: State the decision about the null hypothesis

To make a decision, compare the p-value to the significance level, typically \(\alpha = 0.05\).

- If \( p \leq \alpha \), reject the null hypothesis.
- If \( p > \alpha \), fail to reject the null hypothesis.

In this case, since \( p = 0.0173 \) is less than \( 0.05 \), we reject the null hypothesis.

### Step 5: Conclusion

There **isn't** sufficient evidence to conclude that the mean price is the same as the mean from the month of February 2021.

---

Would you like further details or have any questions?

### Related Questions:
1. What is the importance of the significance level in hypothesis testing?
2. How would the conclusion change if the sample size were smaller?
3. What would be the conclusion if the p-value were greater than 0.05?
4. How does the standard deviation affect the test statistic?
5. Why is a t-test appropriate in this scenario instead of a z-test?
6. What if the sample mean had been 265K instead of 270K?
7. How would the conclusion change if the significance level was set at 0.01?
8. Can you explain the difference between one-tailed and two-tailed tests?

**Tip:** Always check if the test assumptions are met before performing any hypothesis test.

Explore hypothesis testing in statistics with an example from the Orlando housing market. Compare March 2021 home prices with February 2021 averages using t-tests, standard deviations, and sample means.

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