On their website, a property management company states that the mean monthly rent for apartments on the east side of town is $650.
A researcher for a consumer advocacy group believes that, due to the construction of newer apartment complexes in neighboring towns, the mean monthly rent on the east side,, is now lower. A random sample of
18 monthly rents for apartments on the east side has a mean of $639
, with a standard deviation of $19.

Assume that current monthly rents for apartments on the east side are approximately normally distributed. Based on the sample, is there enough evidence to conclude, at the 0.05 level of significance, that the population mean monthly rent is now less than what is stated on the website?

Perform a one-tailed test. Then complete the parts below.
Carry your intermediate computations to three or more decimal places.

(a) State the full hypothesis H0 and the alternative hypothesis H1.
(b)	Determine the type of test statistic to use. (c)Find the value of the test statistic. (Round to three or more decimal places.) (d)Find the p-value. (Round to three or more decimal places.) (e)Can we conclude that the mean monthly rent for apartments on the east side is less than what is stated on the website? yes/no

To solve this problem, we'll perform a one-tailed t-test for the population mean, given the sample data.

### (a) State the Hypotheses

- **Null Hypothesis (H₀):** The population mean monthly rent is equal to $650.
  \[
  H_0: \mu = 650
  \]

- **Alternative Hypothesis (H₁):** The population mean monthly rent is less than $650.
  \[
  H_1: \mu < 650
  \]

### (b) Determine the Type of Test Statistic

Since the sample size is small (n = 18), and the population standard deviation is unknown, we use the **t-statistic** for this hypothesis test.

### (c) Find the Value of the Test Statistic

The test statistic is calculated using the formula:
\[
t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}
\]
where:
- $\bar{x}$ = sample mean = $639
- $\mu_0$ = hypothesized population mean = $650
- $s$ = sample standard deviation = $19
- $n$ = sample size = 18

Substituting the values:
\[
t = \frac{639 - 650}{19/\sqrt{18}} = \frac{-11}{4.478}
\]

Calculating this:
\[
t \approx -2.456
\]

### (d) Find the p-Value

To find the p-value, we'll look up the t-statistic in a t-distribution table with $df = n - 1 = 18 - 1 = 17$ degrees of freedom, or use statistical software.

For $t = -2.456$ and 17 degrees of freedom, the p-value is approximately 0.013.

### (e) Conclusion

Compare the p-value to the significance level ($\alpha = 0.05$):

- If $p \leq 0.05$, reject the null hypothesis $H_0$.
- If $p > 0.05$, fail to reject the null hypothesis $H_0$.

Here, $p = 0.013$ is less than 0.05, so we **reject the null hypothesis**.

**Conclusion:** Yes, there is enough evidence at the 0.05 level of significance to conclude that the mean monthly rent for apartments on the east side is less than what is stated on the website.

---

Would you like a detailed breakdown of any part of this process, or do you have any further questions?

Here are some related questions to consider:
1. How does sample size affect the t-test results?
2. What is the difference between a one-tailed and a two-tailed test?
3. What assumptions are necessary for a t-test to be valid?
4. How would the results change if the standard deviation were larger?
5. What if the significance level were 0.01 instead of 0.05?
6. How can the p-value be interpreted in the context of this problem?
7. What is the role of the degrees of freedom in determining the p-value?
8. How would a larger sample size impact the power of the test?

**Tip:** Always check the normality assumption when conducting a t-test, especially with small sample sizes.

Understand and perform a one-tailed t-test to determine if the mean monthly rent for apartments on the east side of town is less than what is stated on the website. Includes step-by-step calculations with sample mean, standard deviation, and hypothesis testing at 0.05 significance level.

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