Question

A simple random sample of 35 rent prices for a three-bedroom apartment is obtained. The mean of the sample is found to be $985. The population standard deviation is $49. An urban housing committee claims the mean price for a three-bedroom apartment is less than $1000. Assume a confidence level of α = 0.01. Find the P-value you would use to test the claim that the mean price for a three-bedroom apartment is less than $1000.

A) –1.81
B) 0.9649
C) 0.0351
D) 0.0702

Answer 1

To find the P-value for the hypothesis test about the mean rent price of three-bedroom apartments, we can follow these steps:

1. **State the Hypotheses**:
- Null hypothesis (\(H_0\)): \(\mu = 1000\) (the population mean rent price is $1000)
- Alternative hypothesis (\(H_a\)): \(\mu < 1000\) (the population mean rent price is less than $1000)

2. **Calculate the Test Statistic**:
We can use the Z-test for this scenario since the population standard deviation is known. The formula for the Z-test statistic is:
\[
Z = \frac{\bar{X} - \mu_0}{\sigma / \sqrt{n}}
\]
Where:
- \(\bar{X} = 985\) (the sample mean),
- \(\mu_0 = 1000\) (the hypothesized population mean),
- \(\sigma = 49\) (the population standard deviation),
- \(n = 35\) (the sample size).

Plugging in the values:
\[
Z = \frac{985 - 1000}{49 / \sqrt{35}} = \frac{-15}{49 / 5.9161} \approx \frac{-15}{8.2863} \approx -1.81
\]

3. **Find the P-value**:
Since we are conducting a one-tailed test (to the left), we need to find the probability of getting a Z-score less than -1.81. Using the standard normal distribution table or a calculator for the Z-score:
- For \(Z = -1.81\), we find the corresponding P-value.

The P-value for \(Z = -1.81\) is approximately \(0.0351\).

So the P-value we would use to test the claim that the mean price for a three-bedroom apartment is less than $1000 is:

**C) 0.0351**.