Problem 53 - Basic Statistics for Business and Economics with Student

Question

An urban planner claims that, nationally, 20 percent of all families renting condominiums move during a given year. A random sample of 200 families renting condominiums in the Dallas Metroplex revealed that 56 had moved during the past year. At the .01 significance level, does this evidence suggest that a larger proportion of condominium owners moved in the Dallas area? Determine the \(p\) -value.

Step-by-Step Solution

Answer

Reject the null hypothesis; more families moved in Dallas. The p-value is less than 0.005.

Step 1: State the Hypotheses

We need to establish our null and alternative hypotheses. The null hypothesis (\(H_0\)) is that the proportion of families that moved in Dallas is equal to the national proportion (0.20). The alternative hypothesis (\(H_a\)) is that the proportion of families that moved in Dallas is greater than 0.20. Thus, we have:- \(H_0: p = 0.20\)- \(H_a: p > 0.20\)

Step 2: Determine the Test Statistic

For a test of proportion, we use the formula for the z-test statistic: \[z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\]where \(\hat{p}\) is the sample proportion (56/200 = 0.28), \(p_0\) is the hypothesized population proportion (0.20), and \(n\) is the sample size (200).

Step 3: Calculate the Test Statistic

Substitute the given values into your formula:\[z = \frac{0.28 - 0.20}{\sqrt{\frac{0.20(1-0.20)}{200}}} = \frac{0.08}{\sqrt{\frac{0.16}{200}}} = \frac{0.08}{\sqrt{0.0008}} = \frac{0.08}{0.0283} \approx 2.83\]

Step 4: Determine the Critical Value and Region

At a significance level of \(\alpha = 0.01\), we need to find the critical value from the standard normal distribution table for a one-tailed test. The critical value for \(\alpha = 0.01\) is approximately 2.33. Since the test is one-tailed, the critical region is where \(z > 2.33\).

Step 5: Make a Decision

The calculated z-statistic (2.83) is greater than the critical value (2.33). Therefore, we reject the null hypothesis. This suggests that a larger proportion of families renting condominiums in Dallas moved during the past year compared to the national proportion.

Step 6: Calculate the p-value

The p-value is the probability of observing a test statistic as extreme as, or more extreme than, the observed statistic under the null hypothesis. Using a standard normal distribution table, a z-value of 2.83 corresponds to a p-value that is less than 0.005. Since this is less than the 0.01 significance level, it further supports rejecting the null hypothesis.

Key Concepts

Null Hypothesis

In hypothesis testing, the null hypothesis is a statement that we assume to be true until there is sufficient evidence to conclude otherwise. In our example, the urban planner's claim that 20% of families renting condos nationally move each year sets our null hypothesis. This is denoted as \(H_0: p = 0.20\), where \(p\) represents the population proportion of families that move. The null hypothesis functions as the baseline or default assumption, stating that there is no effect or no difference beyond what is observed in the sample.

Alternative Hypothesis

The alternative hypothesis is what you would conclude if you find strong enough evidence against the null hypothesis. It offers a new perspective different from the null assumption. In our scenario, the alternative hypothesis suggests that a larger proportion of families in Dallas move compared to the national average. This is denoted as \(H_a: p > 0.20\). The alternative hypothesis is tested against the null to see if there is statistical significance, meaning an observed effect or difference unlikely due to chance.

Z-test

The z-test is a statistical method used to determine if there is a significant difference between the sample proportion and the hypothesized population proportion. It is primarily used when sample sizes are large (typically \(n > 30\)). For the example of families moving, we calculate the z-test statistic using the formula:

• \(z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\)
• Where \(\hat{p}\) is the sample proportion, \(p_0\) is the hypothesized proportion, and \(n\) is the sample size.

This formula helps us understand if the observed sample deviation is statistically significant. A significant statistic suggests that the sample statistic is not just due to random sample variation but indicates a genuine difference.

P-value

The p-value measures the probability of obtaining a test statistic as extreme as, or more extreme than, the one observed under the null hypothesis. In simpler terms, it tells us how likely it is to get our sample results if the null hypothesis were true.

• A small p-value (typically less than 0.05) suggests that such an extreme result would be unlikely under the null hypothesis.
• In our case, the p-value is less than 0.005, indicating very low probability, which means further supporting the rejection of the null hypothesis.

Understanding the p-value helps determine the statistical significance of your result, offering insight into whether to support or refute the null hypothesis.