Question

According to the almanac, the average sales price of a single-family home in the metropolitan Dallas/Ft. Worth/Irving, Texas, area is \(\$ 215,200 .\) The average home price in Orlando, Florida, is \(\$ 198,000 .\) The mean of a random sample of 45 homes in the Texas metroplex was \(\$ 216,000\) with a population standard deviation of \(\$ 30,000 .\) In the Orlando, Florida, area a sample of 40 homes had a mean price of \(\$ 203,000\) with a population standard deviation of \(\$ 32,500\). At the 0.05 level of significance, can it be concluded that the mean price in Dallas exceeds the mean price in Orlando? Use the \(P\) -value method.

Short answer

Yes, the mean home price in Dallas exceeds the mean home price in Orlando at the 0.05 significance level.

Step-by-step solution

State the Hypotheses

We need to set up our null and alternative hypotheses. The null hypothesis (\(H_0\)) is that the mean price of homes in Dallas is equal to the mean price of homes in Orlando. The alternative hypothesis (\(H_a\)) is that the mean price in Dallas is greater than in Orlando. Mathematically, this is written as: \(H_0: \mu_{Dallas} \leq \mu_{Orlando}\) and \(H_a: \mu_{Dallas} > \mu_{Orlando}\).

Identify the Test Statistic

Since both sample sizes are greater than 30, we will use the standard normal distribution (Z-test) for comparing the two means. The formula for the Z-test statistic in this context is: \[ Z = \frac{(\bar{X}_{Dallas} - \bar{X}_{Orlando}) - (\mu_{Dallas} - \mu_{Orlando})}{\sqrt{\frac{\sigma_{Dallas}^2}{n_{Dallas}} + \frac{\sigma_{Orlando}^2}{n_{Orlando}}}} \] where \(\mu_{Dallas} = \mu_{Orlando} = 0\) under the null hypothesis.

Calculate the Test Statistic

Substitute the given values into the formula: \(\bar{X}_{Dallas} = 216,000\), \(\bar{X}_{Orlando} = 203,000\), \(\sigma_{Dallas} = 30,000\), \(\sigma_{Orlando} = 32,500\), \(n_{Dallas} = 45\), and \(n_{Orlando} = 40\). Calculate: \[ Z = \frac{(216,000 - 203,000) - 0}{\sqrt{\frac{30,000^2}{45} + \frac{32,500^2}{40}}} \] This simplifies to \[ Z = \frac{13,000}{\sqrt{20,000,000 + 26,406,250}} \] \[ Z = \frac{13,000}{7,146.77} \approx 1.819 \].

Find the P-value

Using a Z-table, find the P-value corresponding to \( Z = 1.819 \). The P-value is the probability that a standard normal variable exceeds this calculated Z value. Look up \( Z = 1.819 \) in the Z-table or use a calculator, which gives a P-value of approximately 0.0344.

Make a Decision

Compare the P-value to the level of significance \(\alpha = 0.05\). Since the P-value (0.0344) is less than 0.05, we reject the null hypothesis.

Conclusion

Based on the P-value method, we conclude that there is sufficient evidence at the 0.05 level of significance to support the claim that the mean home price in Dallas exceeds the mean home price in Orlando.

Key concepts

Z-test

The Z-test is a statistical method used to determine whether there is a significant difference between the means of two groups. It is applicable when the sample size is large (usually greater than 30), and the population standard deviation is known. In this exercise, the Z-test was chosen because both sample sizes exceed 30 and the standard deviations for the Dallas and Orlando home prices are provided.

To perform a Z-test, one computes the Z-test statistic which quantifies the distance between the sample mean and the population mean under the null hypothesis, expressed in terms of the number of standard deviations. The formula for the Z-test statistic in this context is:
\[ Z = \frac{(\bar{X}_{Dallas} - \bar{X}_{Orlando}) - (\mu_{Dallas} - \mu_{Orlando})}{\sqrt{\frac{\sigma_{Dallas}^2}{n_{Dallas}} + \frac{\sigma_{Orlando}^2}{n_{Orlando}}}} \]

The numerator represents the difference in sample means, and the denominator adjusts this difference by the standard error of the difference. A calculated Z greater than the critical Z value at a given significance level indicates a statistically significant difference.

P-value method

The P-value method is a widely used approach for hypothesis testing. It helps determine the strength of evidence against the null hypothesis. The P-value is the probability of obtaining an observed result, or more extreme, assuming that the null hypothesis is true.

In this exercise, the P-value was obtained by first calculating the Z-test statistic and then looking up the corresponding probability in a Z-table. For our example with a Z of approximately 1.819, the P-value was found to be 0.0344. This value represents the probability that a standard normal variable exceeds 1.819 if the null hypothesis is true.

If the P-value is lower than the pre-determined significance level (often 0.05), the null hypothesis is rejected. A smaller P-value indicates stronger evidence in favor of the alternative hypothesis. Thus, since the P-value of 0.0344 is less than 0.05, the null hypothesis was rejected in this case.

Mean comparison

Mean comparison involves evaluating the average values of two different groups to understand if there is a statistically significant difference between them. In the exercise, the mean price of homes in the Dallas metropolitan area was compared against those in Orlando.

The procedure starts with defining null and alternative hypotheses. Here, the null hypothesis stated that the mean home price in Dallas is less than or equal to that in Orlando, while the alternative proposed that Dallas has a higher mean price. Such comparisons are essential for analyzing real-world differences, for instance, in economic data like housing prices.

Careful calculation of the sample means, and considering the population standard deviations, allows for a more accurate understanding of whether the observed differences reflect true divergences or if they might have occurred by random chance.

Statistical significance

Statistical significance is a measure of how likely it is that an observed result, such as the difference between two means, is due to something other than mere random chance. It is a key concept in hypothesis testing, providing assurance that the findings of a study or experiment are meaningful.

In this exercise, statistical significance was established by comparing the P-value to the significance level \(\alpha\) set at 0.05. The rule of thumb is that if the P-value is less than \(\alpha\), the result is considered statistically significant.

When a test result is deemed statistically significant, it implies that the null hypothesis can be rejected with confidence, leading to the acceptance of the alternative hypothesis. For this exercise, the statistical significance confirmed that the mean price of homes in Dallas exceeds that in Orlando, supporting the decision to reject the null hypothesis.