Question

The average sales price of new one-family houses in the Midwest is \(\$ 250,000\) and in the South is \(\$ 253,400\). A random sample of 40 houses in each region was examined with the following results. At the 0.05 level of significance, can it be concluded that the difference in mean sales price for the two regions is greater than \(\$ 3400 ?\) $$ \begin{array}{lll} & \text { South } & \text { Midwest } \\ \hline \text { Sample size } & 40 & 40 \\ \text { Sample mean } & \$ 261,500 & \$ 248,200 \\ \text { Population standard deviation } & \$ 10.500 & \$ 12.000 \end{array} $$

Short answer

Yes, the difference in mean sales price is greater than $3,400 with a significant Z-score of 5.96.

Step-by-step solution

Define the null and alternative hypotheses

We need to test if the difference in mean sales price between the South and Midwest is greater than $3,400. The null hypothesis \( H_0 \) states that the mean difference \( \mu_S - \mu_M \leq 3,400 \), where \( \mu_S \) and \( \mu_M \) are the population means for South and Midwest respectively. The alternative hypothesis \( H_a \) states that \( \mu_S - \mu_M > 3,400 \).

Determine the appropriate test statistic

Since the population standard deviations are known, we use the Z-test for the difference between two means. The formula for the Z-test statistic is \[ Z = \frac{(\bar{X}_S - \bar{X}_M) - 3400}{\sqrt{\frac{\sigma_S^2}{n} + \frac{\sigma_M^2}{n}}} \] where \( \bar{X}_S = 261,500 \), \( \bar{X}_M = 248,200 \), \( \sigma_S = 10,500 \), \( \sigma_M = 12,000 \), and \( n = 40 \).

Calculate the Z-score

Substitute the values into the formula: \[ Z = \frac{(261,500 - 248,200) - 3400}{\sqrt{\frac{10,500^2}{40} + \frac{12,000^2}{40}}} \] \[ Z = \frac{13,300 - 3,400}{\sqrt{2,756,250}} \approx \frac{9,900}{1,660} \approx 5.96 \]

Determine the critical value and decision rule

The significance level is 0.05 for a one-tailed test. Looking at standard Z-tables, the critical value for \( \alpha = 0.05 \) is 1.645. We reject \( H_0 \) if \( Z > 1.645 \).

Make a decision

Since the calculated Z-score \( 5.96 \) is greater than the critical value \( 1.645 \), we reject the null hypothesis \( H_0 \). This suggests that the difference in mean sales price is indeed greater than $3,400.

Key concepts

Z-test

A Z-test is a statistical method used to determine whether there is a significant difference between sample and population means. In situations where the population standard deviation is known, and the sample size is large (usually greater than 30), a Z-test is typically applied. In the context of this exercise, a Z-test is employed to evaluate the difference in average sales prices between houses in the South and Midwest.

The formula for a Z-test when comparing two means is:

• \(Z = \frac{(\bar{X}_S - \bar{X}_M) - \text{hypothesized difference}}{\sqrt{\frac{\sigma_S^2}{n} + \frac{\sigma_M^2}{n}}}\)

Here, \(\bar{X}_S\) and \(\bar{X}_M\) represent the sample means for the South and Midwest respectively, while \(\sigma_S\) and \(\sigma_M\) are the population standard deviations for these two regions. \(n\) is the sample size, which is equal for both groups in this exercise.

Using this method allows researchers to draw conclusions about whether any observed differences in the data are simply due to random chance or whether they reflect true differences in the populations.

Significance Level

In hypothesis testing, the significance level (denoted as \(\alpha\)) is a threshold that is set by the researcher to determine when to reject a null hypothesis. It represents the probability of mistakenly rejecting a true null hypothesis, which is an error known as a "Type I error."

In this exercise, the significance level is set at 0.05, or 5%. This means that there is a 5% risk of concluding that there is a significant difference in mean sales prices when, in fact, no such difference exists. The choice of significance level depends on the context of the research and the potential consequences of both Type I and Type II errors (the latter being the error of not rejecting a false null hypothesis).

When the significance level is set to 0.05, it means that if the calculated statistic falls into the region that accounts for the top 5% of the distribution (beyond the critical value), the null hypothesis can be rejected with confidence. This choice assures that the researchers are minimizing the chance of making an incorrect inference from the data.

Critical Value

The critical value is a point on the scale of the test statistic beyond which we reject the null hypothesis. It is determined based on the chosen significance level and the nature of the test (one-tailed or two-tailed).

In the provided exercise, the test is one-tailed because the alternative hypothesis is concerned with whether the difference in sales prices exceeds $3,400, specifically in one direction. At a significance level of 0.05, the critical value can be found using Z-tables or standard normal distribution tables. For a one-tailed test, this critical value is 1.645.

Any Z-score that exceeds the critical value indicates a statistically significant result that leads us to reject the null hypothesis. Here, the computed Z-score was significantly higher than the critical value (5.96 vs. 1.645), reinforcing the decision to reject the null hypothesis. Therefore, in this scenario, the difference in mean sales prices is greater than the hypothesized value of $3,400.