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A two-sample Z-test is a statistical technique used to determine if there is a significant difference between the means of two independent groups. In our context, Fairfield Homes is investigating whether the mean incomes of inquiries at two different developments are different.

A Z-test is applicable when the sample sizes are large, generally assumed to be greater than 30. This test compares the sample means by calculating a Z-score, which will then be used to determine if the observed difference is statistically significant.

The Z-score indicates how many standard deviations the difference between the sample means is from the hypothesized difference (usually zero) under the null hypothesis. A large Z-score suggests the means are different enough that it's unlikely the difference is due to sampling variability alone.

Choosing the correct statistical test is crucial to validate our claims and ensure that our conclusions are based on evidence rather than assumptions.

The significance level, denoted by \( \alpha \), is a threshold used in hypothesis testing to determine whether to reject the null hypothesis. It represents the probability of making a Type I error, which is rejecting the null hypothesis when it is actually true.

In this problem, the significance level is set to 0.05. This means there is a 5% risk of concluding that the population means are different when they are not. A common choice for many statistical analyses, the 0.05 significance level provides a balance between being too lenient and too strict.

When conducting a hypothesis test, if the p-value obtained is less than or equal to the significance level, then the results are considered statistically significant. In our exercise, since the calculated Z-score falls outside the critical value range associated with this \( \alpha \), we have sufficient evidence to reject the null hypothesis.

The foundation of hypothesis testing involves setting up two competing statements: the null hypothesis and the alternative hypothesis.

The null hypothesis, denoted as \( H_0 \), is the assertion that there is no effect or no difference. In the Fairfield Homes case, it suggests that the mean incomes of inquiries at both developments are the same, or mathematically, \( H_0: \mu_1 = \mu_2 \).

Conversely, the alternative hypothesis, \( H_a \), represents what we seek to prove. Here, it states that the mean incomes are different, expressed as \( H_a: \mu_1 eq \mu_2 \). This two-tailed hypothesis test addresses the possibility of the means being either greater or smaller, not specifying the direction of deviation.

Clearly stating these hypotheses is crucial as it guides the testing process and helps in interpreting the results correctly.

The process of calculating the test statistic is a critical step in hypothesis testing. For the two-sample Z-test, we calculate a Z-score to determine the standard distance between the sample means relative to the combined standard deviation.

The formula for the test statistic in the Z-test is:

• \[ z = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]

Plugging the given values from the exercise:

• \( \bar{x}_1 = 150,000 \), \( \bar{x}_2 = 180,000 \)
• \( s_1 = 40,000 \), \( s_2 = 30,000 \)
• \( n_1 = 75 \), \( n_2 = 120 \)

We proceed to calculate the difference in sample means and adjust for the variability of both samples. The computation yields a Z-value of approximately \( -6.42 \).

This score indicates that the observed difference in sample means is approximately 6.42 standard deviations below the hypothesized value under the null hypothesis, providing strong evidence against \( H_0 \) at the chosen significance level.