In Exercises 17 and 18, (e) interpret the decision in the context of the original claim. Assume the samples are random and independent, and the populations are normally distributed.


A real estate agent claims that there is no difference between the mean household incomes of two neighborhoods. The mean income of 12 randomly selected households from the first neighborhood is \$52,750 with a standard deviation of \$2900. In the second neighborhood, 10 randomly selected households have a mean income of \$51,200 with a standard deviation of \$2225. At α=0.01, can you reject the real estate agent’s claim? Assume the population variances are equal.

Verified step by step guidance

1. Step 1: Define the null and alternative hypotheses. The null hypothesis \(H_0\) states that there is no difference between the mean household incomes of the two neighborhoods, so \(H_0: \mu_1 = \mu_2\). The alternative hypothesis \(H_a\) states that there is a difference, so \(H_a: \mu_1 \neq \mu_2\).
2. Step 2: Since the population variances are assumed equal, use the pooled standard deviation to estimate the common variance. Calculate the pooled variance \(s_p^2\) using the formula: \[s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}\] where \(n_1\) and \(n_2\) are the sample sizes, and \(s_1\) and \(s_2\) are the sample standard deviations.
3. Step 3: Calculate the test statistic \(t\) for the difference between two means using the formula: \[t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\] where \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means.
4. Step 4: Determine the degrees of freedom for the test, which is \(df = n_1 + n_2 - 2\). Then, find the critical value(s) of \(t\) from the \(t\)-distribution table at the significance level \(\alpha = 0.01\) for a two-tailed test.
5. Step 5: Compare the calculated test statistic \(t\) with the critical value(s). If \(|t|\) is greater than the critical value, reject the null hypothesis \(H_0\). Otherwise, do not reject \(H_0\). Finally, interpret this decision in the context of the original claim about the mean household incomes.

Key Concepts

Hypothesis Testing

Hypothesis testing is a statistical method used to decide whether there is enough evidence to reject a null hypothesis. In this question, the null hypothesis states that the mean incomes of the two neighborhoods are equal. The alternative hypothesis suggests a difference exists. The test evaluates sample data to make this decision at a given significance level (α).

Two-Sample t-Test for Means with Equal Variances

This test compares the means of two independent samples assuming their population variances are equal. It uses the pooled standard deviation to calculate the test statistic, which follows a t-distribution. This method is appropriate here because the samples are independent, normally distributed, and variances are assumed equal.

Significance Level and Decision Rule

The significance level (α = 0.01) defines the threshold for rejecting the null hypothesis, representing a 1% risk of a Type I error. After calculating the test statistic, it is compared to critical t-values. If the statistic falls in the rejection region, the null hypothesis is rejected, indicating a significant difference in mean incomes.