"In Exercises 17 and 18, (b) find the critical value(s) and identify the rejection region(s), 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."

검증된 단계별 안내

Step 1: Identify the hypotheses for the test. Since the agent claims no difference in mean incomes, the null hypothesis is $H_0: \mu_1 = \mu_2$, and the alternative hypothesis is $H_a: \mu_1 \neq \mu_2$ (a two-tailed test).

Step 2: Determine the significance level $\alpha = 0.01$ and the degrees of freedom for the test. Since population variances are assumed equal, use the pooled variance approach. The degrees of freedom is $df = n_1 + n_2 - 2$, where $n_1$ and $n_2$ are the sample sizes.

Step 3: Find the critical value(s) from the $t$-distribution for a two-tailed test at $\alpha = 0.01$ with $df$ degrees of freedom. The critical values are $\pm t_{\alpha/2, df}$, which define the rejection regions.

Step 4: Define the rejection regions as $t < -t_{\alpha/2, df}$ or $t > t_{\alpha/2, df}$. If the calculated test statistic falls into either region, reject the null hypothesis.

Step 5: (For completeness) To test the claim, calculate the pooled standard deviation, then compute the test statistic using the formula $t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$, where $s_p$ is the pooled standard deviation. Compare this $t$ value to the critical values to decide whether to reject $H_0$.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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 determine if observed differences are statistically significant 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 a pooled variance estimate to calculate the test statistic, which follows a t-distribution with degrees of freedom based on sample sizes. This approach is appropriate here because the problem states equal variances and normal populations.

Critical Value and Rejection Region

The critical value is a threshold from the t-distribution that defines the boundary of the rejection region for the test statistic at a chosen significance level (α=0.01). If the calculated test statistic falls into this region, the null hypothesis is rejected. Identifying these values helps determine whether the observed data provide strong enough evidence against the null hypothesis.