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Ch. 8 - Hypothesis Testing with Two Samples
Larson - Elementary Statistics: Picturing the World 8th Edition
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Larson 8th EditionCh. 8 - Hypothesis Testing with Two Samples문제 8.RE.18e
8장, 문제 8.RE.18e

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.

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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 \(H_a: \mu_1 \neq \mu_2\).
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.
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.
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.
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.

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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 make this decision at a given significance level (α).
추천 영상:
05:52
Performing Hypothesis Tests: Proportions

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.
추천 영상:
가이드 코스
08:24
Difference in Means: Hypothesis Tests

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.
추천 영상:
03:53
Conditional Probability Rule
관련 실천
교과서 질문

In Exercises 11–16, test the claim about the difference between two population means μ1 and μ2 at the level of significance α. Assume the samples are random and independent, and the populations are normally distributed.


Claim: μ1> μ2; α=0.10. Assume (σ1)^2 ≠ (σ2)^2


Sample statistics: x̅1= 520, s1= 25, n1= 7 and x̅2= 500, s2= 55, n2= 6

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교과서 질문

In Exercises 19–22, test the claim about the mean of the differences for a population of paired data at the level of significance α. Assume the samples are random and dependent, and the populations are normally distributed.


Claim: μd≠0; α=0.05.


Sample statistics: d̄=17.5, sd=4.05, n=37



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교과서 질문

In Exercises 1–4, classify the two samples as independent or dependent and justify your answer.


Sample 1: The heights of 37 children

Sample 2: The heights of the same 37 children after 1 year

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교과서 질문

In Exercises 5–8, test the claim about the difference between two population means μ1 and μ2 at the level of significance α. Assume the samples are random and independent, and the populations are normally distributed.


Claim: μ1=μ2; α=0.01


Population statistics: σ1= 52 and σ2= 68


Sample statistics: x̅1 = 5595, n1 = 156, and x̅2= 5575, n2= 216

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교과서 질문

In Exercises 11–16, test the claim about the difference between two population means μ1 and μ2 at the level of significance α. Assume the samples are random and independent, and the populations are normally distributed.


Claim: μ1< μ2; α=0.10. Assume (σ1)^2 ≠ (σ2)^2


Sample statistics: x̅1=0.015, s1=0.011, n1= 8 and x̅2=0.019, s2=0.004, n2= 6

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교과서 질문

"In Exercises 9 and 10, (a) identify the claim and state Ho and Ha , (b) find the critical value(s) and identify the rejection region(s), (c) find the standardized test statistic z, (d) decide whether to reject or fail to reject the null hypothesis, and (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 career counselor claims that the mean annual salaries of entry level paralegals in Dayton, Ohio, and Coventry, Rhode Island, are the same. The mean annual salary of 40 randomly selected entry level paralegals in Dayton is \$58,180. Assume the population standard deviation is \$10,990. The mean annual salary of 35 randomly selected entry level paralegals in Coventry is \$61,120. Assume the population standard deviation is \$11,850. At α=0.10, is there enough evidence to reject the counselor’s claim? (Adapted from Salary.com)"

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