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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: 9780137493470Not the one you use?Change textbook
All textbooksLarson 8th EditionCh. 8 - Hypothesis Testing with Two SamplesProblem 8.RE.7
Chapter 8, Problem 8.RE.7

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=0.10


Population statistics: σ1= 0.11 and σ2= 0.10


Sample statistics: x̅1 = 0.28, n1 = 41, and x̅2= 0.33, n2= 34

Verified step by step guidance
1
Identify the null and alternative hypotheses based on the claim μ1 < μ2. The null hypothesis (H0) is μ1 ≥ μ2, and the alternative hypothesis (H1) is μ1 < μ2.
Determine the significance level α = 0.10, which will be used to find the critical value for the test.
Since the population standard deviations σ1 and σ2 are known, use the z-test for the difference between two means. Calculate the test statistic using the formula: \[\text{z} = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\] Note that under the null hypothesis, (μ1 - μ2) = 0.
Calculate the standard error of the difference between the sample means using: \[\text{SE} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}\]
Compare the calculated z-test statistic to the critical z-value for a left-tailed test at α = 0.10. If the test statistic is less than the critical value, reject the null hypothesis; otherwise, do not reject it.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hypothesis Testing for Two Population Means

This involves formulating null and alternative hypotheses about the difference between two population means (μ1 and μ2). The goal is to determine if there is enough evidence to support the claim (μ1 < μ2) using sample data, by comparing a test statistic to a critical value or p-value at a given significance level.
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Difference in Means: Hypothesis Tests

Sampling Distribution and Test Statistic

The test statistic measures how far the observed difference between sample means (x̅1 - x̅2) is from the hypothesized difference under the null hypothesis. When population standard deviations (σ1, σ2) are known and samples are independent and normal, the test statistic follows a normal distribution, allowing calculation of z-scores.
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Sampling Distribution of Sample Proportion

Significance Level and Decision Rule

The significance level (α = 0.10) defines the probability of rejecting the null hypothesis when it is true (Type I error). It sets the critical value(s) for the test statistic. If the test statistic falls into the rejection region determined by α, the null hypothesis is rejected in favor of the alternative claim.
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Related Practice
Textbook Question

In Exercises 17 and 18, (c) find the standardized test statistic t, 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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Textbook Question

Take this test as you would take a test in class.For each exercise, perform the steps below.

c.Find the critical value(s) and identify the rejection region(s).



A real estate agency says that the mean home sales price in Olathe, Kansas, is greater than in Rolla, Missouri. The mean home sales price for 39 homes in Olathe is \$392,453. Assume the population standard deviation is \$224,902. The mean home sales price for 38 homes in Rolla is \$285,787. Assume the population standard deviation is \$330,578. At α=0.05, is there enough evidence to support the agency’s claim? (Adapted from Realtor.com)

Textbook Question

"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."

1
views
Textbook Question

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=0.05. Assume (σ1)^2 = (σ2)^2


Sample statistics: x̅1=228, s1=27, n1= 20 and x̅2=207, s2=25, n2= 13

1
views
Textbook Question

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

Textbook Question

Take this test as you would take a test in class.For each exercise, perform the steps below.

a. Identify the claim and state and


A real estate agency says that the mean home sales price in Olathe, Kansas, is greater than in Rolla, Missouri. The mean home sales price for 39 homes in Olathe is \$392,453. Assume the population standard deviation is \$224,902. The mean home sales price for 38 homes in Rolla is \$285,787. Assume the population standard deviation is \$330,578. At α=0.05, is there enough evidence to support the agency’s claim? (Adapted from Realtor.com)