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Chapter 10: Q.10.4 (page 404)

Discuss the basic strategy for performing a hypothesis test to compare the means of two populations, based on independent samples.

Short Answer

Expert verified

To compare the means of two populations using an independent sample, a hypothesis test can be used as follows:

Take a sample of data from population 1 and a sample of data from population 2 at random.
Calculate the mean value of the samples from population 1 $(\bar{x_{1}})$ and the mean value of the samples from population 2 $(\bar{x_{2}})$
If the sample means $\bar{x_{1}} and \bar{x_{2}}$ differ by too much, reject the null hypothesis; otherwise, do not reject the null hypothesis.

Step by step solution

01

Given information

Given in the question that, we need to discuss the basic strategy for performing a hypothesis test to compare the means of two populations, based on independent samples.

02

Explanation

The concept of independent and dependent samples is employed when comparing two populations. Independent samples means that a sample taken from one population should have no bearing on samples taken from another. In the case of independent samples, the chance of each conceivable outcome is the same.

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Most popular questions from this chapter

Driving Distances. Refer to Exercise 10.48 and determine a 95% confidence interval for the difference between last year's mean VMTs by midwestern and southern households.

Two-Tailed Hypothesis Tests and CIs. As we mentioned on page $413$ , the following relationship holds between hypothesis tests and confidence intervals: For a two-tailed hypothesis test at the significance level $α$ , the null hypothesis $H_{0} : μ_{1} = μ_{2}$ will be rejected in favor of the alternative hypothesis $H_{2} : μ_{1} \neq μ_{2}$ if and only if the $(1 - α)$ -level confidence interval for $μ_{1} - μ_{2}$ does not contain 0. In each case, illustrate the preceding relationship by comparing the results of the hypothesis test and confidence interval in the specified exercises.

a. Exercises $10.81$ and $10.87$

b. Excrcises $10.86$ and $10.92$

In each of Exercises 10.35-10.38, we have provided summary statistics for independent simple random samples from two populations. Preliminary data analyses indicate that the variable under consideration is normally distributed on each population. Decide, in each case, whether use of the pooled $t$ -lest and pooled $t$ -interval procedure is reasonable. Explain your answer.
10.35 ${\bar{x}}_{1} = 468.3, s_{1} = 38.2, n_{1} = 6$

$x_{2} = 394.6, s_{2} = 84.7, n_{2} = 14$

Right-Tailed Hypothesis Tests and CIs. If the assumptions for a pooled $t$ -interval are satisfied, the formula for a $(1 - α)$ -level lower confidence bound for the difference, $μ_{1} - μ_{2}$ , between two population means is

$({\bar{x}}_{1} - {\hat{x}}_{2}) - t_{a} \cdot S_{p} \sqrt{(1 / n_{1}) + (1 / n_{2})}$

For a right-tailed hypothesis test at the significance level $α$ ,

the null hypothesis $H_{0} : μ_{1} = μ_{2}$ will be rejected in favor of the alternative hypothesis $H_{a} : μ_{1} > μ_{2}$ if and only if the $(1 - α)$ -level lower confidence bound for $μ_{1} - μ_{2}$ is greater than or equal to $0$ . In each case, illustrate the preceding relationship by obtaining the appropriate lower confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.

a. Exercise 10.47

b. Exercise 10.50

Data on household vehicle miles of travel (VMT) are compiled annually by the Federal Highway Administration and are published in the National Household Travel Survey, Summary of Travel Trends. Independent random samples of $15$ midwestern households and $14$ southern households provided the following data on last year's VMT, in thousands of miles.

At the $5 %$ significance level, does there appear to be a difference in last year's mean VMT for midwestern and southern households? (Note: ${\bar{x}}_{1} = 16.23, s_{1} = 4.06, {\bar{x}}_{2} = 17.69$ , and $s_{2} = 4.42$ .)

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