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Why do you need to know the sampling distribution of the difference between two sample means in order to perform a hypothesis test to compare two population means?

Short Answer

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It is important to know the sampling distribution of the difference between two sample means in order to perform a hypothesis test to compare two population means because μ1-μ2 allows researchers to assess if the difference between the sample means is small enough to be attributed to sampling error or high enough to conclude that the population means are different.

Step by step solution

01

Given Information

Given In the question that a hypothesis test is performed to compare two population means. we have to know the sampling distribution of the difference between two sample means in order to perform a hypothesis test to compare two population means.

02

Explanation

A concept that can be tested is known as a hypothesis. To test the hypothesis, many statistical models are used.

The term "generally accepted fact" is used to describe a null hypothesis. The null hypotheses are being challenged, disproved, and nullified by researchers. The researcher offers an alternative hypothesis to refute the null hypothesis.

Assume that μ1is the mean of a variable in population 1, and μ2is the mean of a variable in population 2. The sampling distribution of the difference between two means is designatedμ1-μ2in this case.

This μ1-μ2allows researchers to assess if the difference between the sample means is small enough to be attributed to sampling error or high enough to conclude that the population means are different.

As a result, while running a hypothesis test to compare two population means, it's critical to understand the sampling distribution of the difference between two sample means.

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Most popular questions from this chapter

The intent is to employ the sample data to perform a hypothesis test to compare the means of the two populations from which the data were obtained. In each case, decide which of the procedures should be applied.

Independent: n1=20

n2=15

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 x¯1=468.3,s1=38.2,n1=6

x2=394.6,s2=84.7,n2=14

Ha:μ1≠μ2

In each of exercise 10.13-10.18, we have presented a confidence interval for the difference,μ1andμ2, between two population means. interpret each confidence interval.

95%CI is from-20to-15

Left-Tailed Hypothesis Tests and CIs. If the assumptions for a nonpooled t-interval are satisfied, the formula for a (1-α) level upper confidence bound for the difference, μ1-μ2. between two population means is

f1-f2+t0·s12/n1+s22/n2

For a left-tailed hypothesis test at the significance level α, the null hypothesis H0:μ1=μ2 will be rejected in favor of the alternative hypothesis H2:μ1<μ2 if and only if the (1-α)-level upper confidence bound for μ1-μ2 is less than or equal to 0. In each case, illustrate the preceding relationship by obtaining the appropriate upper confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.

a. Exercise 10.83

b. Exercise 10.84

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