In this section, we introduced the pooled t-test, which provides a method for comparing two population means. In deriving the pooled f-test, we stated that the variable

z=f^1-x^2-μ1-μ2σ1/n1+1/n2

cannot be used as a basis for the required test statistic because σ is unknown. Why can't that variable be used as a basis for the required test statistic?

Short Answer

Expert verified

sp integrates data about the variation of both samples into a single estimate of the population standard deviation's common value. As a result, sp stands for pooled standard deviation.

Step by step solution

01

Given Information

In deriving the pooled t- test, it is stated that the variable z=x¯1-x¯2-μ1-μ2σ1n1+1n2

02

Explanation

A test statistic is considered when performing a hypotheses test based on independent samples to compare the means of two populations with equal and unknown standard deviation.

The variable z=x¯1-x¯2-μ1-μ2σ1n1+1ncannot be used as a basis for the required test.

Because σis unknown,

In this case, the parameter σrepresents the average standard deviation of two populations. Because the standard deviation of the majority of the population is unknown in general, the common standard deviation of the two populations is also unlikely to be known.

As a result, the individual sample variancess12 and s22of two populations are taken and pooled by weighting them based on their sample size or degree of freedom, n1and n2

sp=n1-1s12+n2-1s22n1+n2-2

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Most popular questions from this chapter

The primary concern is deciding whether the mean of Population 1 differs from the mean of Population 2 .

Acute Postoperative Days. Refer to Example 10.6(page 420). The researchers also obtained the following data on the number of acute postoperative days in the hospital using the dynamic and static systems.

At the 5%significance level, do the data provide sufficient evidence to conclude that the mean number of acute postoperative days in the hospital is smaller with the dynamic system than with the static system? (Note: x_{1}=7.36,s_{1}=1.22,x_{2}=10.50and s_{2}=4.59.)

The primary concern is deciding whether the mean of Population 2 differs from the mean of Population 1 .

Stressed-Out Bus Drivers. An intervention program designed by the Stockholm Transit District was implemented to improve the work conditions of the city's bus drivers. Improvements were evaluated by G. Evans et al., who collected physiological and psychological data for bus drivers who drove on the improved routes (intervention) and for drivers who were assigned the normal routes (control). Their findings were published in the article "Hassles on the Job: A Study of a Job Intervention with Urban Bus Drivers" (Journal of Organizational Behavior, Vol. 20, pp. 199-208). Following are data, based on the results of the study, for the heart rates, in beats per minute, of the intervention and control drivers.

a. At the 5%significance level, do the data provide sufficient evidence to conclude that the intervention program reduces mean heart rate of urban bus drivers in Stockholm? (Note; x1=67.90, s1=5.49,x¯2=66.81and s2=9.04.

b. Can you provide an explanation for the somewhat surprising results of the study?

c. Is the study a designed experiment or an observational study? plain your answer.

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, μ1-μ2, between two population means is

x¯1-x^2-ta·Sp1/n1+1/n2

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

the null hypothesis H0:μ1=μ2will be rejected in favor of the alternative hypothesis Ha:μ1>μ2if and only if the (1-α)-level lower confidence bound for μ1-μ2is 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

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