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An experiment is performed to determine whether the average nicotine content of one kind of cigarette exceeds that of another kind by 0.20 milligram. If n1 = 50 cigarettes of the first kind had an average nicotine content of ī¡ = 2.61 milligrams with a standard deviation of s̟ = 0.12 milligram, whereas n2 = 40 cigarettes of the other kind had an average nicotine content of ãz = 2.38 milligrams with a standard deviation of s2 = 0.14 milligram, test the null hypothesis µ1 – µ2 = 0.20 against the alternative hypothesis µ1 – µ2 · 0.20 at the 0.05 level of significance. %3D %3D 1,2,3,4

An experiment is performed to determine whether the average nicotine content of one kind of cigarette exceeds that of another kind by 0.20 milligram. If n1 = 50 cigarettes of the first kind had an average nicotine content of ī¡ = 2.61 milligrams with a standard deviation of s̟ = 0.12 milligram, whereas n2 = 40 cigarettes of the other kind had an average nicotine content of ãz = 2.38 milligrams with a standard deviation of s2 = 0.14 milligram, test the null hypothesis µ1 – µ2 = 0.20 against the alternative hypothesis µ1 – µ2 · 0.20 at the 0.05 level of significance. %3D %3D 1,2,3,4

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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please use procdeure 1 through 4 in the attachment. math11_10 is th question

An experiment is performed to determine whether the average nicotine content of one kind of cigarette exceeds that of another kind by 0.20 milligram. If n1 = 50 cigarettes of the first kind had an average nicotine content of T1 = 2.61 milligrams with a standard deviation of s1 = 0.12 milligram, whereas n2 = 40 cigarettes of the other kind had an average nicotine content of ã2 = 2.38 milligrams with a standard deviation of s2 = 0.14 milligram, test the null hypothesis µ1 – µ2 = 0.20 against the alternative hypothesis µ1 - µ2 - 0.20 at the 0.05 level of significance. " %3D %3D %3D %3D 1,2,3,4 I
Transcribed Image Text:An experiment is performed to determine whether the average nicotine content of one kind of cigarette exceeds that of another kind by 0.20 milligram. If n1 = 50 cigarettes of the first kind had an average nicotine content of T1 = 2.61 milligrams with a standard deviation of s1 = 0.12 milligram, whereas n2 = 40 cigarettes of the other kind had an average nicotine content of ã2 = 2.38 milligrams with a standard deviation of s2 = 0.14 milligram, test the null hypothesis µ1 – µ2 = 0.20 against the alternative hypothesis µ1 - µ2 - 0.20 at the 0.05 level of significance. " %3D %3D %3D %3D 1,2,3,4 I
Traditionally, it has been the custom to outline tests of hypotheses by means of the following steps: 1. Formulate H, and H1, and specify a. 2. Using the sampling distribution of an appropriate test statistic, determine a critical region of size a. 3. Determine the value of the test statistic from the sample data. 4. Check whether the value of the test statistic falls into the critical region and, accordingly, reject the null hypothesis, or accept it, or reserve judgment.
Transcribed Image Text:Traditionally, it has been the custom to outline tests of hypotheses by means of the following steps: 1. Formulate H, and H1, and specify a. 2. Using the sampling distribution of an appropriate test statistic, determine a critical region of size a. 3. Determine the value of the test statistic from the sample data. 4. Check whether the value of the test statistic falls into the critical region and, accordingly, reject the null hypothesis, or accept it, or reserve judgment.
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