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Two types of plastic are suitable for use by an electronics component manufacturer. The breaking strength of this plastic is important. It is known that σ1 = σ2 = 1.0 psi. From a random sample of size of n1 = 10 and n2 = 12, we obtain x̄1 = 162.5 and x̄2 = 155.0. The company will not adopt plastic 1 unless its mean breaking strength exceeds that of plastic 2 by at least 10 psi. Based on the sample information, should it use plastic

Two types of plastic are suitable for use by an electronics component manufacturer. The breaking strength of this plastic is important. It is known that σ1 = σ2 = 1.0 psi. From a random sample of size of n1 = 10 and n2 = 12, we obtain x̄1 = 162.5 and x̄2 = 155.0. The company will not adopt plastic 1 unless its mean breaking strength exceeds that of plastic 2 by at least 10 psi. Based on the sample information, should it use plastic

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Two types of plastic are suitable for use by an electronics component manufacturer. The breaking strength of this plastic is important. It is known that σ1 = σ2 = 1.0 psi. From a random sample of size of n1 = 10 and n2 = 12, we obtain x̄1 = 162.5 and x̄2 = 155.0. The company will not adopt plastic 1 unless its mean breaking strength exceeds that of plastic 2 by at least 10 psi. Based on the sample information, should it use plastic 1? Use α = 0.05 in reaching a decision.

 

State the null and alternative hypotheses.

A B C (D E H0₁=₂ +10 H₁:μ₁ +10<μ₂ Ho" -",=10 H₁:μ₁₂+10 H₁-H₂=10 H₁:₁ +10>₂ H₁μ₁ = μ₂ + 10 H₁₁<₂+10 H₁M₁-M₂=10 H₁:₁-₂10
Transcribed Image Text:A B C (D E H0₁=₂ +10 H₁:μ₁ +10<μ₂ Ho" -",=10 H₁:μ₁₂+10 H₁-H₂=10 H₁:₁ +10>₂ H₁μ₁ = μ₂ + 10 H₁₁<₂+10 H₁M₁-M₂=10 H₁:₁-₂10
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