= 5. Two types of plastic are suitable for use by an electronic calculator manufacturer. The breaking strength of this plastic is important. It is known that 0₁ 02 = 1.0 psi. From random samples of n₁ = 10 and n₂ = 12 we obtain ₁ = 162.5 and y2 = 155.0. The company will not adopt plastic 1 unless its breaking strengths exceeds that of plastic 2 by at least 10 psi. Based on the sample information, should they use plastic 1? In answering this question, set up and test appropriate hypotheses using a = 0.01. Construct a 99 percent confidence interval on the true mean difference in breaking strength.

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5. Two types of plastic are suitable for use by an electronic calculator manufacturer. The breaking
strength of this plastic is important. It is known that 0₁ = 02 = 1.0 psi. From random samples
of n₁ = 10 and n₂ = 12 we obtain y₁ = 162.5 and y2 = 155.0. The company will not adopt plastic
1 unless its breaking strengths exceeds that of plastic 2 by at least 10 psi. Based on the sample
information, should they use plastic 1? In answering this question, set up and test appropriate
hypotheses using a = 0.01. Construct a 99 percent confidence interval on the true mean difference
in breaking strength.
Transcribed Image Text:5. Two types of plastic are suitable for use by an electronic calculator manufacturer. The breaking strength of this plastic is important. It is known that 0₁ = 02 = 1.0 psi. From random samples of n₁ = 10 and n₂ = 12 we obtain y₁ = 162.5 and y2 = 155.0. The company will not adopt plastic 1 unless its breaking strengths exceeds that of plastic 2 by at least 10 psi. Based on the sample information, should they use plastic 1? In answering this question, set up and test appropriate hypotheses using a = 0.01. Construct a 99 percent confidence interval on the true mean difference in breaking strength.
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