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Two types of plastic are suitable for an electronics component manufacturer to use. The breaking strength of this plastic is important. It is known that σ1 =σ2 = 1.0 psi. From a random sample of size n1 =10 and n2 =12, you obtain ¯x1 =162.5 and ¯x2 = 155.0. The company will not adopt plastic 1 unless its mean breaking strength exceeds that of plastic 2 by at least 10 psi (inclusive). (a) Based on the sample information, should it use plastic 1? Test this using α=0.05. Write formal null and alternate hypotheses statements for the test. (b) Find the p-value for this test.

Respuesta :

Answer:

(a) Null Hypothesis (H₀) = Mean of x₁ > Mean of x₂

Alternative Hypothesis (H₁) = Mean of x₁ ≠ Mean of x₂

(b) We will use here t-test for two sample means which is given by,

[tex][tex]t = \frac{\bar_{x_{1}}-\bar_{x_{2}}}{\sqrt{s^2(\frac{1}{n_{1}}+\frac{1}{n_{2}}  )}}[/tex][/tex]

Putting all values we get, t = 17.5 with 20 degree of freedom

At α = 0.05

p-value = 95% Confidence Interval for the Difference ( 6.6068 , 8.3932 )

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