Question

Two types of plastics are suitable for an electronics component manufacturer to use. The breaking strength of this plastic is important. It is known that the standard deviations of the two types of plastics are the same, with a value of 1.0 psi. From a random sample of 10 and 12 for type 1 and type 2 plastics, respectively, we obtain sample means of 162.5 and 155. The company will not adopt plastic 1 unless its mean breaking strength exceeds that of plastic 2 by at least 10 psi.
(a) Based on the sample information, should it use plastic 1? Use α = 0.05 in reaching a decision. find the P-value.

(b) Calculate a 95% confidence interval on the difference in means. Suppose that the true difference in means is really 12 psi.

(c) Find the power of the test assuming that α = 0.05.

(d) If it is really important to detect a difference of 12 psi, are the sample sizes employed in part (a) adequate, in your opinion?

Answer 1

Answer:

a. We fail reject to the null hypothesis because zo = -5.84 < 1.65 = zα and P-value = 1 (approximately)

b. The confidence Interval for u1 - u2 is; 6.79 ≤ u1 - u2

c. The power of the test = 1 -

β = 0.998736

d. The sample size is adequate because the power of the test is approximately 1

Step-by-step explanation:

Given

Standard Deviations; σ1 = σ2 = 1.0 psi

Size: n1 = 10; n2 = 12

X = 162.5; Y = 155.0

Let X1, X2....Xn be a random sample from Population 1

Let Y1, Y2....Yn be a random sample from Population 2

We assume that both population are normal and the two are independent.

Therefore, the test statistic

Z = (X - Y - (u1 - u2))/√(σ1²/n1 + σ2²/n2)

See attachment for explanation