Question

Q3. Two companies manufacture a rubber material intended for use in an automotive application. The part will be subjected to abrasive wear in the field application, so we decide to compare the material produced by each company are tested in an abrasion test, and the amount of wear after 1000 cycles is observed. For company 1, the sample mean and standard deviation of wear are x ̅1 = 20 milligrams/1000 cycles and s1 = 2 milligrams/1000 cycles, while for company 2 we obtain x ̅2 = 15 milligrams/1000 cycles and s2 = 8 milligrams/1000 cycles.
(a) Do the data support the claim that the two companies produce material with different mean wear? Use  = 0.05, and assume each population is normally distributed but that their variances are not equal.
(b) What is P-value for this test?
(c) Do the data support a claim that the material from company 1 has higher mean wear than the material from company 2? Use the same assumptions as in part (a).

Answer 1

$a) H_0: \mu_1=\mu_2; H_1:\mu_1\neq \mu_2\text{ (two-sided)}.\\ \alpha=0.05\\ \overline{x}_1=20\\ \overline{x}_2=15\\ s_1=2\\ s_2=8\\ n=m=1000\\ \text{We will use the following random variable:}\\ Z=\frac{\overline{X}-\overline{Y}}{\sqrt{\frac{D_e(X)}{n}+\frac{D_e(Y)}{m}}}.\\Z=\frac{20-15}{\sqrt{\frac{2^2}{1000}+\frac{8^2}{1000}}}\approx 19.17.\\ \Phi(z_{cr})=\frac{1-\alpha}{2}=0.475.\\ z_{cr}=1.96.\\ (-\infty, -1.96)\cup (1.96,\infty)\text{ --- critical region}.\\ Z \in \text{critical region. So we reject } H_0.\\ b)\text{p-value}=2(0.5-\Phi(19.17))\approx 0\text{ where }\\ \Phi(x)=\frac{1}{\sqrt{2\pi}}\int_0^x e^{-\frac{z^2}{2}}dz.\\ \text{p-value}<\alpha. \text{ So we reject } H_0.\\ c)H_0: \mu_1=\mu_2; H_1:\mu_1>\mu_2\text{ (one-sided)}.\\ \text{We get } Z\approx 19.17.\\ \Phi(z_{cr})=\frac{1-2\alpha}{2}=0.45.\\ z_{cr}=1.64.\\ (-\infty, -1.64)\cup (1.64,\infty)\text{ --- critical region}.\\ Z \in \text{critical region. So we reject } H_0.$