Question

Two types of fibers are suitable for use by a fish net manufacturer. the breaking strength of this fabric is important. it is known that $\sigma$₁₌$\sigma$₂=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 fiber 1 unless its breaking strength exceeds that of fiber 2 by at least 10 psi. based on the the sample information, should they use fiber 1? in answering this question, set up and test appropriate hypotheses using $\alpha$=0.01. construct a 99 percent confidence interval on the true mean difference in breaking strength.

Answer 1

a) The following null and alternative hypotheses need to be tested:

$H_0:\mu_1-\mu_2\geq10$

$H_1:\mu_1-\mu_2<10$

This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.01,$ and the critical value for a left-tailed test is $z_c=--2.3263.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}_1-\bar{x}_2-\mu}{\sqrt{\dfrac{\sigma^2_1}{n_1}+\dfrac{\sigma_2^2}{n_2}}}$$

$$=\dfrac{162.5-155-10}{\sqrt{\dfrac{(1)^2}{10}+\dfrac{(1)^2}{12}}}=-5.8387$$

Since it is observed that $z=-5.8387<-2.3263=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=P(z<-5.8387)\approx0,$ and since $p=0<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is not enough evidence to claim that the breaking strength of fabric1 exceeds the breaking strength of fabric 2 by at least 10 psi, at the $\alpha=0.01$ significance level.

b) The critical value for $\alpha=0.01$ is $z_c=z_{1-\alpha/2}=2.5758.$

The corresponding confidence interval is computed as shown below:

$$CI=\bigg(\bar{x}_1-\bar{x}_2-z_c\sqrt{\dfrac{\sigma^2_1}{n_1}+\dfrac{\sigma_2^2}{n_2}},$$

$$\bar{x}_1-\bar{x}_2+z_c\sqrt{\dfrac{\sigma^2_1}{n_1}+\dfrac{\sigma_2^2}{n_2}}\bigg)$$

$$=\bigg(162.5-155-2.5758\sqrt{\dfrac{(1)^2}{10}+\dfrac{(1)^2}{12}},$$

$$162.5-155+2.5758\sqrt{\dfrac{(1)^2}{10}+\dfrac{(1)^2}{12}}\bigg)$$

$$=(6.397, 8.603)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $6.397<\mu<8.603,$ which indicates that we are 99% confident that the true population mean $\mu$ is contained by the interval $(6.397, 8.603).$