Question

Two extrusion machines that manufacture polyester fibers are being compared. in a sample of 1000 fibers taken from machine 1, 960 met specifications regarding fineness and strength. in a sample of 600 fibers taken from machine 2, 582 met the specifications. machine 2 is more expensive to run, so it is decided that machine 1 will be used unless it can be convincingly shown that machine 2 produces a larger proportion of fibers meeting specifications.

a. state the appropriate null and alternate hypothesis for making decision as to which machine to use.

b. compute the test statistics.

c. which machine should be used?

Answer 1

a. For machine 1, we have that the sample size is $n_1=1000,$ the number of favorable cases is $x_1=960,$ so then the sample proportion is $\hat{p}_1=\dfrac{950}{1000}=0.96.$

For machine 2, we have that the sample size is $n_2=600,$ the number of favorable cases is $x_2=582,$ so then the sample proportion is $\hat{p}_2=\dfrac{582}{600}=0.97.$

The value of the pooled proportion is computed as

$$\bar{x}=\dfrac{x_1+x_2}{n_1+n_2}=\dfrac{960+582}{1000+600}=0.96375$$

Assume that the significance level is $\alpha=0.05.$

The following null and alternative hypotheses for the population proportion needs to be tested:

$H_0:p_1\geq p_2$

$H_1:p_1< p_2$

This corresponds to a left-tailed test, and a z-test for two population proportions will be used.

b. Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a left-tailed test is $z_c=-1.6449.$

The rejection region for this left-tailed test is $R=\{z:z<-1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}_1-\hat{p}_2}{\sqrt{\bar{p}(1-\bar{p})(1/n_1+1/n_2)}}$$

$$=\dfrac{0.96-0.97}{\sqrt{0.96375(1-0.96375)(1/1000+1/600)}}$$

$$\approx-1.0360$$

Since it is observed that $z=-1.0360>-1.6449=z_c,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p_1$ is less than $p_2,$ at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value is $p=P(Z<-1.0360)=0.15009,$ and since $p=0.15009>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p_1$ is less than $p_2,$ at the $\alpha=0.05$ significance level.

c. Since there is not enough evidence to claim that the population proportion $p_1$ for machine 1is less than $p_2$ for machine 2, and machine 2 is more expensive to run, machine 1 should be used.