Answer to Question #200412 in Statistics and Probability for Kamilia

Question

An operations manager of a computer chip maker is in the process of selecting a new machine to replace several older ones. Although technological innovations have improved the production process, it is quite common for the machines to produce defective chips. The operations manager must choose between two machines. The cost of machine A is several thousand dollars greater than the cost of machine B. After an analysis of the costs it was determined that machine A is warranted provided that its defective rate is more than 2% less than that of machine B. To help decide both machines are used to produce 200 chips each. Each chip was examined and whether it was defective (code = 2) or not (code = 1) was recorded. Should the operations manager select machine A

Answer 1

We can apply z-test for proportions:

test statistics:

$z=\frac{p_1-p_2-\Delta}{\sqrt{p(1-p)(1/n_1+1/n_2)}}$

In our case, for example:

$H_0:\ p_1-p_2>0.02$

$H_{\alpha}:\ p_1-p_2\le0.02$

significance level is $\alpha=5\ \%$

$n_1=n_2=200$ are samples sizes,

Let:

$p_1=0.06$ is defective rate of machine B,

$p_2=0.03$ is defective rate of machine A,

$p=\frac{0.06\cdot200+0.03\cdot200}{200+200}=0.045$

$\Delta=0.02$ is estimated difference between defective rates of machines A and B.

Then:

$z=\frac{0.06-0.03-0.02}{\sqrt{0.045(1-0.045)(1/200+1/200)}}=0.48$

p-value:

$P((p_1-p_2)>0.02)=P(z>0.48)=1-0.68439=0.31561$

Since p-value is more than $\alpha$ , we accept the null hypothesis. So, the operations manager will select machine A.