Question

Two different types of injection-molding machines are used to form plastic parts. A part is considered defective if it has excessive shrinkage or is discolored. Two random samples, each of size 300, are selected, and 15 defective parts are found in the sample from machine 1 whereas 8 defective parts are found in the sample from machine 2. Is it reasonable to conclude that both machines produce the same fraction of defective parts, using 0.05? Find the p-value for this test.

Answer 1

To test whether both machines produce the same fraction of defective parts, a two-sample test for proportions can be used. The null hypothesis assumes that the proportions are equal, while the alternative hypothesis assumes they are not equal. The p-value can be calculated to determine the likelihood of observing the test statistic under the null hypothesis.

Step-by-step explanation:

To test whether both machines produce the same fraction of defective parts, we will use a two-sample test for proportions. We will set up the null and alternative hypotheses as follows:

Null Hypothesis (H0): The proportion of defective parts produced by machine 1 is equal to the proportion of defective parts produced by machine 2.

Alternative Hypothesis (HA): The proportion of defective parts produced by machine 1 is not equal to the proportion of defective parts produced by machine 2.

We can calculate the test statistic using the formula:

z = ((p1 - p2) - 0) / sqrt((p1 * (1-p1))/n1 + (p2 * (1-p2))/n2)

where p1 and p2 are the sample proportions of defective parts from machine 1 and machine 2 respectively, and n1 and n2 are the sample sizes.

The p-value can be calculated by finding the probability of getting a test statistic as extreme as the observed value under the null hypothesis.