A firm receives components from a supplier, which it uses in its own production.

Question

A firm receives components from a supplier, which it uses in its own production. The components are delivered in batches of 2000. The supplier claims that there are only 1% defective components on average from its production. However, production occasionally goes out of control and a batch is produced with 10% defective components. The firm wishes to intercept these low-quality batches, so a sample of 50 is taken from each batch and tested. If 2 or more defective components are found in a sample then the batch is rejected.

a. Test the hypothesis that the true proportion is 1%, against the alternative of more than 1%.

b. If instead samples of 30 were taken and the batch rejected if 1 or more defective components were found, how would the error probabilities be altered?

Explanation / Answer

a. State the hypotheses: H0:p=0.01 (proportion of defective component is 1%) versus H1:p>0.01 (proportion of defective componnet is greater than 1%).

Assumption: model: random sampling, level of measurement:nominal, sampling distribution:normal. Perform one-proportion z test.

Test statistic, z=(phat-p)/sqrt[p(1-p)/n], where, phat is sample proportion, p is population proportion and n is sample size.

=(2/50-0.01)/sqrt[0.01(1-0.01)/50]

=2.13

p value: 0.017

Per rejection rule based on p value, reject null hypothesis if p value is less than alpha=0.05. Here, p value (0.017) is less than 0.05, therefore, reject null hypothesis and conclude that proportion of defective component is greater than 1%.

b. The probability of Type II error is as follows:

P[Z<={{(p-phat)+zalpha sqrt{p(1-p)/n}}/sqrt{phat(1-phat)/n}]=P[Z<={{0.01-1/30)+1.65 sqrt{0.01(1-0.01)/30}/}sqrt{1/30(1-1/30)/30}]=P[Z<=-0.54]=0.2946

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