Question

1. Fasten-it-All produces wood screws. The screws must be manufactured to within certain tolerances or they are considered defective. Using 99% confidence, a machine that may be producing more than 1.35% defective screws must be shut down. To test Machine 1, a QC inspector randomly samples 1500 screws. The QC inspector’s random sample of 1500 screws contains 12 defective screws.

1. Based on this sample, compute by hand (using the normal approximation to the binomial) and interpret a 99% confidence interval for Machine 1’s defect rate. Confirm your result with Minitab. (NOTE: You must find the correct multiplier using Minitab.)
2. Based on the 99% confidence interval computed in a, what recommendation would you make regarding whether Machine 1 should be shut down?
3. Suppose the QC inspector decided, instead, that a 90% confidence interval could be used. Compute by hand (using the normal approximation to the binomial) and interpret a 90% confidence interval for Machine 1’s defect rate? Confirm your result with Minitab.
4. Based on the 90% confidence interval computed in c, what recommendation would you make regarding whether Machine 1 should be shut down? Briefly explain, how the same sample produced different recommendations in parts b and d.

(Remember: Do the by hand computations, but they do not need to be included in your paper.)

Answer 1

a)

The confidence interval for the proportion is obtained using the formula,

Where,

Interpretation: There is a 99% chance that Machine 1’s defect rate lies within an interval from 0.2% to 1.4%.

b)

The 99% confidence interval for the Machine 1’s defect rate includes the critical 1.35% defective rate hence the Machine 1 should be shut down

c)

Where,

Interpretation: There is a 90% chance that Machine 1’s defect rate lies within an interval from 0.4% to 1.2%.

d)

The 90% confidence interval for the Machine 1’s defect rate doesn't include the critical 1.35% defective rate hence the Machine 1 should not be shut down

Since the significance level for the 90% confidence interval in part (d) is increased from 0.01 in part (a) to 0.10, the margin of error is decreased. Hence the 90% confidence interval is less wide compared to a 99% confidence interval. This is the reason why the recommendation changed in part (d).