Question

) A quality control engineer is interested in estimating the proportion of defective items coming off a production line. In a sample of 200 items, 28 are defective. a) Find the 95% confidence interval for the proportion of defectives from this production line b) Interpret the results (the interval) you got

Answer 1

Step-by-step explanation:

Here are the solutions:

For the sample size of 300, with 27 defectives, the sample proportion is p-hat = 27/300 = 0.09.

For a 90% confidence interval of a proportion, use z = 1.645.

The limits of the interval are then:

Lower limit = p-hat - z* sqrt(p-hat*(1 - p-hat)/n) = 0.09 - 1.645*sqrt(0.09*0.91/300) = 0.0628

Upper limit = p-hat + z* sqrt(p-hat*(1 - p-hat)/n) = 0.09 - 1.645*sqrt(0.09*0.91/300) = 0.1172

The 90% confidence interval is then (0.0628, 0.1172)

For the larger sample, assuming that the same proportion of defectives is found (p-hat = 0.09), the new confidence interval would be:

Lower limit = p-hat - z* sqrt(p-hat*(1 - p-hat)/n) = 0.09 - 1.645*sqrt(0.09*0.91/20,000) = 0.0867

Upper limit = p-hat + z* sqrt(p-hat*(1 - p-hat)/n) = 0.09 - 1.645*sqrt(0.09*0.91/20,000) = 0.0933

In this case, the 90% confidence interval is (0.0867, 0.0933)

Thanks,