Question

A manufacturer wants to estimate the proportion of defective items in production. However, 18 of the 8,000 items in a random sample were found to be defective.

(a) Establish a 99% confidence interval for the proportion of defective items.

(b) If a confidence interval for the proportion of non-defective items had been constructed from of the same data, would you have gotten a larger margin of error?

(c) A proportion of defective items of 3/1000 is considered acceptable. Are we sure of meet the standard in this factory with a 99% confidence level? Justify it.

Answer 1

(a)

Sample proportion, = 18/8000 = 0.00225

Standard error of sample proportion, SE =

= 0.0005297331

Z value for 99% confidence interval is 2.576

Margin of error, E = Z * SE = 2.576 * 0.0005297331 = 0.001364592

99% confidence interval for the proportion of defective items is,

(0.00225 - 0.001364592,  0.00225 + 0.001364592)

(0.000885408 , 0.003614592)

(b)

Sample proportion of non-defective items = 1 -

Sample proportion of non-defective items =

As, the standard error of Sample proportion of non-defective items is equal to that for proportion of defective items, the margin of error will be same for confidence interval for the proportion of non-defective items had been constructed from of the same data.

(c)

Sample proportion, = 3/1000 = 0.003

Since the given proportion lies in the  99% confidence level calculated in part (a), we are sure of meeting the standard in this factory with a 99% confidence level.