Answer:
(a) The estimated proportion of defective in the population is 0.05.
(b) The 95% confidence interval for the proportion defective cups is (2.5%, 7.5%).
(c) Zack does not needs to return the lots.
Step-by-step explanation:
Let <em>X</em> = number of defective cups.
The random sample of cups selected is of size, <em>n</em> = 300.
The number of defective cps in the sample is, <em>X</em> = 15.
(a)
The proportion of the defective cups in the population can be estimated by the sample proportion because the sample selected is quite large.
The sample proportion of defective cups is:

Thus, the estimated proportion of defective in the population is 0.05.
(b)
The (1 - <em>α</em>)% confidence interval for population proportion is:

Compute the critical value of <em>z</em> for 95% confidence level as follows:

Compute the 95% confidence interval for <em>p</em> as follows:



Thus, the 95% confidence interval for the proportion defective cups is (2.5%, 7.5%).
(c)
It is provided that Zack has an agreement with his supplier that he is to return lots that are 10% or more defective.
The 95% confidence interval for the proportion defective is (2.5%, 7.5%). This implies that 95% of the lots have 2.5% to 7.5% defective items.
Thus, Zack does not needs to return the lots.