Suppose that in sampling for the population proportion, it is found that 20 out of 100 items are defective. construct a 95% confidence interval for the proportion of defective items in the population

(0.1216,0.2784)

Step-by-step explanation:

The confidence interval for population proportion is given by :-

, where = Sample proportion.

n= sample size.

z* = critical z-value.

Given : sample size = 100

Sample proportion of defective items =

Confidence level =95%

We know that , Critical z-value for 95% confidence interval  = 1.96

Then, the confidence interval for the proportion of defective items in the population will be :-

Hence, the 95% confidence interval for the proportion of defective items in the population =  (0.1216,0.2784)

answer is 33

step-by-step explanation: