Question

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

Answer 1

Answer: (0.1216,0.2784)

Explanation:

The confidence interval for population proportion is given by :-

`\hat{p}\pm z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}`

, where
`\hat{p}` = Sample proportion.

n= sample size.

z* = critical z-value.

Given : sample size = 100

Sample proportion of defective items =
`\hat{p}=(20)/(100)=0.2`

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 :-

`0.2\pm (1.96)\sqrt{(0.2(1-0.2))/(100)}`

`0.2\pm (1.96)√(0.0016)`

`0.2\pm (1.96)(0.04)`

`0.2\pm0.0784`

`(0.2-0.0784,\ 0.2+ 0.0784)= (0.1216,\ 0.2784)`

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