Question

Of 118 randomly selected adults, 34 were found to have high blood pressure. construct a 95% confidence interval for the true percentage of all adults that have high blood pressure.

Answer 1

The correct answer is: 20.6% < p < 37.0%

Answer 2

Answer:
`(20.63\%,\ 36.97\% )`

Explanation:

The confidence interval for population proportion(p) is given by :-

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

n= Sample size

z*= Critical z-value.

`\hat{p}` = sample proportion.

Let p be the true proportion of all adults that have high blood pressure.

As per given , we have

n= 118

Number of adults found to have high blood pressure =34

Then,
`\hat{p}=(34)/(118)\approx0.288`

Critical z-value for 95% confidence interval : z* = 1.96

Now , the 95% confidence interval for population proportion will be :

`0.288\pm (1.96)\sqrt{(0.288(1-0.288))/(118)}`

`0.288\pm (1.96)√(0.0017377627)`

`0.288\pm (1.96)(0.04168648)`

`0.288\pm0.0817`

`=(0.288-0.0817,\ 0.288+0.0817) =(0.2063,\ 0.3697 )`

In percentage , this would be
`(0.2063,\ 0.3697 )=(20.63\%,\ 36.97\% )`

Hence, the 95% confidence interval for the true percentage of all adults that have high blood pressure =
`(20.63\%,\ 36.97\% )`