Question

Of 375 randomly selected medical students, 30 said that they planned to work in a rural community. Find a 95% confidence interval for the true population proportion of all medical students who plan to work in a rural community.

Answer 1

Answer: (0.05256, 0.10744) .

Explanation:

We know that the confidence interval for population proportion is given by :-

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

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

n= Sample size

z* = Critical z-value.

Let p be the population proportion of all medical students who plan to work in a rural community.

As per given , we have

n= 375

`\hat{p}=(30)/(375)=0.08`

Critical z-value for 95% confidence interval = 1.96

Put all values in (1) , we get

`0.08\pm(1.96)\sqrt{(0.08(1-0.08))/(375)}`

`0.08\pm(1.96)√(0.000196)`

`0.08\pm(1.96)(0.014)`

`0.08\pm0.02744`

`=(0.08-0.02744,\ 0.08+0.02744) =(0.05256,\ 0.10744)`

Hence, the 95% confidence interval for the true population proportion of all medical students who plan to work in a rural community is (0.05256, 0.10744) .