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) .

Step-by-step explanation:

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

$\hat{p}\pm z^* \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$   (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

$\hat{p}=\dfrac{30}{375}=0.08$

Critical z-value for 95% confidence interval = 1.96

Put all values in (1) , we get

$0.08\pm(1.96)\sqrt{\dfrac{0.08(1-0.08)}{375}}$  

$0.08\pm(1.96)\sqrt{0.000196}$  

$0.08\pm(1.96)(0.014)$  

$0.08\pm0.02744$  

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

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) .