Answer :
A 95% confidence interval for the true proportion of all medical students who plan to work in a rural community is (0.0323, 0.0782)
In this question we have been given of 380 randomly selected medical students , 21 said that they planned to work in a rural community.
We need to find a 95% confidence interval for the true proportion of all medical students who plan to work in a rural community.
Here, n = 380, X = 21
So, p{hat} = X/n
p{hat} = 21/380
p{hat} = 0.0553
The confidence interval for the proportion is:
p{hat} ± z_{1 - α/2} √[p{hat}(1 - p{hat})/n]
here, z_{1 - α/2} the standard normal critical value at a level of significance α would be,
z_{1 - α/2} = z_{0.975}
= 1.96
The lower limit of the 95% confidence interval :
p{hat} - z_{1 - α/2} √[p{hat}(1 - p{hat})/n]
= 0.0553 - (1.96) √[0.0553(1 - 0.0553)/380]
= 0.0323
The upper limit of the 95% confidence interval :
p{hat} + z_{1 - α/2} √[p{hat}(1 - p{hat})/n]
= 0.0553 + (1.96) √[0.0553(1 - 0.0553)/380]
= 0.0782
Therefore, a 95% confidence interval is: (0.0323, 0.0782)
Learn more about confidence interval here:
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