Final answer:
The 90% confidence interval for the true proportion of all medical students who plan to work in a rural community is between 3.65% and 7.41%, implying that there is a 90% confidence that the true proportion falls within this range.
Step-by-step explanation:
We want to find a 90% confidence interval for the true proportion of all medical students who plan to work in a rural community. Given that out of 380 randomly selected medical students, 21 said that they planned to work in a rural community, we will use the formula for a confidence interval for a proportion:
CI = p ± Z*sqrt[(p(1-p)/n)]
Where:
- p is the sample proportion (21/380)
- n is the sample size (380)
- Z* is the Z-value from the standard normal distribution for the given confidence level (In this case, for a 90% confidence interval, Z is approximately 1.645)
Calculating the sample proportion we get p = 21/380 = 0.0553. Then the standard error SE is calculated as:
SE = sqrt[(0.0553*(1-0.0553)/380)] = 0.0114
The margin of error (ME) is:
ME = Z* * SE = 1.645 * 0.0114 = 0.0188
The 90% confidence interval is thus:
CI = 0.0553 ± 0.0188 = (0.0365, 0.0741)
Rounded to four decimal places, the interval is (0.0365, 0.0741).
Interpreting the confidence interval, we can say that we are 90% confident that the true proportion of all medical students who plan to work in a rural community falls between 3.65% and 7.41%.