Question

of 380 randomly selected medical students, 21 said that they planned to work in a rural community. find a 95% confidence interval for the true proportion of all medical students who plan to work in a rural community.
a. 0.0251 < p < 0.0854
b. 0.0323 < p < 0.0782
c. 0.0360 < p < 0.0745
d. 0.0280 < p < 0.0826

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 21$ (number of successes) and $n=380$ (sample size).
$\hat{p}=\frac{21}{380}\approx0.0553$

Step2: Calculate $q$

Since $q = 1-\hat{p}$, then $q=1 - 0.0553=0.9447$

Step3: Find $z$-value for 95% confidence interval

For a 95% confidence interval, the $z$-value $z_{\alpha/2}=1.96$ (from standard normal distribution table)

Step4: Calculate margin of error $E$

The formula for margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}\hat{q}}{n}}$
Substitute the values:
$E=1.96\sqrt{\frac{0.0553\times0.9447}{380}}$
First calculate $\frac{0.0553\times0.9447}{380}=\frac{0.0522}{380}\approx0.000137$
Then $\sqrt{0.000137}\approx0.0117$
$E = 1.96\times0.0117\approx0.0229$

Step5: Calculate confidence interval

The confidence interval is $\hat{p}-ESubstitute $\hat{p}=0.0553$ and $E = 0.0229$
$0.0553- 0.0229=0.0324$
$0.0553+0.0229=0.0782$

Answer:

B. \(0.0323 < p < 0.0782\)