of 380 randomly selected medical students, 21 said that they planned to work in a rural community. find a…

Explanation:

Step1: Calculate sample proportion

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

Step2: Determine z - value for 95% confidence interval

For a 95% confidence interval, the significance level $\alpha = 1 - 0.95=0.05$, and $\alpha/2=0.025$. The $z$-value $z_{\alpha/2}=z_{0.025} = 1.96$.

Step3: Calculate the margin of error

The formula for the margin of error $E$ for a proportion is $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.0553$, $n = 380$ and $z_{\alpha/2}=1.96$ into the formula: [ \begin{align*} E&=1.96\sqrt{\frac{0.0553\times(1 - 0.0553)}{380}}\ &=1.96\sqrt{\frac{0.0553\times0.9447}{380}}\ &=1.96\sqrt{\frac{0.05224291}{380}}\ &=1.96\sqrt{0.000137481342}\ &\approx1.96\times0.011725\ &\approx0.02298 \end{align*} ]

Step4: Calculate the confidence interval

The confidence interval for the population proportion $p$ is $\hat{p}-E<p<\hat{p} + E$. $\hat{p}-E=0.0553- 0.02298=0.03232$ and $\hat{p}+E=0.0553 + 0.02298=0.07828$. So the 95% confidence interval is $0.0323<p<0.0782$.

Answer:

D. $0.0323 < p < 0.0782$