Question

a sociologist is studying the prevalence of crime in one major city. in a sample of 300 randomly selected residents, 96 say that they have been victimized by a criminal. based on this sample, construct a 95% confidence interval for the proportion of all residents in this city who have been victimized by a criminal. then find the lower limit and upper limit of the 95% confidence interval.
carry your intermediate computations to at least three decimal places. round your answers to two decimal places. (if necessary, consult a list of formulas.)
lower limit:
upper limit:

Explanation:

Step1: Calculate sample proportion $\hat{p}$

Sample proportion $\hat{p}=\frac{x}{n}$, where $x = 96$ (number of successes) and $n=300$ (sample size). So, $\hat{p}=\frac{96}{300}=0.32$.

Step2: Find $z$-value for 95% confidence interval

For a 95% confidence interval, the $z$-value $z_{\alpha/2}$ is 1.96.

Step3: Calculate standard error $SE$

Standard error $SE=\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substituting $\hat{p}=0.32$ and $n = 300$, we get $SE=\sqrt{\frac{0.32\times(1 - 0.32)}{300}}=\sqrt{\frac{0.32\times0.68}{300}}\approx\sqrt{\frac{0.2176}{300}}\approx\sqrt{0.000725}\approx0.027$.

Step4: Calculate margin of error $ME$

Margin of error $ME=z_{\alpha/2}\times SE$. Substituting $z_{\alpha/2}=1.96$ and $SE\approx0.027$, we get $ME = 1.96\times0.027\approx0.053$.

Step5: Calculate lower and upper limits

Lower limit $=\hat{p}-ME=0.32 - 0.053 = 0.267\approx0.27$.
Upper limit $=\hat{p}+ME=0.32+ 0.053=0.373\approx0.37$.

Answer:

Lower limit: $0.27$
Upper limit: $0.37$