A random sample of $n=100$ individuals is collected, and a confidence interval for the population proportion was constructed: $(0.12,0.26)$. A. Compute a point estimate of the population proportion. Point estimate $\hat{p}=$ B. Compute the margin of error. Margin of error = C. Compute the number of "successes" $x$. \[ x= \]
Solution
Step 1 ：A random sample of 100 individuals is collected, and a confidence interval for the population proportion was constructed: (0.12,0.26).
Step 2 ：The point estimate of the population proportion is the midpoint of the confidence interval. So, we calculate the point estimate as the average of the lower and upper bounds of the confidence interval. Hence, point estimate \( \hat{p} = \frac{0.12 + 0.26}{2} = \boxed{0.19} \).
Step 3 ：The margin of error is the difference between the point estimate and either end of the confidence interval. So, we calculate the margin of error as the difference between the point estimate and the lower bound of the confidence interval. Hence, margin of error = \( \hat{p} - 0.12 = 0.19 - 0.12 = \boxed{0.07} \).
Step 4 ：The number of 'successes' can be calculated by multiplying the point estimate by the sample size. Hence, number of 'successes' \( x = \hat{p} \times n = 0.19 \times 100 = \boxed{19} \).
