Question

out of 500 people sampled, 380 preferred candidate a. round to three decimals. based on this estimate, what proportion (as a decimal) of the voting population (p) prefers candidate a? 76 compute a 95% confidence interval, and give your answers to 3 decimal places. < p <

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 380$ (number of successes, i.e., people who prefer Candidate A) and $n=500$ (sample size). So, $\hat{p}=\frac{380}{500}=0.760$.

Step2: Calculate standard error

The standard error for a proportion is $SE=\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.760$ and $n = 500$ into the formula: $SE=\sqrt{\frac{0.760\times(1 - 0.760)}{500}}=\sqrt{\frac{0.760\times0.240}{500}}=\sqrt{\frac{0.1824}{500}}\approx\sqrt{0.0003648}\approx0.019$.

Step3: Find z - value for 95% confidence interval

For a 95% confidence interval, the z - value $z = 1.96$.

Step4: Calculate lower limit of confidence interval

The lower limit $LL=\hat{p}-z\times SE$. Substitute $\hat{p}=0.760$, $z = 1.96$ and $SE\approx0.019$: $LL=0.760-1.96\times0.019=0.760 - 0.03724\approx0.723$.

Step5: Calculate upper limit of confidence interval

The upper limit $UL=\hat{p}+z\times SE$. Substitute $\hat{p}=0.760$, $z = 1.96$ and $SE\approx0.019$: $UL=0.760 + 1.96\times0.019=0.760+0.03724\approx0.797$.

Answer:

The proportion of the voting population that prefers Candidate A is $0.760$.
$0.723 < p < 0.797$