Question

8.2 one - sample interval for the proportion. construct and interpret a confidence interval for the population proportion. out of 200 people sampled, 112 preferred candidate a. based on this, estimate what proportion of the voting population (p) prefers candidate a. use a 90% confidence level, and give your answers as decimals, to three places. < p < question help: message instructor post to forum

Explanation:

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

$\hat{p}=\frac{x}{n}=\frac{112}{200} = 0.56$

Step2: Determine $z$-score for 90% confidence level

For a 90% confidence level, $\alpha=1 - 0.90=0.10$, and $\alpha/2=0.05$. The $z$-score $z_{\alpha/2}=z_{0.05}\approx1.645$

Step3: Calculate margin of error $E$

$E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=1.645\sqrt{\frac{0.56\times(1 - 0.56)}{200}}=1.645\sqrt{\frac{0.56\times0.44}{200}}\approx1.645\times0.035 = 0.0576$

Step4: Calculate confidence - interval

Lower limit: $\hat{p}-E=0.56 - 0.0576 = 0.502$
Upper limit: $\hat{p}+E=0.56 + 0.0576 = 0.618$

Answer:

$0.502 < p < 0.618$