Answer to Question #193591 in Statistics and Probability for Jack Walker

Question

For a sample size of 100 people, what is the margin of error that 53% of consumers will vote “yes” on a new product line from Company A to Z?

Answer 1

Given,

n = 100

Point estimate = sample proportion = $\hat p = 0.53$

$1-\hat p=1-0.53=0.47$

At 95% confidence level

$\alpha =1-95\%\\\alpha=1-0.95=0.05\\\alpha/2=0.025\\Z_{\alpha/2}=Z_{0.025}=1.96$

Margin of error = $E=Z_{\alpha/2}\sqrt{\dfrac{\hat p\cdot (1-\hat p)}{n}}$

$E= 1.96\sqrt{\dfrac{0.53\times 0.47}{100}}\\\Rightarrow E=1.96\times0.04991\\\Rightarrow E=0.0978$

$\boxed{\text{Margin of error = 0.0978}}$