Use the given data to find the minimum sample size required to estimate the population proportion.
-Margin of error: 0.008; confidence level $95 \% ; \hat { \mathrm { p } } \text { and } \hat { \mathrm { q } } \text { unknown }$
A) 15,007
B) 15,098
C) 14,488
D) 5045

Question

Quizplus · Accepted Answer

The formula for the minimum sample size required to estimate the population proportion is:n = (Z^2 * p * q) / e^2where:n is the minimum sample sizeZ is the z-score for the desired confidence levelp is the estimated population proportionq is the estimated population proportione is the margin of errorIn this case, we are given the following information:Margin of error: 0.008Confidence level: 95%p and q are unknownThe z-score for a 95% confidence level is 1.96.We can estimate p and q as 0.5, since we have no prior information to suggest otherwise.Plugging these values into the formula, we get:n = (1.96^2 * 0.5 * 0.5) / 0.008^2n = 15,007Therefore, the minimum sample size required to estimate the population proportion with a margin of error of 0.008 and a confidence level of 95% is 15,007.

Use the Given Data to Find the Minimum Sample Size 95%;p^ and q^ unknown 95 \% ; \hat { \mathrm { p } } \text { and } \hat { \mathrm { q } } \text { unknown }95%;p^​ and q^​ unknown

Use the Given Data to Find the Minimum Sample Size $95 \% ; \hat { \mathrm { p } } \text { and } \hat { \mathrm { q } } \text { unknown }$