a private opinion poll is conducted for a politician to determine what proportion of the population favors…

a private opinion poll is conducted for a politician to determine what proportion of the population favors decriminalizing marijuana possession. how large a sample is needed in order to be 95% confident that the sample proportion will not differ from the true proportion by more than 5%?\n\na. 271\nb. 385\nc. 10\nd. 769

a private opinion poll is conducted for a politician to determine what proportion of the population favors decriminalizing marijuana possession. how large a sample is needed in order to be 95% confident that the sample proportion will not differ from the true proportion by more than 5%?\n\na. 271\nb. 385\nc. 10\nd. 769

Answer

Explanation:

Step1: Identify the formula

For sample - size calculation in estimating a proportion, the formula is $n=\frac{z^{2}\cdot p(1 - p)}{E^{2}}$. For a 95% confidence interval, the critical value $z$ is approximately 1.96. Since we have no prior estimate of the proportion $p$, we use $p = 0.5$ (this gives the maximum value of $p(1 - p)$). The margin of error $E=0.05$.

Step2: Calculate $p(1 - p)$

When $p = 0.5$, $p(1 - p)=0.5\times(1 - 0.5)=0.5\times0.5 = 0.25$.

Step3: Calculate $n$

Substitute $z = 1.96$, $p(1 - p)=0.25$ and $E = 0.05$ into the formula $n=\frac{z^{2}\cdot p(1 - p)}{E^{2}}$. $n=\frac{(1.96)^{2}\times0.25}{(0.05)^{2}}=\frac{3.8416\times0.25}{0.0025}=\frac{0.9604}{0.0025}=384.16\approx385$.

Answer:

B. 385