Question

question 13
a researcher would like to estimate p, the proportion of u.s. adults who support raising the federal minimum wage.
if the researcher would like to be 95% sure that the obtained sample proportion would be within 2.4% of p (the proportion in the entire population of u.s. adults), what sample size should be used?
a. 6,945
b. 1,737
c. 435
d. 42

Explanation:

Step1: Identify the z - value for 95% confidence level

For a 95% confidence level, the critical value $z_{\alpha/2}= 1.96$. The margin of error $E = 0.024$. When the proportion $p$ is unknown, we use $p = 0.5$ (this gives the maximum variance and thus maximum sample - size).

Step2: Use the sample - size formula for proportion

The formula for sample size $n$ in estimating a proportion is $n=\frac{z_{\alpha/2}^{2}\cdot p(1 - p)}{E^{2}}$. Substitute $z_{\alpha/2}=1.96$, $p = 0.5$, and $E=0.024$ into the formula.
\[

$$\begin{align*} n&=\frac{(1.96)^{2}\times0.5\times(1 - 0.5)}{(0.024)^{2}}\\ &=\frac{(1.96)^{2}\times0.25}{(0.024)^{2}}\\ &=\frac{3.8416\times0.25}{0.000576}\\ &=\frac{0.9604}{0.000576}\\ &\approx1667 \end{align*}$$

\]
Since we made an approximation in the calculation process and considering the closest value among the options, when we calculate more precisely with the formula $n=\frac{z^{2}\cdot p(1 - p)}{E^{2}}$, and $z = 1.96$, $E=0.024$, $p = 0.5$, we have $n=\frac{(1.96)^{2}\times0.5\times0.5}{(0.024)^{2}}=\frac{3.8416\times0.25}{0.000576}\approx1667$. The closest value to this in the options is 1737.

Answer:

B. 1,737