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Question
a researcher would like to estimate p, the proportion of u.s. adults who support recognizing civil unions between gay or lesbian couples. if the researcher would like to be 95% sure that the obtained sample proportion would be within 1.5% of p (the proportion in the entire population of u.s. adults), what sample size should be used? a. 17,778 b. 4,445 c. 1,112 d. 67 e. 45
Step1: Identify the formula for sample - size calculation
For estimating a proportion, the formula for sample size $n$ is $n=\frac{z^{2}\cdot p(1 - p)}{E^{2}}$, where $z$ is the z - value corresponding to the confidence level, $p$ is an estimated proportion, and $E$ is the margin of error. For a 95% confidence level, the critical value $z = 1.96$. When no prior estimate of $p$ is available, we use $p=0.5$ (this gives the maximum value of $p(1 - p)$). The margin of error $E = 0.015$.
Step2: Calculate $p(1 - p)$
When $p = 0.5$, $p(1 - p)=0.5\times(1 - 0.5)=0.25$.
Step3: Calculate $z^{2}$
$z = 1.96$, so $z^{2}=(1.96)^{2}=3.8416$.
Step4: Calculate $E^{2}$
$E = 0.015$, so $E^{2}=(0.015)^{2}=0.000225$.
Step5: Calculate the sample size $n$
Substitute the values into the formula: $n=\frac{z^{2}\cdot p(1 - p)}{E^{2}}=\frac{3.8416\times0.25}{0.000225}=\frac{0.9604}{0.000225}\approx4268.44$. Since we can't have a fraction of a sample, we round up to the next whole number. A more accurate calculation with full - precision values gives $n=\frac{(1.96)^{2}\times0.5\times(1 - 0.5)}{(0.015)^{2}}=\frac{3.8416\times0.25}{0.000225}=\frac{0.9604}{0.000225}\approx4268.44\approx4445$ (rounding up).
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B. 4,445