Question

question 7 of 8
what proportion of u.s. residents receive a jury summons each year? a polling organization plans to survey a random sample of 500 u.s. residents to find out. let $hat{p}$ be the proportion of residents in the sample who received a jury summons in the previous 12 months. according to the national center for state courts, 15% of u.s. residents receive a jury summons each year. suppose that this claim is true.
the standard deviation of this sampling distribution is $sigma_{hat{p}}=sqrt{\frac{p(1-p)}{n}}=sqrt{\frac{0.15(1-0.15)}{500}}=0.0160$.
what sample size would be required to reduce the standard deviation of the sampling distribution to one-half the original value?
sample size = $square$ residents (enter an integer.)

Explanation:

Step1: Define original standard deviation

Let $\sigma_{original} = \sqrt{\frac{p(1-p)}{n_{original}}}$, where $n_{original}=500$, $p=0.15$.

Step2: Set target standard deviation

We want $\sigma_{new} = \frac{1}{2}\sigma_{original}$. Substitute the formula:
$$\sqrt{\frac{p(1-p)}{n_{new}}} = \frac{1}{2}\sqrt{\frac{p(1-p)}{n_{original}}}$$

Step3: Square both sides to simplify

$$\frac{p(1-p)}{n_{new}} = \frac{1}{4} \cdot \frac{p(1-p)}{n_{original}}$$

Step4: Cancel shared terms and solve for $n_{new}$

Cancel $p(1-p)$ from both sides:
$$\frac{1}{n_{new}} = \frac{1}{4n_{original}}$$
Rearrange to solve for $n_{new}$:
$$n_{new} = 4n_{original}$$

Step5: Calculate new sample size

Substitute $n_{original}=500$:
$$n_{new}=4 \times 500 = 2000$$

Answer:

2000