Question

question 6 of 8 >
a usa today poll asked a random sample of 1012 u.s. adults what they do with the milk in their cereal bowl after they have eaten. let $hat{p}$ be the proportion of people in the sample who drink the cereal milk. a spokesman for the dairy industry claims that 70% of all u.s. adults drink the cereal milk. suppose this claim is true.
what sample size would be required to reduce the standard deviation of the sampling distribution to one-half the original value?
sample size = $square$ adults enter an integer.

Explanation:

Step1: Recall standard deviation formula

The standard deviation of the sampling distribution of a sample proportion $\hat{p}$ is given by:
$$\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$$
where $p$ is the population proportion, and $n$ is the sample size.

Step2: Define original and new values

Let the original standard deviation be $\sigma_1 = \sqrt{\frac{p(1-p)}{n_1}}$, where $n_1=1012$. We want a new standard deviation $\sigma_2 = \frac{1}{2}\sigma_1$, with sample size $n_2$.

Step3: Set up the equality

$$\sqrt{\frac{p(1-p)}{n_2}} = \frac{1}{2}\sqrt{\frac{p(1-p)}{n_1}}$$

Step4: Solve for $n_2$

Square both sides to eliminate square roots:
$$\frac{p(1-p)}{n_2} = \frac{1}{4} \cdot \frac{p(1-p)}{n_1}$$
Cancel $p(1-p)$ from both sides (since $p
eq 0,1$):
$$\frac{1}{n_2} = \frac{1}{4n_1}$$
Rearrange to solve for $n_2$:
$$n_2 = 4n_1$$

Step5: Calculate the new sample size

Substitute $n_1=1012$:
$$n_2 = 4 \times 1012 = 4048$$

Answer:

4048