Chapter 8.6, Problem 32SE

Find the margin of error for a poll, assuming that π = .50.

1. a. n = 50
2. b. n = 200
3. c. n = 500
4. d. n = 2,000

To determine

Find the margin of error for a poll assuming $\pi =0.50$.

Answer to Problem 32SE

The margin of error for a poll assuming $\pi =0.50$ is 0.1386.

Explanation of Solution

Calculation:

The given information is that $n=50\text{ and }\pi =0.50$.

Margin of error for a poll:

The formula for the Margin of error for a poll typically based on a 95% confidence level is, $z\sqrt{\frac{\pi \left(1-\pi \right)}{n}}$ .

From TABLE 8.9: Common Confidence Levels and z-Values, the z-value at 95% confidence level is 1.96.

Substitute $z=1.96,\pi =0.50\text{ and }n=50$ in the formula for Margin of error.

$\begin{array}{c}z\sqrt{\frac{\pi \left(1-\pi \right)}{n}}=1.96\sqrt{\frac{0.50\left(1-0.50\right)}{50}}\\ =1.96\sqrt{\frac{0.25}{50}}\\ =1.96\sqrt{0.005}\\ =1.96\left(0.0707\right)\\ \approx 0.1386\end{array}$

Thus, the margin of error for a poll assuming $\pi =0.50$ is 0.1386.

To determine

Find the margin of error for a poll assuming $\pi =0.50$.

Answer to Problem 32SE

The margin of error for a poll assuming $\pi =0.50$ is 0.0694.

Explanation of Solution

Calculation:

The given information is that $n=200\text{ and }\pi =0.50$.

Substitute $z=1.96,\pi =0.50\text{ and }n=200$ in the formula for Margin of error.

$\begin{array}{c}z\sqrt{\frac{\pi \left(1-\pi \right)}{n}}=1.96\sqrt{\frac{0.50\left(1-0.50\right)}{200}}\\ =1.96\sqrt{\frac{0.25}{200}}\\ =1.96\sqrt{0.00125}\\ =1.96\left(0.0354\right)\\ \approx 0.0694\end{array}$

Thus, the margin of error for a poll assuming $\pi =0.50$ is 0.0694.

To determine

Find the margin of error for a poll assuming $\pi =0.50$.

Answer to Problem 32SE

The margin of error for a poll assuming $\pi =0.50$ is 0.0439.

Explanation of Solution

Calculation:

The given information is that $n=500\text{ and }\pi =0.50$.

Substitute $z=1.96,\pi =0.50\text{ and }n=500$ in the formula for Margin of error.

$\begin{array}{c}{z}_{0.025}\sqrt{\frac{\pi \left(1-\pi \right)}{n}}=1.96\sqrt{\frac{0.50\left(1-0.50\right)}{500}}\\ =1.96\sqrt{\frac{0.25}{500}}\\ =1.96\sqrt{0.0005}\\ =1.96\left(0.02236\right)\\ \approx 0.0439\end{array}$

Thus, the margin of error for a poll assuming $\pi =0.50$ is 0.0439.

To determine

Find the margin of error for a poll assuming $\pi =0.50$.

Answer to Problem 32SE

The margin of error for a poll assuming $\pi =0.50$ is 0.022.

Explanation of Solution

Calculation:

The given information is that $n=2,000\text{ and }\pi =0.50$.

Substitute $z=1.96,\pi =0.50\text{ and }n=2,000$ in the formula for Margin of error.

$\begin{array}{c}z\sqrt{\frac{\pi \left(1-\pi \right)}{n}}=1.96\sqrt{\frac{0.50\left(1-0.50\right)}{2,000}}\\ =1.96\sqrt{\frac{0.25}{2,000}}\\ =1.96\sqrt{0.000125}\\ =1.96\left(0.01118\right)\\ \approx 0.022\end{array}$

Thus, the margin of error for a poll assuming $\pi =0.50$ is 0.022.