Question

A pollster wants to construct an 80% confidence interval for the proportion of adults who believe that economic conditions are getting better. A poll taken in July 2010 estimates this proportion to be 0.41. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.04?

Answer 1

(b) Estimate the sample size needed if no estimate of p is available.

Answer 2

To determine the required sample size for an 80% confidence interval with a specified margin of error, we can use the formula for the sample size \( n \) needed to achieve a margin of error \( E \):

\[
n = \left( \frac{Z \cdot \sqrt{p(1-p)}}{E} \right)^2
\]

Where:
- \( Z \) is the z-score corresponding to the desired confidence level,
- \( p \) is the estimated proportion,
- \( E \) is the margin of error.

### Step 1: Identify the parameters
- The estimated proportion \( p = 0.41 \).
- The margin of error \( E = 0.04 \).

### Step 2: Find the z-score for an 80% confidence level
For an 80% confidence level, the z-score can be found using standard z-tables or a calculator. The critical value \( Z \) for 80% confidence is approximately:

\[
Z \approx 1.28
\]

### Step 3: Plug the values into the sample size formula
Now we substitute the values we have into the formula:

\[
n = \left( \frac{1.28 \cdot \sqrt{0.41(1 - 0.41)}}{0.04} \right)^2
\]

Calculate \( p(1-p) \):

\[
p(1 - p) = 0.41 \cdot (1 - 0.41) = 0.41 \cdot 0.59 = 0.2419
\]

Now compute \( \sqrt{0.2419} \):

\[
\sqrt{0.2419} \approx 0.4918
\]

Next, calculate the numerator:

\[
1.28 \cdot 0.4918 \approx 0.6291
\]

Now divide it by the margin of error \( E = 0.04 \):

\[
\frac{0.6291}{0.04} \approx 15.7275
\]

Finally, square this result to find \( n \):

\[
n = (15.7275)^2 \approx 247.24
\]

### Step 4: Round up to the nearest whole number
Since sample sizes must be whole numbers, we round up to the next whole number:

\[
n \approx 248
\]

### Conclusion
The required sample size needed to obtain an 80% confidence interval with a margin of error of \( 0.04 \) is **248**.

Answer 3

When no prior estimate of the proportion \( p \) is available, a conservative approach is to use \( p = 0.5 \). This is because the product \( p(1 - p) \) reaches its maximum when \( p = 0.5 \), resulting in the largest sample size needed for a given margin of error.

### Step 1: Use the conservative estimate for \( p \)

Using \( p = 0.5 \), we can proceed with the same formula for sample size:

\[
n = \left( \frac{Z \cdot \sqrt{p(1-p)}}{E} \right)^2
\]

Where:
- \( E = 0.04 \) (the margin of error),
- \( Z \approx 1.28 \) (the z-score for an 80% confidence interval),
- \( p = 0.5 \).

### Step 2: Calculate \( p(1 - p) \)

\[
p(1 - p) = 0.5 \cdot (1 - 0.5) = 0.5 \cdot 0.5 = 0.25
\]

### Step 3: Plug the values into the formula

Now we can calculate \( n \):

\[
n = \left( \frac{1.28 \cdot \sqrt{0.25}}{0.04} \right)^2
\]

Calculating \( \sqrt{0.25} \):

\[
\sqrt{0.25} = 0.5
\]

Now substitute this back into the equation:

\[
n = \left( \frac{1.28 \cdot 0.5}{0.04} \right)^2
\]

Calculating the numerator:

\[
1.28 \cdot 0.5 = 0.64
\]

Now divide by the margin of error:

\[
\frac{0.64}{0.04} = 16
\]

Finally, square this result:

\[
n = 16^2 = 256
\]

### Conclusion

The required sample size needed when no estimate of \( p \) is available, using \( p = 0.5 \) for a conservative estimate, is **256**.