Question

Assuming that sample proportion is 0.35, what sample size is needed to have a 5% margin of error and 99% confidence?

Answer 1

To calculate the required sample size for a given margin of error and confidence level, we can use the formula for the sample size \( n \) when estimating a proportion:

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

where:
- \( Z \) = Z-score corresponding to the desired confidence level
- \( p \) = sample proportion
- \( E \) = margin of error

### Step 1: Determine the Z-score

For a 99% confidence level, the Z-score can be found using a Z-table or calculator. The Z-score for a 99% confidence level is approximately:

\[
Z \approx 2.576
\]

### Step 2: Plug in the values

Given:
- \( p = 0.35 \)
- \( E = 0.05 \)

We can now substitute these values into the formula:

\[
n = \left( \frac{(2.576)^2 \cdot 0.35 \cdot (1 - 0.35)}{(0.05)^2} \right)
\]

Calculating \( 1 - p \):

\[
1 - p = 1 - 0.35 = 0.65
\]

Calculating \( n \):

\[
n = \left(\frac{(2.576)^2 \cdot 0.35 \cdot 0.65}{(0.05)^2}\right)
\]

Now calculating each part:

1. \( (2.576)^2 \approx 6.635776 \)
2. \( 0.35 \cdot 0.65 = 0.2275 \)
3. \( (0.05)^2 = 0.0025 \)

Substituting these values back into the formula:

\[
n = \left( \frac{6.635776 \cdot 0.2275}{0.0025} \right)
\]

Calculating the numerator:

\[
6.635776 \cdot 0.2275 \approx 1.507299
\]

Now divide by the denominator:

\[
n = \frac{1.507299}{0.0025} \approx 602.9196
\]

### Step 3: Round up

Since sample size must be a whole number, we round up to the next whole number:

\[
n \approx 603
\]

### Conclusion

The required sample size needed to achieve a 5% margin of error and 99% confidence level is **603**.