Question

Sample size is 45
O=40
Find the margin of error for a 95% confidence level

Answer 1

To calculate the margin of error (ME) for a population proportion, we'll need to use the formula:

\[
ME = z^* \cdot \sqrt{\frac{p(1-p)}{n}}
\]

Where:
- \( z^* \) is the z-score corresponding to the desired confidence level (for a 95% confidence level, \( z^* \approx 1.96 \)).
- \( p \) is the sample proportion, which is calculated as \( p = \frac{O}{n} \) (where \( O \) is the number of successes and \( n \) is the sample size).
- \( n \) is the sample size.

Given:
- \( O = 40 \)
- \( n = 45 \)

First, we calculate the sample proportion \( p \):

\[
p = \frac{O}{n} = \frac{40}{45} \approx 0.8889
\]

Now, we can calculate \( p(1-p) \):

\[
p(1-p) = 0.8889 \times (1 - 0.8889) = 0.8889 \times 0.1111 \approx 0.0988
\]

Next, we compute the standard error (SE):

\[
SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.0988}{45}} \approx \sqrt{0.002196} \approx 0.0468
\]

Now, we calculate the margin of error (ME):

\[
ME = z^* \cdot SE = 1.96 \cdot 0.0468 \approx 0.0918
\]

To conclude, the margin of error for a 95% confidence level is approximately **0.0918**.