Question
A medical researcher wants to construct a 99% confidence interval for the proportion of knee replacement surgeries that result in complications.
(a) An article in a medical journal suggested that approximately 16% of such operations result in complications. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of .04?
(a) An article in a medical journal suggested that approximately 16% of such operations result in complications. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of .04?
Answers
GPT-4o mini
To determine the sample size needed for a certain margin of error in a confidence interval for a population proportion, we can use the formula for the sample size \( n \) required for estimating proportions:
\[
n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right)
\]
where:
- \( Z \) is the Z-value corresponding to the desired confidence level,
- \( p \) is the estimated proportion,
- \( E \) is the desired margin of error.
### Given:
- Confidence level = 99%
- Corresponding Z-value \( Z \) for 99% is approximately \( 2.576 \) (you can find this value from Z-tables).
- Estimated proportion \( p = 0.16 \)
- Desired margin of error \( E = 0.04 \)
### Calculating Sample Size:
1. Substitute the values into the formula:
\[
n = \left( \frac{(2.576)^2 \cdot 0.16 \cdot (1 - 0.16)}{(0.04)^2} \right)
\]
2. Calculate \( (1 - p) \):
\[
1 - p = 1 - 0.16 = 0.84
\]
3. Now calculate the components:
\[
n = \left( \frac{(2.576)^2 \cdot 0.16 \cdot 0.84}{(0.04)^2} \right)
\]
4. Calculate \( (2.576)^2 \):
\[
(2.576)^2 \approx 6.636
\]
5. Now plug in the values:
\[
n = \left( \frac{6.636 \cdot 0.16 \cdot 0.84}{(0.04)^2} \right)
\]
6. Calculate the product \( 6.636 \cdot 0.16 \cdot 0.84 \):
\[
6.636 \cdot 0.16 \approx 1.06176
\]
\[
1.06176 \cdot 0.84 \approx 0.8911
\]
7. Then divide by \( (0.04)^2 = 0.0016 \):
\[
n = \frac{0.8911}{0.0016} \approx 556.9375
\]
8. Since we cannot have a fraction of a sample, we round up to the next whole number:
\[
n \approx 557
\]
### Conclusion:
The required sample size to achieve a 99% confidence interval with a margin of error of 0.04 for the proportion of knee replacement surgeries that result in complications is **557**.
\[
n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right)
\]
where:
- \( Z \) is the Z-value corresponding to the desired confidence level,
- \( p \) is the estimated proportion,
- \( E \) is the desired margin of error.
### Given:
- Confidence level = 99%
- Corresponding Z-value \( Z \) for 99% is approximately \( 2.576 \) (you can find this value from Z-tables).
- Estimated proportion \( p = 0.16 \)
- Desired margin of error \( E = 0.04 \)
### Calculating Sample Size:
1. Substitute the values into the formula:
\[
n = \left( \frac{(2.576)^2 \cdot 0.16 \cdot (1 - 0.16)}{(0.04)^2} \right)
\]
2. Calculate \( (1 - p) \):
\[
1 - p = 1 - 0.16 = 0.84
\]
3. Now calculate the components:
\[
n = \left( \frac{(2.576)^2 \cdot 0.16 \cdot 0.84}{(0.04)^2} \right)
\]
4. Calculate \( (2.576)^2 \):
\[
(2.576)^2 \approx 6.636
\]
5. Now plug in the values:
\[
n = \left( \frac{6.636 \cdot 0.16 \cdot 0.84}{(0.04)^2} \right)
\]
6. Calculate the product \( 6.636 \cdot 0.16 \cdot 0.84 \):
\[
6.636 \cdot 0.16 \approx 1.06176
\]
\[
1.06176 \cdot 0.84 \approx 0.8911
\]
7. Then divide by \( (0.04)^2 = 0.0016 \):
\[
n = \frac{0.8911}{0.0016} \approx 556.9375
\]
8. Since we cannot have a fraction of a sample, we round up to the next whole number:
\[
n \approx 557
\]
### Conclusion:
The required sample size to achieve a 99% confidence interval with a margin of error of 0.04 for the proportion of knee replacement surgeries that result in complications is **557**.