To construct a 99.9% confidence interval for the proportion of people who performed volunteer work, we can use the formula for the confidence interval for a proportion:
\[
\hat{p} \pm z^* \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]
Where:
- \( \hat{p} \) is the sample proportion
- \( n \) is the sample size
- \( z^* \) is the z-value corresponding to the desired confidence level (for a 99.9% confidence level, \( z^* \approx 3.291 \))
From the previous calculation, we found:
- \( \hat{p} = 0.403 \)
- \( n = 1302 \)
Step 1: Calculate the standard error (SE)
\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.403(1 - 0.403)}{1302}} = \sqrt{\frac{0.403 \times 0.597}{1302}}
\]
Calculating this step-by-step:
-
Calculate \( \hat{p}(1 - \hat{p}) \):
\[
0.403 \times 0.597 \approx 0.240501
\]
-
Divide by \( n \):
\[
\frac{0.240501}{1302} \approx 0.000185
\]
-
Take the square root:
\[
SE \approx \sqrt{0.000185} \approx 0.0136
\]
Step 2: Calculate the margin of error (ME)
Now we compute the margin of error using the z-value for 99.9% confidence:
\[
ME = z^* \cdot SE \approx 3.291 \cdot 0.0136 \approx 0.0448
\]
Step 3: Calculate the confidence interval
Now we can construct the confidence interval:
\[
\hat{p} - ME \quad \text{to} \quad \hat{p} + ME
\]
So the interval is:
\[
0.403 - 0.0448 \quad \text{to} \quad 0.403 + 0.0448
\]
Calculating this gives:
\[
0.3582 \quad \text{to} \quad 0.4478
\]
Step 4: Round the results
Rounding each end of the interval to at least three decimal places, we obtain:
\[
0.358 \quad \text{to} \quad 0.448
\]
Final Result
Thus, a 99.9% confidence interval for the proportion of people who performed volunteer work during the past year is:
\[
\text{CI: } 0.358 < p < 0.448
\]
