To construct a 99% confidence interval for the difference in mean BMI between women and men with diabetes, we can use the formula for the confidence interval for the difference between two means. The formula is given by:
\[ \text{CI} = (\bar{x}_1 - \bar{x}_2) \pm z^* \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]
Where:
- \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means,
- \(s_1\) and \(s_2\) are the sample standard deviations,
- \(n_1\) and \(n_2\) are the sample sizes,
- \(z^*\) is the z-score for the desired confidence level.
Given values:
-
For women:
- \(\bar{x}_1 = 31.30\)
- \(s_1 = 0.3\)
- \(n_1 = 1928\)
-
For men:
- \(\bar{x}_2 = 30.50\)
- \(s_2 = 0.6\)
- \(n_2 = 1559\)
Step 1: Calculate the difference in means
\[ \bar{x}_1 - \bar{x}_2 = 31.30 - 30.50 = 0.80 \]
Step 2: Calculate the standard error (SE) of the difference
\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{(0.3)^2}{1928} + \frac{(0.6)^2}{1559}} \]
Calculating each part:
\[ \frac{(0.3)^2}{1928} = \frac{0.09}{1928} \approx 0.0000466 \] \[ \frac{(0.6)^2}{1559} = \frac{0.36}{1559} \approx 0.000230 \]
Adding them together:
\[ SE = \sqrt{0.0000466 + 0.000230} = \sqrt{0.0002766} \approx 0.0166 \]
Step 3: Find the z-score for a 99% confidence interval
For a 99% confidence level, we find the critical z-value. From standard normal distribution tables, the z-value for a two-tailed test at 99% confidence is approximately:
\[ z^* \approx 2.576 \]
Step 4: Construct the confidence interval
\[ \text{CI} = (0.80) \pm 2.576 \cdot (0.0166) \]
Calculating the margin of error:
\[ 2.576 \cdot 0.0166 \approx 0.0427 \]
Now calculate the confidence interval:
\[ \text{CI} = (0.80 - 0.0427, 0.80 + 0.0427) = (0.7573, 0.8427) \]
Conclusion
The 99% confidence interval for the difference in mean BMI between women and men with diabetes is approximately:
\[ (0.7573, 0.8427) \]
This indicates that, with 99% confidence, the mean BMI of women with diabetes is estimated to be between 0.7573 and 0.8427 units higher than that of men with diabetes.
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