To construct a 99.9% confidence interval for the difference between the mean weights of one-year-old boys and girls, we can use the following formula for the confidence interval for the difference between two means:
\[ (\bar{x}_1 - \bar{x}_2) \pm z^* \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]
where:
- \(\bar{x}_1\) = mean weight of boys = 25.3 pounds
- \(\bar{x}_2\) = mean weight of girls = 24.0 pounds
- \(s_1\) = standard deviation of boys = 3.5 pounds
- \(s_2\) = standard deviation of girls = 3.7 pounds
- \(n_1\) = number of boys = 312
- \(n_2\) = number of girls = 290
- \(z^\) = z-score for the desired confidence level (for a 99.9% confidence level, \(z^\) is approximately 3.291)
Step 1: Calculate the Difference in Sample Means
\[ \bar{x}_1 - \bar{x}_2 = 25.3 - 24.0 = 1.3 \text{ pounds} \]
Step 2: Calculate the Standard Error
The standard error (SE) of the difference can be calculated as follows:
\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]
\[ SE = \sqrt{\frac{(3.5)^2}{312} + \frac{(3.7)^2}{290}} \]
Calculating each term:
\[ \frac{(3.5)^2}{312} = \frac{12.25}{312} \approx 0.0392 \]
\[ \frac{(3.7)^2}{290} = \frac{13.69}{290} \approx 0.0472 \]
Now, combining them:
\[ SE = \sqrt{0.0392 + 0.0472} = \sqrt{0.0864} \approx 0.2939 \]
Step 3: Calculate the Margin of Error
Now, we need to compute the margin of error:
\[ ME = z^* \cdot SE \]
\[ ME = 3.291 \cdot 0.2939 \approx 0.9671 \]
Step 4: Construct the Confidence Interval
Now we can construct the confidence interval:
\[ (1.3 - 0.9671, 1.3 + 0.9671) = (0.3329, 2.2671) \]
Final Result
Thus, the 99.9% confidence interval for the difference between the mean weights of one-year-old boys and girls is approximately:
\[ (0.33, 2.27) \text{ pounds} \]
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