Question

he National Health Statistics Reports reported that a sample of 312 one-year-old boys had a mean weight of 25.3 pounds with a standard deviation of 3.5 pounds. In addition, a sample of 290 one-year-old girls had a mean weight of 24.0 pounds with a standard deviation of 3.7 pounds.

Construct a 99.9% confidence interval for the difference between the mean weights. Let uv1 denote the mean weight of one-year-old boys.

Answer 1

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:

\[
(0.33, 2.27) \text{ pounds}
\]