1. **Stating the problem:** We are given the weights of 12 newborn babies and asked to find the confidence interval for the mean weight with a significance level $\alpha = 2\%$. The data are: 2491, 2603, 3249, 3012, 2472, 3211, 3000, 2533, 2602, 2459, 3100, 3260. 2. **Formula and rules:** The confidence interval for the mean when the population standard deviation is unknown and sample size is small is given by: $$\bar{x} \pm t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}}$$ where: - $\bar{x}$ is the sample mean - $s$ is the sample standard deviation - $n$ is the sample size - $t_{\alpha/2, n-1}$ is the t-score from the t-distribution with $n-1$ degrees of freedom 3. **Calculate the sample mean $\bar{x}$:** $$\bar{x} = \frac{2491 + 2603 + 3249 + 3012 + 2472 + 3211 + 3000 + 2533 + 2602 + 2459 + 3100 + 3260}{12}$$ $$= \frac{33092}{12} = 2757.67$$ 4. **Calculate the sample standard deviation $s$:** First, calculate each squared deviation $(x_i - \bar{x})^2$, sum them, then divide by $n-1=11$, and take the square root. Sum of squared deviations: $$\sum (x_i - \bar{x})^2 = (2491-2757.67)^2 + (2603-2757.67)^2 + \cdots + (3260-2757.67)^2$$ Calculating each: $$(-266.67)^2 = 71111.11$$ $(-154.67)^2 = 23922.11$ $(491.33)^2 = 241405.37$ $(254.33)^2 = 64691.37$ $(-285.67)^2 = 81600.11$ $(453.33)^2 = 205509.11$ $(242.33)^2 = 58728.11$ $(-224.67)^2 = 50472.11$ $(-155.67)^2 = 24233.11$ $(-298.67)^2 = 89205.11$ $(342.33)^2 = 117191.11$ $(502.33)^2 = 252337.11$ Sum: $$71111.11 + 23922.11 + 241405.37 + 64691.37 + 81600.11 + 205509.11 + 58728.11 + 50472.11 + 24233.11 + 89205.11 + 117191.11 + 252337.11 = 1,254,406.74$$ Sample variance: $$s^2 = \frac{1,254,406.74}{11} = 114,036.07$$ Sample standard deviation: $$s = \sqrt{114,036.07} = 337.66$$ 5. **Find the t-score for $\alpha/2 = 0.01$ and $df=11$:** From t-distribution tables or calculator, $t_{0.01,11} \approx 2.718$. 6. **Calculate the margin of error:** $$E = t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}} = 2.718 \times \frac{337.66}{\sqrt{12}}$$ $$= 2.718 \times 97.46 = 264.82$$ 7. **Construct the confidence interval:** $$2757.67 \pm 264.82$$ $$= (2492.85, 3022.49)$$ **Final answer:** The 98% confidence interval for the mean weight of newborn babies is approximately $$\boxed{(2492.85, 3022.49)}$$.