To solve this problem, let’s go through the information provided and find the margin of error for the estimate of the mean $\mu$ with a 98% confidence level.

Given Data:

• $X \sim N(\mu, 0.25)$ (normally distributed with mean $\mu$ and variance $0.25$)
• Sample size, $n = 12$
• Sample mean, $\bar{X} = 15.5$
• Sample standard deviation, $S_X = 0.62$
• Confidence level: 98%
• Given t-values and z-values for specific degrees of freedom and significance levels:
  • $t_{0.02}(10) = 2.764$
  • $t_{0.02}(11) = 2.718$
  • $z_{0.01} = 2.33$
  • $z_{0.02} = 2.055$

Solution:

Since we have a small sample size $n = 12$, we should use the t-distribution for calculating the confidence interval.

1. Determine the correct t-value:
   For a 98% confidence level, the significance level $\alpha$ is 0.02. Since $n = 12$, the degrees of freedom $df = n - 1 = 11$. Therefore, we will use $t_{0.02}(11) = 2.718$.