To construct a 98% confidence interval for the population mean from a simple random sample, we typically use the formula for the confidence interval when the population is normally distributed and the sample size is small:

$\text{Confidence Interval} = \bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}$

Where:

• $\bar{x}$ is the sample mean.
• $t_{\alpha/2}$ is the critical value from the $t$-distribution for the desired confidence level.
• $s$ is the sample standard deviation.
• $n$ is the sample size.

Steps:

1. Identify the sample mean $\bar{x}$, sample size $n$, and sample standard deviation $s$. (You should provide these values for calculation.)
2. Find the critical value $t_{\alpha/2}$ for a 98% confidence level with $n - 1$ degrees of freedom.
   • For a 98% confidence level, $\alpha = 1 - 0.98 = 0.02$, so $\alpha/2 = 0.01$.
   • Using $n - 1 = 16$ degrees of freedom, we find $t_{\alpha/2}$ from the $t$-distribution table or a calculator.
3. Plug values into the formula to calculate the margin of error and the confidence interval.

Could you provide the sample mean and standard deviation for the calculation, or would you like me to demonstrate with assumed values?