Final answer:
In interval estimation for a population proportion, the value of z at a 98.2% confidence level is not standard and would be less than 2.326 but more than 1.96. The exact value would need to be calculated using statistical software or a normal distribution calculator.
Step-by-step explanation:
In interval estimation for a proportion of a population, the value of z at 98.2% confidence level refers to the number of standard deviations that a proportion value should be from the mean so that there is a 98.2% chance that the true population proportion lies within that range. This value is not explicitly provided in common z-tables, as they typically show standard values such as 90%, 95%, or 99% confidence levels. However, you can calculate this value using a calculator or statistics software that allows for the computation of z-scores corresponding to non-standard confidence levels. For a 98.2% confidence level, you would use the provided information that for a 99% confidence level, the z-score is approximately 2.326. Since a 98.2% confidence level is less than 99%, the corresponding z-score will be less than 2.326 but more than the z-score for a 95% confidence level, which is 1.96. To find the exact value, using the normal distribution probability tools available would be necessary.