Assume an auditor has 50,000 accounts and wants to know the average account payable’s value. They decide to pull a sample and want the results to be within $46 of the correct figure. The standard deviation they are expecting from the sample is 250 dollars. How many accounts do they need to check? Round up to whole number of cases.

To determine the sample size needed for the auditor's requirements, we can use the formula for calculating the required sample size for a given margin of error and standard deviation. The formula is: $$\[ n = \left( \frac{{Z \times \sigma}}{{E}} \right)^2 \]$$ Where: $$- \( n \) is the required sample size.$$ $$- \( Z \) is the Z-score corresponding to the desired level of confidence.$$ For a 95% confidence level, \( Z = 1.96 \). $$- \( \sigma \) is the standard deviation.$$ $$- \( E \) is the desired margin of error.$$ Given the values: $$- \( Z = 1.96 \)$$ $$- \( \sigma = 250 \) dollars$$ $$- \( E = 46 \) dollars$$ Plug in these values into the formula: $$\[ n = \left( \frac{{1.96 \times 250}}{{46}} \right)^2 \]$$ Calculating this: $$\[ n = (10.652)^2 \approx 113.47 \]$$ Since the sample size needs to be a whole number, we'll round up to the nearest whole number, resulting in a required sample size of 114 accounts. Therefore, the auditor should check at least 114 accounts from the total of 50,000 accounts to achieve a margin of error within $46 of the correct average account payable's value with a standard deviation of $250 and a 95% confidence level.