Answer:
To calculate the minimum sample size needed to estimate the population mean with a margin of error of no more than 2 dollars and 99% confidence when you don't know the standard deviation, you can use the worst-case scenario by assuming the standard deviation is at its maximum, which is $50 (the highest possible spending).
The formula for sample size in this case is:
\[n = \frac{Z^2 * σ^2}{E^2}\]
Where:
- \(n\) is the sample size.
- \(Z\) is the Z-score corresponding to the desired confidence level. For 99% confidence, the Z-score is approximately 2.576.
- \(σ\) is the assumed maximum standard deviation, which is $50 in this case.
- \(E\) is the maximum acceptable margin of error, which is $2.
Plugging in these values:
\[n = \frac{2.576^2 * 50^2}{2^2}\]
Calculate:
\[n ≈ \frac{6.647776 * 2500}{4}\]
\[n ≈ 13,219.44\]
Since you can't have a fraction of a person in your sample, you would need a minimum sample size of 13,220 to estimate the population mean with a margin of error of no more than $2 with 99% confidence, assuming the worst-case scenario for the standard deviation.