The general manager for Gordon's Self Service Gasoline is interested in estimating the mean number of gallons of gasoline that are purchased by customers at their Philadelphia location. He would like his estimate to be within plus or minus 0.50 gallons, and he would like the estimate to be at a 99% confidence level. Past studies have shown that the standard deviation for purchase amount is 4.0 gallons. Find the required sample size.
Formula:
n = [(z-value * sd)/E]^2
...where n = sample size, z-value will be 2.58 using a z-table to represent the 99% confidence interval, sd = 4, E = 0.5, ^2 means squared, and * means to multiply.
Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.
To find the required sample size, we need to use the formula for sample size calculation for estimating the population mean using a known standard deviation:
n = (Z * σ / E)²
Where:
n = required sample size
Z = z-score corresponding to the desired confidence level (in this case, 99%)
σ = standard deviation of the population
E = desired margin of error
Let's calculate the required sample size step by step:
1. Find the z-score corresponding to a 99% confidence level. This is the z-score that leaves 1% in the tails of the standard normal distribution. Using a z-score table or a calculator, we find that the z-score for a 99% confidence level is approximately 2.576.
2. Plug in the known values into the formula:
n = (2.576 * 4.0 / 0.5)²
3. Calculate the result:
n = (10.304 / 0.5)²
n = 409.6
4. Round up to the nearest whole number since you cannot have a fraction of a person:
n = 410
Therefore, the required sample size to estimate the mean number of gallons of gasoline with a margin of error of ±0.50 gallons at a 99% confidence level is 410.