A provincial politician would like to determine the average amount earned during summer employment by the provinces teenagers during the past summer’s vacation period. She wants to have 95% confidence that the sample mean is within $25 of the actual population mean. Bases on past studies, she knows the amount earned is normally distributed with a standard deviation of $400. What sample size must she take?
95% = mean ± 1.96 SEm
SEm = SD/√n
25 = 1.96 (400/√n)
Calculate n.
To determine the sample size required to estimate the average amount earned during summer employment by teenagers with a certain level of confidence and a given margin of error, you can use the formula:
n = (Z * σ / E)²
Where:
n = sample size
Z = Z-score corresponding to the desired confidence level (in this case, 95%)
σ = standard deviation of the population
E = margin of error (in this case, $25)
Given that the standard deviation of the population (σ) is $400, and the desired margin of error (E) is $25, we can substitute these values into the formula and solve for the sample size.
n = (Z * σ / E)²
n = (1.96 * 400 / 25)²
Calculating this expression, we find that:
n ≈ 245.76
Since sample size must be a whole number, we can round up to the nearest whole number to ensure an adequate sample size. Therefore, the provincial politician must take a sample size of 246 to estimate the average amount earned during summer employment by teenagers with 95% confidence and a margin of error of $25.