95% = mean ± 1.96 SEm
1.96 SEm = $500
SEm = SD/√n
graduates who majored in biology. How many such incomes must be found if the
economist wants to be 95% confident that the sample mean is within $500 of the true
population mean? Assume that research has found that the population standard
deviation for such incomes is equal to $6,250.
1.96 SEm = $500
SEm = SD/√n
n = (Z * σ / E)^2
Where:
n = sample size
Z = Z-score corresponding to the desired confidence level (in this case, 95% confidence level can be approximated to 1.96)
σ = population standard deviation
E = margin of error
In this case, the margin of error (E) is given as $500, and the population standard deviation (σ) is $6,250.
Plugging in the values into the formula:
n = (1.96 * 6,250 / 500)^2
Simplifying the calculation:
n = (12,250 / 500)^2
n = (24.5)^2
n ≈ 598
Therefore, the economist needs to find approximately 598 such incomes to be 95% confident that the sample mean is within $500 of the true population mean.