An economist wants to estimate the mean income for the first year of work for college

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.

95% = mean ± 1.96 SEm

1.96 SEm = $500

SEm = SD/√n

To determine the sample size needed to estimate the mean income for college graduates who majored in biology, we can use the formula for sample size:

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.