A researcher is interested in estimating the average salary of police officers in a large city. She wants to be 95% confident that her estimate is correct. If the standard deviation is $1050,how
large a sample is needed to get the desired information and to be accurate within $200.
95% = mean ± 1.96 SEm
1.96 SEm = 200
SEm = SD/√n
To determine the sample size needed to estimate the average salary of police officers with the desired confidence level and margin of error, we 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, for a 95% confidence level, Z is approximately 1.96)
σ = standard deviation of the population
E = desired margin of error
In this case, the standard deviation (σ) is given as $1050, and the desired margin of error (E) is $200.
Plugging in the values into the formula:
n = (1.96 * $1050 / $200)^2
Calculating this expression will give us the required sample size.