A researcher is interested in estimating the average salary of Police officers in a large city. she want 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.
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To determine the sample size needed to estimate the average salary of police officers with a desired level of confidence and a specified margin of error, we can use the formula for calculating the sample size for estimating a population mean:
n = (Z * σ / E)²
Where:
n = sample size
Z = Z-score representing the desired level of confidence (in this case, we want a 95% confidence level, which corresponds to a Z-score of 1.96)
σ = standard deviation
E = margin of error
In this case, the standard deviation (σ) is $1050, and we want the estimate to be accurate within $200 (E = $200).
Plugging the values into the formula:
n = (1.96 * 1050 / 200)²
n = 3.8416 * 5.25²
n = 3.8416 * 27.5625
n ≈ 105.9821
Since we can't have a fraction of a sample, we need to round up to the nearest whole number. Therefore, the researcher would need a sample size of approximately 106 police officers to estimate the average salary with 95% confidence and an accuracy of within $200.