A researcher is interested in
estimating the average salary of teachers in a large urban
school district. She wants to be 95% confident that her
estimate is correct. If the standard deviation is $1050, how
large a sample is needed to be accurate within $200?
Formula:
n = [(z-value * sd)/E]^2
...where n = sample size, z-value will be 1.96 using a z-table to represent the 95% confidence interval, sd = 1050, E = 200, ^2 means squared, and * means to multiply.
Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.
10
106
To determine the sample size needed to estimate the average salary accurately within a certain margin of error, we can use the formula for sample size calculation:
n = (Z * σ / E)²
Where:
n = sample size
Z = Z-score (corresponding to the desired confidence level)
σ = standard deviation of the population
E = desired margin of error
In this case, the researcher wants to be 95% confident that her estimate is correct, so the Z-score corresponding to a 95% confidence level is approximately 1.96.
Using the given values:
Z = 1.96
σ = $1050
E = $200
Now we can substitute these values into the formula and solve for n:
n = (1.96 * 1050 / 200)²
Calculating this equation gives:
n ≈ 89.6716
Since you cannot have a fraction of a person, you would round up to the nearest whole number. Therefore, a sample size of 90 would be needed to estimate the average salary of teachers accurately within $200 with 95% confidence.