confidence intervals: A researcher is interested in estimating the average salary of teachers in a large 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.
To determine the sample size needed, we can use the formula for calculating the sample size for estimating a population mean with a specified margin of error:
\[n = \left(\frac{z \cdot \sigma}{E}\right)^2\]
Where:
- \(n\) is the sample size we need to determine
- \(z\) is the z-score corresponding to the desired confidence level. For a 95% confidence level, the z-score is approximately 1.96.
- \(\sigma\) is the standard deviation of the population. In this case, it is given as $1050.
- \(E\) is the desired margin of error for the estimate. In this case, it is $200.
Using these values, we can substitute them into the formula:
\[n = \left(\frac{1.96 \cdot 1050}{200}\right)^2\]
Calculating the expression inside the parentheses:
\[n = \left(\frac{2058}{200}\right)^2\]
Dividing to get the squared value:
\[n = (10.29)^2\]
Finally:
\[n \approx 106\]
Therefore, a sample size of at least 106 teachers is needed for the researcher to estimate the average salary of teachers in the large school district with a margin of error of $200 and a 95% confidence level.