An economist found that a random sample of 42 high school teachers had an average annual salary of $56,490. A previous study showed that the population standard deviation is $4150. Calculate the margin of error, E, for a 99% confidence interval for µ, the average annual salary of all high school teachers.
99% = mean ± 2.575 SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score. To include each end, double proportion.
To calculate the margin of error (E) for a 99% confidence interval for the average annual salary of all high school teachers (µ), we need to use the formula:
E = Z * (σ / √n)
Where:
- E is the margin of error.
- Z is the Z-score corresponding to the desired confidence level (in this case, 99%).
- σ is the population standard deviation.
- n is the sample size.
First, let's calculate the Z-score. The Z-score can be found using a Z-table or a statistical calculator. For a 99% confidence level, the Z-score is approximately 2.576.
Next, we substitute the values into the formula:
E = 2.576 * (4150 / √42)
Now, we can calculate the margin of error:
E ≈ 2.576 * (4150 / √42) ≈ 1177.17
Therefore, the margin of error (E) for a 99% confidence interval for the average annual salary of all high school teachers is approximately $1177.17.