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. Find the 99% confidence interval for µ, the average annual salary of all high school teachers.
99% = mean ± 2.575 SEm
SEm = SD/√n
I'll let you do the calculations.
To find the confidence interval for the average annual salary of all high school teachers, we'll use the formula for a confidence interval for a population mean:
Confidence Interval = X̄ ± Z * (σ / √n),
where X̄ is the sample mean, Z is the critical value corresponding to the desired confidence level, σ is the population standard deviation, and n is the sample size.
In this case, X̄ = $56,490 (the sample mean), σ = $4,150 (the population standard deviation), n = 42 (the sample size), and the desired confidence level is 99%.
Step 1: Find the critical value (Z) for a 99% confidence level. The critical value can be obtained from a standard normal distribution table or by using a statistical calculator. For a 99% confidence level, the critical value is approximately 2.576.
Step 2: Substitute the values into the formula to calculate the confidence interval:
Confidence Interval = $56,490 ± 2.576 * ($4,150 / √42)
Step 3: Calculate the margin of error:
Margin of Error = 2.576 * ($4,150 / √42)
Step 4: Calculate the lower and upper limits of the confidence interval:
Lower Limit = $56,490 - Margin of Error
Upper Limit = $56,490 + Margin of Error
Finally, substitute the values and calculate the confidence interval.
Let's calculate the confidence interval:
Margin of Error = 2.576 * ($4,150 / √42)
Margin of Error ≈ $982.13
Lower Limit = $56,490 - $982.13 ≈ $55,507.87
Upper Limit = $56,490 + $982.13 ≈ $57,472.13
Therefore, the 99% confidence interval for the average annual salary of all high school teachers is approximately $55,507.87 to $57,472.13.