I need to verify one of my answers quickly...the problem is:

The mean annual salary for classroom teachers is $43,658. Assume a standard deviation of $8000.
1) Determine the sampling distribution of the sample mean for a sample of size 256. Interpret your answer in terms of the distribution of of all possible sample mean salaries for samples of 256 teachers.
2) Determine the percentage of all samples of 256 public school teachers that have mean salaries within $1000 of the population mean salary of $43,658. Interpret your answer in terms of sampling error.
Are my answers correct? :
1)SE within samples = 8,000/√256 =500
2) 42.1% of samples are within $500 of the mean $8,000.
or...
is it 84.2% that're 1/in $1000?
or...
am I way off?

1) looks correct

For 2) You have an standard error of 500, and are asked what is likelihood of similar sample being within 1000 or 2.0 standard deviations away from the mean. Look up 2.0 in your cumulative normal distribution table (probably in the back of your stats book). I get .9772 -- meaning 97.72% of such samples with be within $1000 of the mean of $43,658

To verify your answers, let's go through the calculations step by step:

1) To determine the sampling distribution of the sample mean for a sample size of 256, we need to calculate the standard error (SE) first. The formula for standard error is the population standard deviation divided by the square root of the sample size:

SE = σ / √n

where σ is the population standard deviation (in this case $8000) and n is the sample size (in this case 256).

SE = 8000 / √256
SE = 8000 / 16
SE = 500

So, your calculation for the standard error within samples is correct.

Interpretation: The sampling distribution of the sample mean salaries for samples of 256 teachers has a standard error of $500. This means that on average, the sample mean salaries will deviate from the population mean salary by approximately $500.

2) To determine the percentage of all samples of 256 public school teachers that have mean salaries within $1000 of the population mean salary, we need to calculate the z-score for this range. The formula for z-score is the difference between the sample mean and the population mean divided by the standard error:

z = (x̄ - μ) / SE

where x̄ is the sample mean, μ is the population mean (in this case $43,658), and SE is the standard error (in this case $500).

For a range within $1000 of the population mean, we need to find the z-scores for $1000/$500 = 2 standard errors above and below the population mean.

z = 2

To find the percentage of samples within this range, we can refer to the standard normal distribution table. Looking up the z-score of 2 in the table, we find that the area under the curve to the left of the z-score is approximately 0.9772. Since we want the area between 2 standard deviations on both sides, we can subtract 1 - 0.9772 to get the central area.

Central area = 1 - 0.9772 = 0.0228

Interpretation: Approximately 2.28% of all samples of 256 public school teachers will have mean salaries within $1000 of the population mean salary. This indicates that there is a small sampling error, and the vast majority of samples will have mean salaries further away from the population mean.

Therefore, your second interpretation (2.28%) is correct, while your first interpretation (42.1%) and third interpretation (84.2%) are off.