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 (σ = 1050), how large of a sample is needed to be accurate within $200 (E = 200)? That is, find n.

105.88

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 calculate the sample size. Round your answer to the next highest whole number.

I hope this will help get you started.

n= 106

Why did the math book become friends with the history book? Because they figured they could always count on each other! Now, let's solve this problem together.

To find the sample size needed to estimate the average salary within a certain margin of error, we can use the formula:

n = (Z * σ / E)²

Where:
n = Sample size
Z = Z-value (confidence level)
σ = Standard deviation
E = Margin of error

In this case, the researcher wants to be 95% confident and the margin of error is $200. The Z-value for a 95% confidence level is approximately 1.96. Plugging in the values:

n = (1.96 * 1050 / 200)²

Calculating the expression in brackets:
n = (2.058)*²
n = 4.235

Rounding up to the nearest whole number (since we can't have a fraction of a person for our sample size), the researcher would need a sample size of 5 to estimate the average salary within $200 with 95% confidence.

To calculate the sample size (n) needed to estimate the average salary of teachers in a large urban school district with a given confidence level (95%) and a certain margin of error ($200), we can use the following formula:

n = (Z * σ / E)²

Where:
n = sample size
Z = z-score corresponding to the desired confidence level (95% confidence corresponds to a z-score of 1.96)
σ = standard deviation of the population
E = margin of error

Plugging in the values given in the question:
Z = 1.96
σ = $1050
E = $200

n = (1.96 * 1050 / 200)²
n = (2058 / 200)²
n ≈ 10.29²
n ≈ 106

So, a sample size of at least 106 teachers is needed in order to estimate the average salary within $200 with 95% confidence.