A government survey conducted to estimate the mean of houses in a metropolitan area is designed to have a margin of error of 10,000. Pilot studies suggest that the population standard deviation is 70,000. Estimate the minimum sample size needed to estimate the population mean with the stated accuracy.
Which is the best estimate for 48% of 31?
To estimate the minimum sample size needed to estimate the population mean with a specific margin of error, we can use the formula:
n = (Z * σ / E)²
Where:
n = sample size
Z = Z-score/confidence level (corresponding to desired level of confidence, usually denoted by 1 - α)
σ = population standard deviation
E = margin of error
In this case, the margin of error is 10,000, and the population standard deviation is 70,000. However, we need to determine the Z-score for the desired confidence level.
Assuming we want a 95% confidence level (corresponding to a significance level of α = 0.05), the Z-score is 1.96. This value can be looked up in a standard normal distribution table.
Substituting the given values into the formula, we get:
n = (1.96 * 70,000 / 10,000)²
n = (137,200 / 10,000)²
n = 13.72²
n ≈ 188
Therefore, the minimum sample size needed to estimate the population mean with a margin of error of 10,000 is approximately 188.