How large a sample should be taken if the population mean is to be estimated with 90% confidence to within $75? The population has a standard deviation of $892.
So far I have n= (.2)(5)^2, but I don't think that's right....
To determine the sample size needed to estimate the population mean with a given level of confidence and margin of error, we can use the formula:
n = (Z * σ / E)^2
Where:
n is the required sample size
Z is the z-score corresponding to the desired confidence level
σ is the population standard deviation
E is the desired margin of error
In this case, you want to estimate the population mean with 90% confidence to within $75. The population standard deviation is given as $892.
Step 1: Find the z-score corresponding to a 90% confidence level
The z-score for a 90% confidence level is found using a standard normal distribution table or calculator. The z-score for a 90% confidence level is approximately 1.645.
Step 2: Calculate the required sample size
Plug in the values into the formula:
n = (1.645 * 892 / 75)^2
Calculating that expression gives:
n ≈ (1441.54)^2
n ≈ 2,080,204.48
Since the required sample size must be a whole number, we round up to the nearest whole number:
n ≈ 2,080,205
Therefore, the sample size that should be taken to estimate the population mean with 90% confidence to within $75, with a population standard deviation of $892, is approximately 2,080,205.