How large a sample should be taken if the population mean is to be estimated with 99% confidence to within $72? The population has a standard deviation of $800
Please Help, im stuck
To determine the sample size needed to estimate the population mean with a given level of confidence and margin of error, you can use the following formula:
n = (Z * σ / E)²
Where:
n = sample size
Z = Z-score corresponding to the desired confidence level (in this case, 99% confidence corresponds to a Z-score of 2.576)
σ = standard deviation of the population
E = desired margin of error
Given:
Z = 2.576
σ = $800
E = $72
Plugging in the values into the formula:
n = (2.576 * 800 / 72)²
Simplifying the equation:
n = (2060.8 / 72)²
n = 28.622222²
n ≈ 821
Therefore, you should take a sample size of at least 821 to estimate the population mean with 99% confidence and a margin of error of $72.
To determine the sample size required to estimate the population mean with a specified level of confidence, margin of error, and known population standard deviation, you can use the formula:
n = (Z * σ / E)²
Where:
n = required sample size
Z = Z-score corresponding to the desired level of confidence
σ = population standard deviation
E = margin of error
In this case:
Z = Z-score corresponding to 99% confidence level. Since it is not provided, you can look it up in a standard normal distribution table or use a statistical calculator. For a 99% confidence level, the Z-score is approximately 2.576.
σ = population standard deviation = $800
E = margin of error = $72
Now let's substitute these values into the formula:
n = (2.576 * 800 / 72)²
Calculating this equation will give you the required sample size.
n ≈ (2060.8 / 72)²
n ≈ 667.88²
n ≈ 445,108.6544
Since you cannot have a fraction of a sample, you would round up to the nearest whole number. Therefore, you would need a sample size of at least 445,109 to estimate the population mean with a 99% confidence level and a margin of error of $72, assuming the known population standard deviation is $800.