A private medical clinic wants to estimate the true mean annual income of its patients. The clinic needs to be within $100 of the true mean. The clinic estimates that the true population standard deviation is around $2,100. If the confidence level is 95%, find the required sample size in order to meet the desired accuracy.
Use a formula to find sample size.
Here is one:
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 = 2100, E = 100, ^2 means squared, and * means to multiply.
Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.
Hope this helps.
A population is normally distributed with mean 26.8 and standard deviation 2.5. Find p( 26.8 < x < 29.3 ). (Round your answer to FOUR decimal places.)
To find the required sample size, we can use the formula for the sample size of a mean:
n = (Z * σ / E)^2
Where:
n = required sample size
Z = Z-score corresponding to the desired confidence level
σ = population standard deviation
E = desired margin of error
In this case, the desired margin of error is $100, and the confidence level is 95%. To find the Z-score corresponding to a 95% confidence level, we can refer to a Z-table or use a statistical calculator.
The Z-score for a 95% confidence level is approximately 1.96.
Plugging these values into the formula:
n = (1.96 * 2100 / 100)^2
n = (4104 / 100)^2
n = 41.04^2
n ≈ 1688.81
Since we can't have a fraction of a sample, we round up to the nearest whole number.
Therefore, the required sample size needed to meet the desired accuracy is approximately 1689.