The required sample size is 15 to achieve a margin of error within ±1.5 at a 90% confidence level.
To calculate the required sample size for estimating a population mean with a given margin of error and confidence level, you can use the formula:
n=( Z^2 ×σ^2/ E^2 )
where:
n is the required sample size,
Z is the Z-score corresponding to the desired confidence level,
σ is the standard deviation of the population,
E is the margin of error.
For a 90% confidence level, the Z-score is approximately 1.645 (you can find this value in a standard normal distribution table).
Given:
Z=1.645 (for 90% confidence),
σ=3.5 (standard deviation),
E=1.5 (margin of error).
Substitute these values into the formula:
n=( 1.645^2 ×3.5^2 /1.5^2)
Calculating this expression will give you the required sample size. Round up to the nearest integer.
n=( 2.70225×12.25/2.25)
n= 33.105812 / 2.25
n≈14.71325
Rounding up to the nearest integer:
n≈15
Therefore, the required sample size is 15 to achieve a margin of error within ±1.5 at a 90% confidence level.
Question
What sample size is needed to give a margin of error within ±1.5 in estimating a population mean with 90%confidence, assuming a previous sample had s=3.5.Round your answer up to the nearest integer.
sample size =