3.1k views
1 vote
A sample of size n=56 is drawn from a population whose standard deviation is σ= 4.5. Find the margin of error for a 95% confidence interval for μ.

User BraggPeaks
by
7.8k points

1 Answer

1 vote

Final answer:

The margin of error for a 95% confidence interval for μ with n=56 and σ=4.5 is approximately 1.179, which is found using the z-score for 95% confidence and the formula for margin of error.

Step-by-step explanation:

The margin of error for a 95% confidence interval for μ with n=56 and σ=4.5 is approximately 1.179, which is found using the z-score for 95% confidence and the formula for margin of error.

To find the margin of error for a 95% confidence interval for μ when the sample size is n=56 and the population standard deviation is σ = 4.5, we need to use the formula for the margin of error in a normal distribution which is E = Z * (σ/√n), where Z is the z-score corresponding to the 95% confidence level.

For a 95% confidence interval, the z-score is commonly 1.96. Applying the formula, we get E = 1.96 * (4.5/√56), which we need to calculate.

First, find the standard error by dividing the standard deviation by the square root of the sample size:

Standard Error (SE) = 4.5 / √56 = 4.5 / 7.4833 = 0.6015

Then, calculate the margin of error:

Margin of Error (E) = 1.96 * SE = 1.96 * 0.6015 = 1.179

Therefore, the margin of error for a 95% confidence interval for μ is approximately 1.179.

User Umer Farooq
by
7.8k points