Question

A random sample of 84 students at a university showed an average age of 22 years and a sample standard deviation of 3 years. Find the margin of error for the 94% confidence interval. (Assume that the population of students is large relative to the sample size. Round your solution to 4 decimal places)

Answer 1

Answer:

The margin of error for the 94% confidence interval is 0.6154.

Step-by-step explanation:

The (1 - α)% confidence interval for population mean is:

$CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}$

The margin of error of this interval is:

$MOE=z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}$

The critical value of z for 94% confidence level is, z = 1.88.

Compute the margin of error for the 94% confidence interval as follows:

$MOE=z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}$

          $=1.88\times\frac{3}{\sqrt{84}}\\\\=0.6154$

Thus, the margin of error for the 94% confidence interval is 0.6154.