Question

For a normal distribution, assume the 'sample' standard deviation is 8. What is the maximal margin of error associated with a 99% confidence interval for the true population mean if n = 25. ?

(Round to two decimal places)

Answer 1

The critical value for $\alpha = 0.01$ and $df = n-1 = 24$ is $t_c = z_{1-\alpha/2; n-1} = 2.79694.$

$$E=t_c\times\dfrac{s}{\sqrt{n}}=2.79694\times\dfrac{8}{\sqrt{25}}=4.48$$

The maximal margin of error is $4.48.$