Question

by confidence intervals, normal distributed data, known variance

Equation 1: If is the sample mean of a random sample of size n from a normal population with known variance o2, a 100 (1- a)% CI on u is given by HIZa/2 n SHST+/2 Vn is the upper 100g percentage point of the standard normal distribution. a/2 where 17. If the sample size n is doubled, by how much is the length of the CI on u in Equation 1 reduced? What happens to the length of the interval if the sample size is increased by a factor of four?

Answer 1

17)

Original length of the CI= $2\times Z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

If sample size becomes '2n', length of the CI= $2\times Z_{\alpha/2}\frac{\sigma}{\sqrt{2n}}=\frac{1}{\sqrt{2}}\times [2\times Z_{\alpha/2}\frac{\sigma}{\sqrt{n}}]$

i.e. the length gets reduced by a factor of $\frac{1}{\sqrt{2}}$ .

If sample size becomes '4n', length of the CI= $2\times Z_{\alpha/2}\frac{\sigma}{\sqrt{4n}}=\frac{1}{2}\times [2\times Z_{\alpha/2}\frac{\sigma}{\sqrt{n}}]$

i.e. the length gets reduced by a factor of $\frac{1}{2}$.