Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
Assume $95\text{%}$ confidence interval for a population mean based on sample size of $n_1 = 40$
If you wish to have a confidence interval of the same confidence level but with a length which is one fourth of the one you already have, then what would be the sample size of $n_2$
I don't understand this question, can someone give me some pointers to start off? What does "but with a length which is one fourth of the one you already have" mean?
Let the first C.I be $(\bar{x} - E_1, \bar{x} + E_1)$, then the length of this interval is $2E_1$,
and let the second C.I be $(\bar{x} - E_2, \bar{x} + E_2)$, then similarly the length of this interval is $2E_2$, and we are given that: $2E_1 = 4(2E_2)$. Thus: $E_2 = \dfrac{E_1}{4}$.
We have: $n = \dfrac{K}{E^2} \Rightarrow \dfrac{n_2}{n_1} = \dfrac{E_1^2}{E_2^2} = 16 \Rightarrow n_2 = 16n_1 = 16(40) = 640$.
