Question

Suppose that\({X_1},...,{X_n}\)form a random sample from the normal distribution with unknown mean μ and known variance\({\sigma ^2}\). How large a random sample must be taken in order that there will be a confidence interval for μ with confidence coefficient 0.95 and length less than 0.01σ?

Short answer

153665

Step-by-step solution

Given information

\({X_1},...,{X_n}\) form a random sample from the normal distribution with unknown mean μ and known variance \({\sigma ^2}\).We need to calculate the minimum sample size that must be taken in order that there will be a confidence interval for μ with confidence coefficient 0.95 and length less than 0.01σ.

Calculation of the sample size

Limits for confidence coefficient \(\gamma \) in the interval \(\left( {A,B} \right)\) is given by

\(\begin{align}A &= \bar X - {\Phi ^{ - 1}}\left( {\frac{{1 + \gamma }}{2}} \right)\frac{\sigma }{{\sqrt n }}\\B &= \bar X + {\Phi ^{ - 1}}\left( {\frac{{1 + \gamma }}{2}} \right)\frac{\sigma }{{\sqrt n }}\end{align}\)

Where \(\gamma \in \left( {0,1} \right)\)

For \(\gamma \)=0.95 we have

\(\begin{align}{\Phi ^{ - 1}}\left( {\frac{{1 + 0.95}}{2}} \right)\\ &= {\Phi ^{ - 1}}\left( {0.975} \right)\\ &= 1.96\end{align}\)

The length of the confidence interval is

\(\begin{align}L &= B - A\\ &= 2{\Phi ^{ - 1}}\left( {\frac{{1 + \gamma }}{2}} \right)\frac{\sigma }{{\sqrt n }}\\ &= 2 \times 1.96\frac{\sigma }{{\sqrt n }}\\ &= 3.92\frac{\sigma }{{\sqrt n }}\end{align}\)

According to the question

\(\begin{align}3.92\frac{\sigma }{{\sqrt n }} < 0.01\sigma \\n > 153665\end{align}\)

Hence the minimum sample size is 153665