91影视

For large n, the standardized random variable

\({\rm{Z - }}\frac{{{\rm{\bar X - \mu }}}}{{{\rm{S/}}\sqrt {\rm{n}} }}\)has approximately a normal distribution with expectation \(0\)and standard deviation \(1\) .

Therefore, alarge-sample confidence interval for \({\rm{\mu }}\)is

\({\rm{\bar x \pm }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}\)

with confidence level of approximately \({\rm{100(1 - \alpha )}}\).This stands regardless the population distribution.

Given

\(\begin{array}{l}{\rm{n = 56}}\\{\rm{\bar x = 8}}{\rm{.17 - 1}}{\rm{.42}}\end{array}\)

a \(95\% \)confidence interval for the true average \({\rm{\mu }}\)is

\(\left( {{\rm{\bar x - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}} \right){\rm{ - }}\left( {{\rm{8}}{\rm{.17 - 1}}{\rm{.96 \times }}\frac{{{\rm{1}}{\rm{.42}}}}{{\sqrt {{\rm{56}}} }}{\rm{,8}}{\rm{.17 + 1}}{\rm{.96 \times }}\frac{{{\rm{1}}{\rm{.42}}}}{{\sqrt {{\rm{56}}} }}} \right){\rm{ - = 7}}{\rm{.798,8}}{\rm{.542)}}\)

where

\(\begin{array}{*{20}{r}}{100(1 - \alpha ) - 95}\\{\alpha - 0.05}\end{array}\)

and

\(\begin{array}{l}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - }}{{\rm{z}}_{{\rm{0}}{\rm{.05/2}}}}{\rm{ - }}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}\mathop {\rm{ - }}\limits^{{\rm{(1)}}} {\rm{1}}{\rm{.96}}\\{\rm{ this is obtained from }}\\{\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}} \right){\rm{ - 0}}{\rm{.025}}\end{array}\)

And from the normal probability table in the appendix. The probability can also be computed with a software.

Hence There were no assumptions made about the distribution.