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Denote with

\(\begin{array}{l}{\rm{\bar p - }}\left( {\frac{{{\rm{\hat p + z}}_{{\rm{\alpha /2}}}^{\rm{2}}}}{{{\rm{2n}}}}} \right){\rm{/}}\left( {{\rm{1 + }}\frac{{{\rm{z}}_{{\rm{\alpha /2}}}^{\rm{2}}}}{{\rm{n}}}} \right)\\{\rm{\hat q - 1 - \hat p}}{\rm{.}}\end{array}\)

A confidence interval for a

population proportion

p with confidence level approximately\({\rm{100(1 - \alpha )}}\)is

\(\left( {{\rm{\tilde p - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\sqrt {{\rm{\hat p\hat q/n + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/4}}{{\rm{n}}^{\rm{2}}}} }}{{{\rm{1 + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/n}}}}{\rm{,\bar p + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\sqrt {{\rm{\hat p\hat q/n + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/4}}{{\rm{n}}^{\rm{2}}}} }}{{{\rm{1 + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/n}}}}} \right)\)

This is also called a score\({\rm{Cl for p}}\)

By replacing\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{with }}{{\rm{z}}_{\rm{\alpha }}}{\rm{ and in \pm }}\) only 鈥� or + stand, the large-sample upper confidence bound for\({\rm{\mu }}\)and the large-sample lower confidence bound for\({\rm{\mu }}\)are obtained.

The upper confidence bound at the\(95\% \)confidence level for the true proportion can be computed using

\({\rm{\bar p + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{,}}\frac{{\sqrt {{\rm{\hat p\hat q/n + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/4}}{{\rm{n}}^{\rm{2}}}} }}{{{\rm{1 + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/n}}}}{\rm{.}}\)

The values in the formula are

\(\begin{array}{l}{\rm{\bar p - }}\frac{{{\rm{0}}{\rm{.25 + 1}}{\rm{.64}}{{\rm{5}}^{\rm{2}}}{\rm{/2000}}}}{{{\rm{1 + 1}}{\rm{.64}}{{\rm{5}}^{\rm{2}}}{\rm{/1000}}}}{\rm{ - 0}}{\rm{.2507,}}\\{\rm{\hat p - }}\frac{{{\rm{250}}}}{{{\rm{1000}}}}{\rm{ - 0}}{\rm{.25,}}\\{\rm{\hat q - 1 - \hat p - 0}}{\rm{.75,}}\end{array}\)

where

\(\begin{array}{*{20}{r}}{{\rm{100(1 - \alpha ) - 95}}}\\{{\rm{\alpha - 0}}{\rm{.05}}}\end{array}\)

and

\({{\rm{z}}_{\rm{\alpha }}}{\rm{ - }}{{\rm{z}}_{{\rm{0}}{\rm{.05}}}}\mathop {\rm{ - }}\limits^{{\rm{(1)}}} {\rm{1}}{\rm{.645}}\)

(1) : this is obtained from

\({\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.05}}}}} \right){\rm{ - 0}}{\rm{.05}}\)

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

Therefore, the upper confidence bound at the\(95\% \)confidence level for the true proportion of such consumers is

\(\begin{array}{l}{\rm{\bar p + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\sqrt {{\rm{\hat p\hat q/n + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/4}}{{\rm{n}}^{\rm{2}}}} }}{{{\rm{1 + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/n}}}}{\rm{ - \& 0}}{\rm{.2507 + }}\frac{{{\rm{1}}{\rm{.645 \times }}\sqrt {{\rm{0}}{\rm{.25 \times 0}}{\rm{.75/1000 + 1}}{\rm{.64}}{{\rm{5}}^{\rm{2}}}{\rm{/}}\left( {{\rm{4 \times 100}}{{\rm{0}}^{\rm{2}}}} \right)} }}{{{\rm{1 + 1}}{\rm{.64}}{{\rm{5}}^{\rm{2}}}{\rm{/1000}}}}\\{\rm{ - 0}}{\rm{.2507 + 0}}{\rm{.0225 - 0}}{\rm{.2732}}{\rm{.}}\end{array}\)

There is\(95\% \)confidence that the true proportion is less than\(0.2732\),which indicated that there is compelling evidence that the true proportion is smaller than\({\rm{1 / 3}}\)The answer is\(0.2732\)

Yes, there is compelling evidence.