Question

A state legislator wishes to survey residents of her district to see what proportion of the electorate is aware of her position on using state funds to pay for abortions.

a. What sample size is necessary if the \({\rm{95\% }}\)CI for p is to have a width of at most . \({\rm{10}}\)irrespective of p?

b. If the legislator has strong reason to believe that at least \({\rm{2/3}}\)of the electorate know of her position, how large a sample size would you recommend?

Short answer

a) The sample size is \({\rm{n = 381}}\)

b) I recommend the sample size of \({\rm{n = 339}}\)

Step-by-step solution

To find the sample size

(a):

The expression for the necessary sample size for a confidence interval to have width w is

\({\rm{n - }}\frac{{{\rm{2z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{\hat p\hat q - z}}_{{\rm{\alpha /2}}}^{\rm{2}}{{\rm{w}}^{\rm{2}}}{\rm{ \pm }}\sqrt {{\rm{4z}}_{{\rm{\alpha /2}}}^{\rm{4}}{\rm{\hat p\hat q}}\left( {{\rm{\hat p\hat q - }}{{\rm{w}}^{\rm{2}}}} \right){\rm{ + }}{{\rm{w}}^{\rm{2}}}{\rm{z}}_{{\rm{\alpha /2}}}^{\rm{4}}} }}{{{{\rm{w}}^{\rm{2}}}}}\)

Therefore, for \({\rm{width w = 0}}{\rm{.1}}\)and\({\rm{\hat p - \hat q - 0}}{\rm{.5}}\) (irrespective of p ) the necessary sample size becomes

\(\begin{array}{l}{\rm{n = }}\frac{{{\rm{2z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{\hat p\hat q - z}}_{{\rm{\alpha /2}}}^{\rm{2}}{{\rm{w}}^{\rm{2}}}{\rm{ \pm }}\sqrt {{\rm{4z}}_{{\rm{\alpha /2}}}^{\rm{4}}{\rm{\hat p\hat q}}\left( {{\rm{\hat p\hat q - }}{{\rm{w}}^{\rm{2}}}} \right){\rm{ + }}{{\rm{w}}^{\rm{2}}}{\rm{z}}_{{\rm{\alpha /2}}}^{\rm{4}}} }}{{{{\rm{w}}^{\rm{2}}}}}\\{\rm{ - }}\frac{{{\rm{2 \times 1}}{\rm{.9}}{{\rm{6}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.5 \times 0}}{\rm{.5 - 1}}{\rm{.9}}{{\rm{6}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}{\rm{ \pm }}\sqrt {{\rm{4 \times 1}}{\rm{.9}}{{\rm{6}}^{\rm{4}}}{\rm{ \times 0}}{\rm{.5 \times 0}}{\rm{.5 \times }}\left( {{\rm{0}}{\rm{.5 \times 0}}{\rm{.5 - 0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}} \right){\rm{ + 0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}{\rm{ \times 1}}{\rm{.9}}{{\rm{6}}^{\rm{4}}}} }}{{{\rm{0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}}}\\{\rm{ = 381}}\end{array}\)

Hence the sample size is \({\rm{n = 381}}\)

To find how large a sample size would you recommend

(b):

For \({\rm{width w = 0}}{\rm{.1}}\),and \({\rm{\hat p - 1/3,\hat q - 2/3}}\)the necessary sample size becomes

\(\begin{array}{l}{\rm{n = }}\frac{{{\rm{2z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{\hat p\hat q - z}}_{{\rm{\alpha /2}}}^{\rm{2}}{{\rm{w}}^{\rm{2}}}{\rm{ \pm }}\sqrt {{\rm{4z}}_{{\rm{\alpha /2}}}^{\rm{4}}{\rm{\hat p\hat q}}\left( {{\rm{\hat p\hat q - }}{{\rm{w}}^{\rm{2}}}} \right){\rm{ + }}{{\rm{w}}^{\rm{2}}}{\rm{z}}_{{\rm{\alpha /2}}}^{\rm{4}}} }}{{{{\rm{w}}^{\rm{2}}}}}\\{\rm{ = }}\frac{{{\rm{2 \times 1}}{\rm{.9}}{{\rm{6}}^{\rm{2}}}{\rm{ \times 1/3 \times 2/3 - 1}}{\rm{.9}}{{\rm{6}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}{\rm{ \pm }}\sqrt {{\rm{4 \times 1}}{\rm{.9}}{{\rm{6}}^{\rm{4}}}{\rm{ \times 1/3 \times 2/3 \times }}\left( {{\rm{1/3 \times 2/3 - 0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}} \right){\rm{ + 0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}{\rm{ \times 1}}{\rm{.9}}{{\rm{6}}^{\rm{4}}}} }}{{{\rm{0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}}}\\{\rm{ = 339}}{\rm{.}}\end{array}\)

Hence I recommend the sample size of \({\rm{n = 339}}\)

Where in both a and b

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

and

\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - }}{{\rm{z}}_{{\rm{0}}{\rm{.05/2}}}}{\rm{ - }}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}\mathop {\rm{ - }}\limits^{{\rm{(1)}}} {\rm{1}}{\rm{.96}}\)

(1) : this is obtained from

\({\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}} \right){\rm{ - 0}}{\rm{.025}}\)

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

Hence There were no assumptions made about the distribution.