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Chapter 7: Q23E (page 294)

The Pew Forum on Religion and Public Life reported on \({\rm{Dec}}{\rm{. 9, 2009}}\), that in a survey of \({\rm{2003}}\) American adults, \({\rm{25\% }}\)said they believed in astrology.

a. Calculate and interpret a confidence interval at the \({\rm{99\% }}\)confidence level for the proportion of all adult Americans who believe in astrology.

b. What sample size would be required for the width of a \({\rm{99\% }}\)CI to be at most .05 irrespective of the value of p?

Short Answer

Expert verified

a) The boundaries of the confidence interval are then\((0.2251,0.2749)\)

b) The sample size is \({\rm{n = 2653}}\)

Step by step solution

01

To Calculate and interpret a confidence interval

Given:

\(\begin{array}{l}{\rm{n = 2003}}\\{\rm{\hat p = 25\% = 0}}{\rm{.25}}\\{\rm{c = 99\% = 0}}{\rm{.99}}\end{array}\)

(a) For confidence level \({\rm{1 - \alpha - 0}}{\rm{.99}}\),determine \({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - }}{{\rm{z}}_{{\rm{0}}{\rm{.005}}}}\)

\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - 2}}{\rm{.575}}\)

Note: We take the average of \({\rm{2}}{\rm{.57 and 2}}{\rm{.58}}\)because \(0.005\)lies exactly in the middle between \({\rm{0}}{\rm{.0049 and 0}}{\rm{.0051}}\)

The margin of error is then:

\({\rm{E - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\sqrt {\frac{{{\rm{\hat p(1 - \hat p)}}}}{{\rm{n}}}} {\rm{ - 2}}{\rm{.575 \times }}\sqrt {\frac{{{\rm{0}}{\rm{.25(1 - 0}}{\rm{.25)}}}}{{{\rm{2003}}}}} {\rm{\gg 0}}{\rm{.0249}}\)

The boundaries of the confidence interval are then:

\(\begin{array}{l}{\rm{\hat p - E - 0}}{\rm{.25 - 0}}{\rm{.0249 - 0}}{\rm{.2251}}\\{\rm{\hat p + E - 0}}{\rm{.25 + 0}}{\rm{.0249 - 0}}{\rm{.2749}}\end{array}\)

Hence The boundaries of the confidence interval are then\((0.2251,0.2749)\)

02

To find the sample size

(b) Given:

\({\rm{ Width - 0}}{\rm{.05}}\)

The margin of error E is half the width of the confidence interval :

\({\rm{E - }}\frac{{{\rm{ Width }}}}{{\rm{2}}}{\rm{ - }}\frac{{{\rm{0}}{\rm{.05}}}}{{\rm{2}}}{\rm{ - 0}}{\rm{.025}}\)

Formula sample size:

\({\rm{\hat p known: n - }}\frac{{{{\left( {{{\rm{z}}_{{\rm{\alpha /2}}}}} \right)}^{\rm{2}}}{\rm{\hat p\hat q}}}}{{{{\rm{E}}^{\rm{2}}}}}{\rm{ - }}\frac{{{{\left( {{{\rm{z}}_{{\rm{\alpha /2}}}}} \right)}^{\rm{2}}}{\rm{\hat p(1 - \hat p)}}}}{{{{\rm{E}}^{\rm{2}}}}}\)

\({\rm{\hat p}}\)unknown: \({\rm{n - }}\frac{{\left( {{{\left. {{{\rm{z}}_{{\rm{\alpha /2}}}}} \right|}^{\rm{2}}}{\rm{0}}{\rm{.25}}} \right.}}{{{{\rm{E}}^{\rm{2}}}}}\)

For confidence level. \({\rm{1 - \alpha - 0}}{\rm{.99}}\), determine \({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - }}{{\rm{z}}_{{\rm{0}}{\rm{.005}}}}\):

\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - 2}}{\rm{.575}}\)

\({\rm{\hat p}}\)is assumed to be unknown (as we are interested in the sample size irrespective of the value of \({\rm{\hat p}}\)), then the sample size is (round up!):

\({\rm{n - }}\frac{{{{\left( {{{\rm{z}}_{{\rm{\alpha /2}}}}} \right)}^{\rm{2}}}{\rm{\hat p(1 - \hat p)}}}}{{{{\rm{E}}^{\rm{2}}}}}{\rm{ - }}\frac{{{\rm{2}}{\rm{.57}}{{\rm{5}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.5(1 - 0}}{\rm{.5)}}}}{{{\rm{0}}{\rm{.02}}{{\rm{5}}^{\rm{2}}}}}{\rm{\gg 2653}}\)

Hence the sample size is \({\rm{n = 2653}}\)

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