91影视

(a) \(\begin{aligned}n &= 2343 \hfill \\\hat p &= 53\% = 0.53 \hfill \\c &= 99\% = 0.99 \hfill \\ \end{aligned} \)

For confidence level \({\text{1 - \alpha - 0}}{\text{.99}}\) , determine \({{\text{z}}_{{\text{\alpha /2}}}}{\text{ - }}{{\text{z}}_{{\text{0}}{\text{.005}}}}\) using the normal probability table in the appendix (look up 0.005 in the table, the z-score is then the found z-score with opposite sign):

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

The margin of error is then:

\({\text{E - }}{{\text{z}}_{{\text{\alpha /2}}}}{\text{ \times }}\sqrt {\frac{{{\text{\hat p(1 - \hat p)}}}}{{\text{n}}}} {\text{ - 2}}{\text{.575 \times }}\sqrt {\frac{{{\text{0}}{\text{.53(1 - 0}}{\text{.53)}}}}{{{\text{2343}}}}} {\text{\gg 0}}{\text{.0266}}\)

The boundaries of the confidence interval are then:

\(\begin{aligned}\hat p - E - 0.53 - 0.0266 - 0.5034 \hfill \\\hat p + E - 0.53 + 0.0266 - 0.5566 \hfill \\\end{aligned} \)

Hence the boundaries of the confidence interval are then \((0.5034,0.5566)\)

(b) Given:

\(\begin{aligned}w&=0.05\hfill\\c&=99\%=0.99\hfill\\\hat p&= 53\%= 0.53 \hfill \\\end{aligned} \)

Formula sample size:

\(\begin{aligned}n\gg &\frac{{4{{\left[ {{z_{\alpha /2}}} \right]}^2}\hat p\hat q}}{{{w^2}}} -\frac{{4{{\left[ {{z_{\alpha /2}}} \right]}^2}\hat p(1 - \hat p)}}{{{w^2}}} - \frac{{4{{\left[ {{z_{\alpha /2}}}\right]}^2}\hat p(1 - \hat p)}}{{{w^2}}} \hfill \\{z_{\alpha /2}} - 2.575 \hfill \\\end{aligned} \)

Note: We take the average of 2.57 and 2.58, because 0.005 is exactly in the middle between 0.0049 and 0.0051

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

\({\text{n = }}\frac{{{\text{4}}{{\left[ {{{\text{z}}_{{\text{\alpha /2}}}}} \right]}^{\text{2}}}{\text{\hat p(1 - \hat p)}}}}{{{{\text{w}}^{\text{2}}}}}{\text{ - }}\frac{{{\text{4 \times 2}}{\text{.57}}{{\text{5}}^{\text{2}}}{\text{ \times 0}}{\text{.5(1 - 0}}{\text{.5)}}}}{{{\text{0}}{\text{.0}}{{\text{5}}^{\text{2}}}}}{\text{ = 2653}}\)

Note: If you use the critical value \({{\text{z}}_{{\text{\alpha /2}}}}{\text{ - 2}}{\text{.58}}\) , then you obtain \({\text{n - 2663}}\)

Hence the sample size is \({\text{n - 2653}}\)