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The margin of the error is then:

\(E = {Z_{\alpha /2}}\sqrt {\frac{{\widehat p(1 - \widehat p)}}{n}} \)

Given that,

\(\begin{array}{l}x = 31\\n = 88\\Width = 0.04\\c = 95\% \end{array}\)

The sample proportion is the number of successes divided by the sample size:

\(\begin{array}{l}\widehat p = \frac{x}{n}\\\widehat p = \frac{{31}}{{88}}\\\widehat p = 0.3523\end{array}\)

For confidence level \(1 - \alpha = 0.95\), determine \({Z_{\alpha /2}} = {Z_{0.025}}\)using the normal probability table in the appendix ( look up \(0.025\)in the table, the z-score is then the found z-score with opposite sign):

\({Z_{\alpha /2}} = 1.96\)

The margin of the error is then:

\(\begin{array}{l}E = {Z_{\alpha /2}}\sqrt {\frac{{\widehat p(1 - \widehat p)}}{n}} \\E = 1.96\sqrt {\frac{{0.3523(1 - 0.3523)}}{{88}}} \\E \approx 0.0998\end{array}\)

The boundaries of the confidence interval are then:

\(\begin{array}{l}\widehat p - E = 0.3523 - 0.0998\\\widehat p - E = 0.2525\\\widehat p + E = 0.3523 + 0.0998\\\widehat p + E = 0.4521\end{array}\)

We are \(95\% \) confident that the true proportion of all athletes found to have the EILO condition under these circumstance is between \(0.2525\) and \(0.4521\)

Hence, The boundaries of the confidence interval is \((0.2525,0.4521)\).

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

\(\begin{array}{l}E = \frac{{Width}}{2}\\E = \frac{{0.04}}{2}\\E = 0.02\end{array}\)

The sample proportion is the number of successes divided by the sample size:

\(\begin{array}{l}\widehat p = \frac{x}{n}\\\widehat p = \frac{{31}}{{88}}\\\widehat p = 0.3523\end{array}\)

Formula sample size:

\(\widehat p\)known:

\(\begin{array}{l}n = \frac{{{{\left( {{Z_{\alpha /2}}} \right)}^2}\widehat p\widehat q}}{{{E^2}}}\\n = \frac{{{{\left( {{Z_{\alpha /2}}} \right)}^2}\widehat p(1 - \widehat p)}}{{{E^2}}}\end{array}\)

\(\widehat p\)unknown:

\(n = \frac{{{{\left( {{Z_{\alpha /2}}} \right)}^2}0.25}}{{{E^2}}}\)

For confidence level \(1 - \alpha = 0.95\), determine \({Z_{\alpha /2}} = {Z_{0.025}}\)using the normal probability table in the appendix ( look up \(0.025\)in the table, the z-score is then the found z-score with opposite sign):

\({Z_{\alpha /2}} = 1.96\)

\(\widehat p\)is known, then the sample size is (round up to the nearest integral);

\(\begin{array}{l}n = \frac{{{{\left( {{Z_{\alpha /2}}} \right)}^2}\widehat p(1 - \widehat p)}}{{{E^2}}}\\n = \frac{{{{(1.96)}^2}(0.3523)(1 - 0.3523)}}{{{{0.02}^2}}}\\n = 2192\end{array}\)

Hence, the sample size, \(n = 2192\).

We are \(95\% \) confident that the true population mean is between \(0.2525\) and \(0.4521\)thus we are \(2.5\% \) sure that the true population mean is below \(0.2525\) and \(2.5\% \) sure that the true population mean is above \(0.4521\). In total there is then\(95\% + 2.5\% = 97.5\% \) confidence that the true population mean is below \(0.4521\).

The upper limit \(0.4521\)is NOT a \(95\% \) upper confidence bound, but is a \(97.5\% \) upper confidence bound.

\(95\% \)upper confidence bound:

The sample proportion is the number of successes divided by the sample size:

\(\begin{array}{l}\widehat p = \frac{x}{n}\\\widehat p = \frac{{31}}{{88}}\\\widehat p = 0.3523\end{array}\)

For confidence level \(1 - \alpha = 0.95\), determine \({Z_{\alpha /2}} = {Z_{0.05}}\)using the normal probability table in the appendix ( look up \(0.025\)in the table, the z-score is then the found z-score with opposite sign):

\({Z_{\alpha /2}} = 1.645\)

The margin of the error is then:

\(\begin{array}{l}E = {Z_{\alpha /2}}\sqrt {\frac{{\widehat p(1 - \widehat p)}}{n}} \\E = 1.645\sqrt {\frac{{0.3523(1 - 0.3523)}}{{88}}} \\E \approx 0.0838\end{array}\)

The upper boundary of the confidence interval is then:

\(\begin{array}{l}\widehat p + E = 0.3523 + 0.0838\\\widehat p + E = 0.4361\end{array}\)

Thus, The \(95\% \)upper confidence bound is \(0.4361\).