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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,

\(\begin{aligned}{\rm{Sample size}}:n &= 55\\{\rm{Confidence interval: }}c &= 90\% \end{aligned}\)

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

\(\begin{aligned}p &= \frac{x}{n}\\p &= \frac{{11}}{{55}}\\\ p &= 0.2\end{aligned}\)

For confidence level \(1 - \alpha = 0.90\), determine \({Z_\alpha } = {Z_{0.1}}\)using the normal probability table in the appendix ( look up \(0.10\)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:

\(\begin{aligned}E &= {Z_{\alpha /2}}\sqrt {\frac{{\widehat p(1 - \widehat p)}}{n}} \\E &= 1.645\sqrt {\frac{{0.2(1 - 0.2)}}{{55}}} \\E \approx 0.0887\end{aligned}\)

The upper boundary of the confidence interval is then:

\(\begin{aligned} p + E &= 0.2 + 0.0887\\p + E &= 0.2887\end{aligned}\)

Hence, The upper boundary of the confidence interval is \(0.2887\).