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It is given that a survey was sponsored by a chocolate company.

Out of 1708 women who were surveyed, 85% of them said that chocolate made them happier.

The formula to construct the confidence interval for the population proportion of women who said that chocolate made them happier is written below:

\(CI = \hat p \pm E\)

Where,

\(\hat p\)be the sample proportion of women who said that chocolate made them happier;

Eis the margin of error and has the given formula:

\(E = {z_{\frac{\alpha }{2}}}\sqrt {\frac{{\hat p\hat q}}{n}} \)

Here,

\[\begin{aligned}{c}\hat p = 85\% \\ = 0.85\end{aligned}\]

And,

\(\begin{aligned}{c}\hat q = 1 - \hat p\\ = 1 - 0.85\\ = 0.15\end{aligned}\)

The sample size,n is equal to 1708.

The confidence level is given to be equal to 99%.

Thus, the level of significance is computed as shown:

\(\begin{aligned}{c}{\rm{Confidence}}\;{\rm{Level}} = 100\left( {1 - \alpha } \right)\\100\left( {1 - \alpha } \right) = 99\\1 - \alpha = 0.99\\\alpha = 0.01\end{aligned}\)

From the standard normal table, the two-tailed critical value of z at 0.01 level of significance is equal to 2.5758.

The 99% confidence interval is computed as follows:

\(\begin{aligned}{c}CI = \hat p \pm E\\ = 0.85 \pm {z_{\frac{{0.01}}{2}}}\sqrt {\frac{{0.85\left( {0.15} \right)}}{{1708}}} \\ = 0.85 \pm \left( {2.5758} \right)\left( {0.0086} \right)\\ = \left( {0.828,0.872} \right)\end{aligned}\)

Therefore, the 99% confidence interval for the percentage of women who feel chocolate makes them happier is equal to (82.8%,87.2%).

We are 99% of the time confident that the actual percentage of women who feel chocolate makes them happier will lie between 82.8% and 87.2%.