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In a sample of 100 candies, 19 are green. It is claimed that 16% of the candies are green.

The confidence interval for the population proportion is shown below:

$\hat{p}-E<p<\hat{p}+E$

Here, E is the margin of error. It has the following formula:

$E=z_{\frac{伪}{2}}脳\sqrt{\frac{\hat{p}\hat{q}}{n}}$where

$\hat{p}$ is the sample proportion of adults who have Facebook pages

$\hat{q}$is the sample proportion of adults who do not have Facebook pages

n is the sample size

$z_{\frac{伪}{2}}$is the one-tailed critical value of z.

The sample size (n) is equal to 100.

The sample proportion of green candiesis equal to:

$$\begin{gathered} \hat{p}=\frac{19}{100} \\ =0.19 \end{gathered}$$

The sample proportion of green candies is equal to:

$$\begin{gathered} \hat{q}=1-\hat{p} \\ =1-0.19 \\ =0.81 \end{gathered}$$

The value of $z_{\frac{伪}{2}}$ for $伪=0.05$ is equal to 1.96.

The margin of error can be computed as follows:

$$\begin{gathered} E=z_{\frac{伪}{2}}脳\sqrt{\frac{\hat{p}\hat{q}}{n}} \\ =1.96脳\sqrt{\frac{0.19脳0.81}{100}} \\ =0.0769 \end{gathered}$$

The confidence interval estimate of the population proportion of green candies is computed as follows:

$$\begin{gathered} \hat{p}-E<p<\hat{p}+E \\ 0.19-0.0769<p<0.19+0.0769 \\ 0.113<p<0.267 \\ 11.3\%<p<26.7\% \end{gathered}$$

The 95% confidence interval is equal to (11.3%, 26.7%).

Since the interval contains the value of 16%, it can be said the claim of 16% green candies is supported.