Using Appendix B Data Sets. In Exercises 29 and 30, use the indicated data set in Appendix B. Green M&Ms Data Set 27 “M&M Weights” in Appendix B includes data from 100 M&M plain candies, and 19 of them are green. The Mars candy company claims that 16% of its M&M plain candies are green. Use the sample data to construct a 95% confidence interval estimate of the percentage of green M&Ms. What do you conclude about the claim of 16%?

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:

$E=z_{\frac{\alpha }{2}}\times \sqrt{\frac{\hat{p}\hat{q}}{n}}$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

$z_{\frac{\alpha }{2}}$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 $z_{\frac{\alpha }{2}}$ for $\alpha =0.05$ is equal to 1.96.

The margin of error can be computed as follows:

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

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.