Answer
There is not sufficient evidence to reject that 16% of plain M&Ms candies are green.
Work Step by Step
$H_{0}:p=16$%=0.16. $H_{a}:p\ne0.16.$ $\hat{p}$ is the number of objects with a specified value divided by the sample size. Hence $\hat{p}=\frac{x}{n}=\frac{19}{100}=0.19.$ The test statistic is:$z=\frac{\hat{p}-p}{\sqrt{p(1-p)/n}}=\frac{0.19-0.16}{\sqrt{0.16(1-0.16)/100}}=0.82.$ The P is the probability of the z-score being more than 0.82 or less than -0.82 is the sum of the probability of the z-score being less than -0.82 plus 1 minus the probability of the z-score being less than 0.82, hence:P=0.2061+1-0.7939=0.4122. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P=0.4122 is more than $\alpha=0.1$, hence we fail to reject the null hypothesis. Hence we can say that there is not sufficient evidence to reject that 16% of plain M&Ms candies are green.
