The candy company that makes M&M's claims that 10% of the M&M's it produces is green.Suppose that the candies are packaged at random in large bags of 200 M&M's.When we randomly pick a bag of M&M's,we may assume that this represents a simple random sample of size n = 200.Suppose we wish to test H₀: p = 0.10 versus H_a: p $\neq$ 0.10.Suppose that in the randomly selected bag of M&M's there are only 12 green M&M's.Calculate the value of the large-sample z statistic.If a 90% confidence interval were also calculated from the data,would it contain the value 0.10? A)Yes B)No C)This cannot be determined from the information given.

Question

Quizplus · Accepted Answer

The large-sample z statistic is calculated using the formula $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$, where $\hat{p}$ is the sample proportion, $p_0$ is the hypothesized population proportion, and $n$ is the sample size. Here, $\hat{p} = \frac{12}{200} = 0.06$, $p_0 = 0.10$, and $n = 200$. Plugging these values into the formula gives $z = \frac{0.06 - 0.10}{\sqrt{\frac{0.10(1-0.10)}{200}}} = \frac{-0.04}{\sqrt{\frac{0.09}{200}}} = \frac{-0.04}{0.0212} \approx -1.89$. This z-value indicates that the observed proportion of green M&M's (0.06) is significantly lower than the hypothesized proportion (0.10) at a 90% confidence level. Therefore, a 90% confidence interval calculated from this data would not contain the value 0.10, indicating that the true proportion of green M&M's is significantly different from 0.10.

The Candy Company That Makes M&M's Claims That 10% of the M&M's