12. M\&Ms Data Set 27 "M\&M Weights" in Appendix B lists data from \( 100 \mathrm{M} \& M s \), and \( 27 \% \) of them are blue. The Mars candy company claims that the percentage of blue M\&Ms is equal to \( 24 \% \). Use a 0.05 significance level to test that claim. Should the Mars company take correc- tive action?

To test the claim that the percentage of blue M\&Ms is equal to 24%, we can use a hypothesis testing approach. Let's denote: - \( p \) as the true proportion of blue M\&Ms in the population. - \( \hat{p} \) as the sample proportion of blue M\&Ms. Given: - Sample size, \( n = 100 \) - Sample proportion of blue M\&Ms, \( \hat{p} = 0.27 \) (27% of M\&Ms are blue) - Claimed proportion of blue M\&Ms, \( p = 0.24 \) (24% of M\&Ms are blue) We want to test the null hypothesis: \[ H_0: p = 0.24 \] Against the alternative hypothesis: \[ H_1: p \neq 0.24 \] The significance level is given as \( \alpha = 0.05 \). To perform the hypothesis test, we can use the z-test for proportions. The formula for the z-test statistic is: \[ z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \] Let's calculate the z-test statistic and determine the critical region to decide whether the Mars company should take corrective action. The z-test statistic is approximately 0.702439. Next, we need to determine the critical region for the z-test. Since we are testing a two-tailed hypothesis, the critical region is the area outside of the z-scores corresponding to the 0.025 probability in each tail. Using a standard normal distribution table or calculator, we find that the critical z-scores for a two-tailed test at a 0.05 significance level are approximately -1.96 and 1.96. Since the calculated z-test statistic (0.702439) falls within the range of -1.96 to 1.96, we fail to reject the null hypothesis. This means that there is insufficient evidence to conclude that the percentage of blue M\&Ms is not equal to 24%. Therefore, the Mars company does not need to take corrective action based on this hypothesis test.