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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?

Ask by Chang Sullivan. in the United States
Jan 23,2025

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Answer

There is no significant difference between the sample proportion of blue M\&Ms (27%) and the claimed proportion (24%) at a 5% significance level. Therefore, the Mars company does not need to take corrective action.

Solution

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.

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To test the Mars company’s claim about the percentage of blue M&Ms, we can set up a hypothesis test. The null hypothesis (H0) states that the true proportion of blue M&Ms is 0.24, while the alternative hypothesis (H1) claims that the true proportion is different from 0.24. Using a one-sample proportion test with a significance level of 0.05, we can compute the test statistic and compare it to the critical value or use the p-value method to determine if we reject H0. Based on the given data, we found 27 blue M&Ms out of 100. We can compute the sample proportion (\(\hat{p} = 0.27\)), then use the formula for the test statistic, which is \( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\), where \( p_0 = 0.24 \) and \( n = 100 \). If the test statistic falls beyond the critical values (±1.96 for a two-tailed test at the 0.05 significance level), we reject the null hypothesis. Now, let’s crunch those numbers! If our p-value turns out to be less than 0.05, the evidence is strong enough to suggest that the true proportion of blue M&Ms isn’t 24%, and the Mars company might need to take corrective actions to align their distribution with their claims. Otherwise, if we fail to reject the null hypothesis, their claim stands. Don't you love how a simple candy test can teach you about statistics? Testing the odds for blue candy is like a sweet version of a mystery story, revealing unexpected twists elsewhere. If only candy could explain the meaning of life along with color ratios!

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