Question

Question 6) Company claims that more than 19% skittles are green. Do you agree with this claim? Use α=0.05.

48 students participated in evaluating small package of skittles. The following table shows the result of this study. Assume the weight of skittles are normally distributed with standard deviation of 2 grams. Net Weight Green Red Yellow Brown Orange Count 2923 Total 2978.4 576 589 641 531 586 Sample Mean 62.05

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Answer 1

$\hat{p}$ = 576 / 2923 = 0.197

p = 0.19

1 - p = 0.81

The Hypothesis:

H0: p $\leq$ 0.19

Ha: p > 0.19

This is a Right tailed Test.

The Test Statistic:

Z = \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}} = \frac{0.197-0.19}{\sqrt{\frac{0.19*0.81}{2923}}} = 0.96

The p Value:    The p value (Right tail) for Z = 0.96, is; p value = 0.1685

The Critical Value:   The critical value (Right tail) at $\alpha$ = 0.05, Zcritical = +1.96

The Decision Rule: If Zobserved is > Zcritical Then Reject H0.

Also If the P value is < $\alpha$ , Then Reject H0

The Decision:    Since Z observed (0.96) is < Zcritical (1.96), We Fail to Reject H0.

Also since P value (0.1685) is > $\alpha$ (0.05), We Fail to Reject H0.

The Conclusion: There isn't sufficient evidence at the 95% significance level to conclude that the the number of green skittles is greater than 19%.

Therefore cannot agree with the company claim.