Question

2. (25 points) According to the manufacturer of M&M candies, a bag of M&Ms contains 465 M&Ms with each M&M weighing 0.8535 grams. In order for the bag to contain the net total weight of 396.9 grams, the manufacturer must guarantee that each M&M's mean weight is at least 0.8535 grams. Assume the standard deviation is 0.0576. A random sample of 100 M&Ms were selected from a bag of 465 M&Ms and the mean value of the sample=0.8635. At the 0.05 level of significance, can the production manager claim that the M&Ms have a mean that is actually greater than 0.8535 so consumers are being given more than the amount indicated on the label? A. State H: B. State Ha: C. Is this a one- or two-tail test? Why? D. What is(are) the critical value(s)? E. What is the decision rule? F. Calculate the statistic: G. What is your decision regarding H, and why? H. Interpretation of the results:

Answer 1

Below are the null and alternative Hypothesis,
A. Null Hypothesis, H0: μ = 0.8535
B. Alternative Hypothesis, Ha: μ > 0.8535

C. This is right tailed test, for α = 0.05 and df = 99

D. Critical value of t is 1.66.

E. Hence reject H0 if t > 1.66

F.
Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (0.8635 - 0.8535)/(0.0576/sqrt(100))
t = 1.736

G. Reject H0 as test statistic greater than critical value

H. There is sufficient evidence to conclude that the mean is actually greater than 0.8535