Study

 Our Sampling Method:

In order to complete our study we needed a large sample of M&Ms so that we could be sure that there would be no influential outliers. The population of interest was all 1.69 Chocolate M&M bags manufactured before the proportions were last adjusted. We obtained 80 1.69 oz. bags of M&Ms that contained 4436 M&Ms in total from Gordan Food Service. We counted the individual M&Ms, keeping track of how many of each color were in each of the 80 bags. After counting all of the 4436 M&Ms, we were able to run the tests that compared proportions of each color.

Hypothesis Tests

Is the true proportion of blue M&Ms in a bag greater than the true proportion of orange M&Ms in a bag?

1. π₁ = the true proportion of blue M&Ms per bag

     π₂ = the true proportion of orange M&Ms per bag

2. H₀: π₁ = π₂

3. Ha: π₁ > π₂

4. α = 0.05

5. Assumptions: Samples must be random and independently selected.

    np₁ = 4436(744/4436) = 744 > 10         n(1 - p₁) = 4436(3692/4436) = 3692 > 10

    np₂ = 4436(747/4436) = 747 > 10         n(1 - p₂) = 4436(3689/4436) = 3689 > 10

    Since all assumptions are met, a two proportion z-test is appropriate.

6.

7. z = -0.09

8. Since this is a upper-tailed test, p-value is the area under the curve to the right of z = -0.09, p-value = 0.543.

 

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Is the true proportion of red M&Ms in a bag less than the true proportion of blue M&Ms in a bag?

 

1. π₁ = the true proportion of red M&Ms per bag

     π₂ = the true proportion of blue M&Ms per bag

2. H₀: π₁ = π₂

3. Ha: π₁ < π₂

4. α = 0.05

5. Assumptions: Samples must be random and independently selected.

    np₁ = 4436(534/4436) = 534 > 10       n(1 - p₁) = 4436(3902/4436) = 3902 > 10

    np₂ = 4436(744/4436) = 744 > 10       n(1 - p₂) = 4436(3692/4436) = 3692 > 10

    Since all assumptions are met, a two proportion z-test is appropriate.

6.

 

7. z = -6.36

8. Since this is a lower-tailed test, p-value is the area under the curve to the left of z = -6.36, p-value = 0.000.

 

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Is the true proportion of brown M&Ms in a bag different than the true proportion of green M&Ms in a bag?

 

 1. π₁ = the true proportion of green M&Ms per bag

      π₂ = the true proportion of brown M&Ms per bag

2. H₀: π₁ = π₂

3. Ha: π₁ ≠ π₂

4. α = 0.05

5. Assumptions: Samples must be random and independently selected.

     np₁ = 4436(984/4436) = 984 > 10           n(1 - p₁) = 4436(3452/4436) = 3452 > 10

     np₂ = 4436(712/4436) = 712 > 10           n(1 - p₂) = 4436(3724/4436) = 3724 > 10

     Since all assumptions are met, a two proportion z-test is appropriate.

6.

7. z = -7.37

8. Since this is a two-tailed test, p-value is the area under the curve to the left of z = -7.37 and to the right of z = 7.37, p-value = 0.000.

 

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Do the proportions reported by the M&M company differ from the sample proportions per bag?

 

1. π₁-π₆ = the true proportion of M&Ms per bag of brown, yellow, orange, red, green, and blue respectively.

2. H0: π₁= 0.13, π₂ = 0.14, π₃ = 0.20, π₄ = 0.13, π₅ = 0.16, π₆ = 0.24

3. Ha: At least one differs.

4. α = 0.05

5. Assumption: Samples are randomly selected and all expected values are > 5.

 

Expected Brown	Expected Yellow	Expected Orange	Expected Red	Expected Green	Expected Blue
576.68	621.04	887.20	576.68	709.76	1064.64

Since the assumptions are met, a chi-square test is appropriate.

 

6.

7. X² = 273.183

8. The p-value = 0.000 with 5 degrees of freedom.