Question

Background

You will need to purchase a regular-sized bag of milk chocolate M&M’s for this project.

Let’s work to answer a very important question, vexing humankind for decades: Are the colors evenly distributed in a bag of M&M's????   Is the company reporting accurate color percentages for their product? This activity will guide you through constructing a confidence interval for the proportion of M&M's in a particular color.

M&M says:

On average, our mix of colors for M&M'S CHOCOLATE CANDIES is:

M&M'S MILK CHOCOLATE: 24% cyan blue, 20% orange, 16% green, 14% bright yellow, 13% red, 13% brown.

Each large production batch is blended to those ratios and mixed thoroughly. However, since the individual packages are filled by weight on high-speed equipment, and not by count, it is possible to have an unusual color distribution.

Part 1: Confidence Interval for Single Bag (20 points)

Milk Chocolate M&M’s come in 6 colors; blue, orange, green, yellow, red, and brown.

1. Choose your favorite color of M&M’s you will be working with for this project. State the color and give the counts below.

Color of choice:

Number of M&M's in your color:

Total number of M&M's:

Proportion of M&M's in your color:

2. Construct a 95% confidence interval for the proportion of M&M’s one can expect to find in the color of your choice.

3. Give an interpretation of your interval.

4. Find the margin of error for your interval.

5. Check the requirements for constructing a confidence interval for the proportion are satisfied. Show your work. (See the note in the gray/blue box on page 426)

6. The conditions might not be satisfied, depending on how many candies were in your bag. If the conditions are not met, what could you do? (Note—you don’t have to redo the experiment to meet the conditions, just say what you would do.)

7. Does your interval contain the stated proportion by M&Ms for the color you chose?

8. If you had a larger sample, how would you expect the confidence interval to be affected? Provide justification.

Answer 1

Color of choice: Green

Number of M&M's in your color:

Total number of M&M's:100

Proportion of M&M's in your color: 0.12

Color	Observed	Expected
Cyan blue	24	24
Orange	15	20
Green	12	16
Bright yellow	14	14
Red	19	13
Brown	16	13

(a)

n = 100   

p = 0.12    

% = 95    

Standard Error, SE = √{p(1 - p)/n} =    √(0.12(1 - 0.12))/100 = 0.032496154

z- score = 1.959963985    

Width of the confidence interval = z * SE =     1.95996398454005 * 0.0324961536185438 = 0.06369129

Lower Limit of the confidence interval = P - width =     0.12 - 0.0636912907284269 = 0.05630871

Upper Limit of the confidence interval = P + width =     0.12 + 0.0636912907284269 = 0.18369129

The confidence interval is [0.0563, 0.1837]

(b)

If samples of size   100 are repeatedly drawn from the population and the confidence intervals constructed such intervals will contain the true proportion 95% of the time.

(c) Margin of error = 0.0637

(d) Conditions to be met are

(i) Conditions of the central limit theorem must be met in order to use the normal model

(ii) The data must be sampled randomly

(iii) The sample values must be independent of each other.

(iv) The sample size must be sufficiently large

All these conditions are met in the present case

(e) Yes, the above confidence interval contains 0.12

(f) If we had a larger sample, the width of the confidence interval would be reduced. For example, if we had a sample size of 200 instead of 100, the margin of error would become 0.045, and confidence interval would become [0.075, 0.165]