Polling a Population of 100 Jelly Beans

Description of Activity: A bowl filled with 100 Jelly Beans represents a population. The color of the candy represents the opinion on the Jelly Bean selected. A red Jelly Bean approves of George W. Bush's performance as president, while a black Jelly Bean disapproves of Bush's performance. Unknown to the students, the population is set in advance by the teacher so that it contains 60 red Jelly Beans (for a 60% approval rating) and 40 black Jelly Beans (for a 40% disapproval rating). The class samples the population in an attempt to draw conclusions about Bush's approval rating.

This activity can be accomplished in a large class setting.

Topics Covered: populations, samples, sampling with replacement, binomial random variable, calculating binomial probabilities, hypothesis testing, p-values

Time: 30 min with additional time for discussion

Materials needed:

A clear bowl or bag

Jelly Beans: 60 red, 40 black

Dry Erase Board & marker, or overhead projector & pen

An overhead of binomial probability table

Instructions for Activity:

1. Prior to class, create the population by putting 60 red Jelly Beans and 40 black Jelly Beans into a bowl or bag.

2. When class starts describe the population to the students. Tell them that each of the 100 Jelly Beans has an opinion about Bush's performance. A red Jelly Bean approves of Bush, while a black Jelly Bean disapproves of Bush. Do not tell the students how many of each type is contained in the population. Only tell them that it is our job to try to draw conclusions about he population.

3. Describe the hypothesis test of interest: An avid Republican, Bush-supporter, such as Carl Rove, claims that Bush's approval rating is as high as 70%, while a Democrat, such as Al Gore believes that it is much lower. Ask the students to describe the null and alternative hypotheses in statistical notation, namely Ho: p=.70 versus H1: p<.70.

4. Ask the students how we might test the hypothesis. Work the students toward a random sample of 10 people with replacement, in which we let X=the number of people in the sample who approve of Bush. Then, X will take on values 0,1, . . . 10.

5.Ask the students to argue that X is a binomial random variable:

a. The number sampled, n= 10, is fixed in advance.

b. There are two outcomes: approve or disapprove.

c. p = P(a Jelly Bean approves) is unknown, but assumed to be 0.70, under the null hypothesis. If we sample with replacement, we have independent trials and p is constant.

d. We let X = the number in the sample who approve of Bush.

6. Now, it is time to take the random sample. Ask a student to volunteer to record the data on the board. The student should create two columns, one labeled Approve (Red) and one labeled Disapprove (black). As the sample is selected, the student should tally the Jelly Beans' opinion under the appropriate column. Then, walk around the room with the bowl in hand, and ask 10 different students to randomly select a Jelly Bean from the population. Make sure that students cannot see inside the bowl. After each student selects a Jelly Bean, the color should be recorded, the opinion tallied, and the Jelly Bean returned to the bowl before the next student selects. Continue until all 10 Jelly Beans are selected.

7. Ask the students what values of X support the alternative hypothesis. Work the students towards "small values."

8. Once the random sample has been selected, count the number of approvers, 6 say. Ask the students to define the p-value. Work the students towards "how likely is that 6 (or fewer) Jelly Beans would approve of Bush if the true population of Jelly Beans in the population who approve of Bush is in fact 0.70?"

9. Calculate the p-value by looking up P(X<=6) on a binomial table with n=10 and p=0.70.

10. Argue that since it is unlikely that we would observe the sample we did if the null hypothesis were indeed true, we will reject the null hypothesis.