Chance

October 20, 2000

Let

$$X$$

be a random quantity determined by some experiment with expected value denoted

$$\mu$$

and standard deviation given by

$$\sigma.$$

For today, c will denote a positive number. In lecture we explored (and in X-session proved) the following fact:

FACT 1: The probability that X is within c units of its expected value is greater than or equal to

$$1 - \frac{\sigma^2}{c^2}.$$

The Fundamental Mysteries of Probability are that

$$Exp(Y+Z) = Exp(Y)+Exp(Z)$$

and that if $Y$ and $Z$ are not dependent on each other, then

$$(Sd(Y+Z))^2=(Sd(Y))^2+(Sd(Z))^2.$$

These mysteries, the above fact, and the less mysterious facts that

$$Exp(cY) =c Exp(Y)$$

and

$$Sd(cY) = \vert c\vert Sd(X)$$

give us the following fact:

FACT 2: Assuming each

$$X_i$$

is an independent run of the X experiment, then the

$$Prob\left( \mu - c < \frac{X_1 + \cdots + X_n}{n} < \mu +c \right) \geq 1 - \frac{\sigma^2}{n c^2}.$$

We also will explore today the following fact:

FACT 3: X is a heads or tails experiment, then

$$Prob\left( \mu - c < \frac{X_1 + \cdots + X_n}{n} < \mu +c \right) \geq 1 - \frac{1}{4 n c^2}.$$

In class together, we will answer:

1. If each $X_i$ is one with probability $p = 0.2$ and zero otherwise, how many trials should we perform so that we can say with 95 percent confidence that the average
   
   $$\frac{X_1 + \cdots + X_n}{n}$$
   
   
   is within 5 percent of $p$.

2. If each $X_i$ is one with some unknown probability $p$ and zero otherwise, how many trials should we perform so that we can say with 95 percent confidence that the average
   
   $$\frac{X_1 + \cdots + X_n}{n}$$
   
   
   is within 5 percent of $p$.

In groups discuss:

1. Discuss how to translate the second and third facts into English. What are they telling you?

Consider Gallup's poll from October 17 (see handout). To make the poll a binomial experiment, Gallup considers each outcome (i.e. Bush more likeable, Gore more likeable, no opinion, etc.) as its own experiment. For example, to calculate the percentage of people who see Gore as more likeable, Gallup uses a binomial distribution where the two outcomes are ``Gore is seen as more likeable'' and ``Gore is not seen as more likeable''.

2.

Using the above facts, with what confidence should the Gallup poll have reported that 46 percent of people, with $\pm5$ error, think that Gore is more likeable than Bush?

3.

Suppose that Gallup's claim of 95 percent confidence with $\pm5$ percent error can be justified. Notice that the difference between Bush's percentage and Gore's percentage is two. With what percent confidence and how much error should we report concerning this difference?

4.

Given your analysis in question 3, what aspect of Gallup's commentary is particularly misleading?

Here are your homework and Journal question for this weekend.

1. (Homework) Explain why the second fact justifies the terminology ``expected value''.

2. (Homework) If each $Y_i$ is the value we see when rolling a six-sided die one time, how many trials should we perform so that the average
   
   $$\frac{Y_1 + \cdots + Y_n}{n}$$
   
   
   is within 8 percent of the expected value of $Y_i$, with 90 percent confidence.

3. (Journal) Articulate the issue arising in question 3, above, and find another example of an actual poll (at Gallup or somewhere else) where a misleading conclusion is presented due to this confusion.