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Parameters

It is important to see the distinction between a probability model and a sample drawn from it. In this course, for the most part, we will be dealing with a sample. But this sample is generated by a probability model. In this section, we will discuss this for a very simple model. This also motivates parameters  which are characteristics of the probability model.

Here is the probability model. Consider the fair spinner with the numbers 1, 2, and 3 on it. Let X denote the number spun. Then the probability model for X is

Range           1       2       3
Probabilities  1/3     1/3     1/3

In practice we won't know the probability model for X. In this case, we won't know if the spinner is fair or not. But in practice we can take a sample from the probability model. Based on the sample, perhaps we can say something about the probability model. Now it is very important that the sample is a random sample.

A sample is a random sample  if:

1.
The items in the sample are drawn independently of one another.
2.
Conditions do not change as the sample is drawn.

In this case, the spins of the spinner are independent of one another and we are not changing the chances of a 1, 2 or 3 from spin to spin.

Here is a Sample drawn from the probability model. Suppose we decide on a sample size of 100. The following is a sample of the probability model; i.e., I spun a fair spinner 100 times: So here are the results of my random sample:

    2    2    1    1    2    2    1    2    1    1    1    1    1    1    1
    1    2    3    1    3    1    3    1    1    2    2    3    3    2    1
    2    3    2    3    3    1    2    1    2    1    3    2    1    3    3
    1    2    2    2    1    3    3    1    1    1    1    3    1    1    1
    1    2    1    3    2    2    1    2    2    2    3    2    3    2    3
    3    3    3    1    1    1    1    2    2    1    1    3    2    3    2
    1    1    3    2    1    3    3    1    1    2

We of course tally up the sample and get the sample distribution:

Range                 1       2       3
Frequency            43      31      26
Relative Frequency  .43     .31     .26

This sample distribution is an estimate  of the probability model for X. That is, .43 is our estimate of p(1), .31 is our estimate of p(2), and .26 is our estimate of p(3). The histogram of the sample

 Each dot represents 3 points
             .
             :
             :                        .
             :                        :                        .
             :                        :                        :
             :                        :                        :
             :                        :                        :
             :                        :                        :
           -------+---------+---------+---------+---------+-------
             1                        2                        3
is our estimate of the graph of the probability model
            1/3                      1/3                      1/3

             :                        :                        :
             :                        :                        :
             :                        :                        :
             :                        :                        :
             :                        :                        :
           -------+---------+---------+---------+---------+-------
             1                        2                        3

What's that you're thinking? It seems a little off if the spinner is fair? Be careful, you are starting to think statistically. You may be even thinking of a formal test statistic to see if such a sample could be generated by the fair spinner probability model. Hey, we'll get there soon!

Suppose you compute the sample mean of sample distribution. You get the value $\bar{x} = 183/100 = 1.83$. Now since the histogram is an estimate of the probability model, what is 1.83 an estimate of?

It's not hard to see. There are two ways to calculate $\bar{x}$, here. One way is to add up the 100 numbers and divide by the sample size 100. However, in adding up these numbers you have added 3 to itself 26 times. Hence a much easier way is to use the tallying and add as follows:

\begin{displaymath}\bar{x} = \frac{(1 \times 43) + (2 \times 31) + (3 \times 26)}{100}
\end{displaymath}


\begin{displaymath}\bar{x} = \bigg(1 \times \frac{43}{100} \bigg) + \bigg(2 \times \frac{31}{100} \bigg) + \bigg(3 \times \frac{26}{100} \bigg)
\end{displaymath}

By the last line it is easy to see what $\bar{x}$ is estimating. Again, .43 is our estimate of p(1), .31 is our estimate of p(2), and .26 is our estimate of p(3). Hence, $\bar{x}$ is estimating

\begin{displaymath}\mu = \big(1 \times p(1) \big) + \big(2 \times p(2) \big) + \big(3 \times p(3) \big)
\end{displaymath}

That's the Greek letter mu.  $\mu$ is called the mean  of the probability model. It is the center of gravity along the horizontal axis of the graph of the probability model. So in this example 1.83 is an estimate of $\mu$.

This estimate is the result of one sample! If I spin the spinner another 100 times, I am going to get a different estimate of $\mu$. If you spin it 100 times you are going to get a different estimate, too. In fact, if everyone in class spins the spinner 100 times we are going to get different estimates, (there may be a few ties because the probability model is discrete).

So the important thing to determine is: "How much does $\bar{x}$ miss $\mu$ by?" That's the way to think. Keep it up!

In the spinner example , if the spinner is fair then p(1) = p(2) = p(3) = 1/3 and $\mu = 2$. So $\bar{x} = 1.83$ is an estimate of $\mu = 2$, in this case. We missed by .17.

In practice, we will not know the population mean. But, hopefully, we will have a random sample. We will calculate $\bar{x}$. We will estimate with a degree of confidence, "How much does $\bar{x}$ miss $\mu$ by?"

In general, the probability model mean, $\mu$, is called a parameter  of the probability model. For a discrete random variable X, to determine the mean, as in the spinner problem, we simply cross multiply the range values by the associate probabilities and total it up; that is,

\begin{displaymath}\mu = \textrm{Sum}\{ x \times p(x) \}, \quad \textrm{ over {\it x} in the range of {\it X}}.
\end{displaymath}


Exercise 5.3.1  
1.
Let S denote the sum of two numbers spun on a fair spinner with the numbers 1, 2, and 3 on it. The range of S is 2, 3, 4, 5, 6. Determine the probability model mean of S.
2.
Repeat the last problem if the spinner is spun 3 times.
3.
Let X denote the number of aces in a 2 card hand drawn without replacement from a well shuffled standard deck 0f 52 cards. Then the range of X is {0, 1, 2}. Determine the probability model mean of X.
4.
Repeat the last problem under sampling with replacement.
5.
Let X denote the number of hearts in a 2 card hand drawn without replacement from a well shuffled standard deck 0f 52 cards. Then the range of X is {0, 1, 2}. Determine the probability model mean of X.
6.
In the urn problem (with the balls well mixed ) discussed above, let Z denote the number of red balls in the sample (without replacement) of size 3. Determine the probability model mean of Z.


next up previous contents index
Next: More Parameters Up: Discrete Populations (Probability Models) Previous: Discrete Populations (Probability Models)

2001-01-01