See if the arithmetic mean number of greens is less than
58/6
A Better Way to Proceed:
Work with a sample (i.e., a part of the population)
and make inferences about the population
A Limitation:
We will have to make some kind of probabilistic
statement about the population
Making Probabilistic Statements about the Population
What We Need:
A probability density function
Thus Far:
We have assigned probabilities to sample points
and used them to find the probability density function
Now:
We want to make a probabilistic statement about a sample mean
To do so, we need the distribution of sample means
(i.e., we need the probability that the mean number of
green M&Ms in any sample is, say 2, or any other
number)
So, we have to calculate this distribution
The Central Limit Theorem
Definition of the (Arithmetic) Sample Mean:
\(\overline{x} = \frac{x_1+x_2+\cdots+x_n}{n}\)
An Observation:
Each time we take a sample it will have a different
mean
The random variables do not have to be normally
distributed -- as long as the random variables are
independent and all have the same distribution, the
distribution of sample means will tend to be normal
Back to the Simple Example
A Single Simulation:
To simulate the purchase of 10 of our small bags of
M&Ms I rolled a 16-sided die 10 times
To See the Value of the Central Limit Theorem:
My first simulated purchase of 10 bags had a mean
of 1.2 greens
My second simulated purchase of 10 bags had a mean
of 1.5 greens
I did this 998 more times, calculating the mean each time
Back to the Simple Example (cont.)
The Frequencies
Back to the Simple Example (cont.)
The Relative Frequencies are "Bell Shaped"
The Distribution of Sample Means
The Mean of this Distribution (\(\mu_{\overline{x}}\)):
\(\mu_{\overline{x}} = \mu\)
The Standard Deviation of this Distribution (\(\sigma_{\overline{x}}\)):
There are 9.667 green M&Ms in a bag (i.e., the population
mean is 9.667)
Testing the Hypothesis at the 0.95 Level:
\(n-1=19\) implies that the critical \(t\)
values is 1.729
This means that the probability is less than 0.05 that
a sample of 20 bags will have a mean number of greens
less than \(9.667 - 0.593 \cdot 1.729 = 8.642\)
Since our actual sample mean is 7.2 we can reject the hypothesis