Hypothesis Testing Part I

Hypothesis Testing Part II

Hypothesis Testing Part III

Binomial Distribution Part I

Binomial Distribution Part II

Binomial Distribution Part III

Binomial Distribution Part IV

Hypothesis Testing Using Binomial Distribution Part I

Hypothesis Testing Using Binomial Distribution Part II

Hypothesis Testing Using Binomial Distribution Part III

An Explanation of Binomial Distribution Part I

An Explanation of Binomial Distribution Part II

Another Example Of Hypothesis Testing With Binomial Distribution

Homework Assignment #1 Questions

Homework Assignment #1 Answers

Binomial Distribution Part III

Example

For a family of size 8 we want to find the coefficient for families with seven ‘successes’ and one ‘failure’. Let Aa be a ‘success’ and aa be a ‘failure’.
Then n = 8, r = 7, n-r = 1.
Now:

8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320
7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040
1! = 1

To find the probability of seven Aa and one aa individuals in family of size eight, ignoring the order of progeny:

Let:
x = probability of ‘success’ for a single event;
y = probability of ‘failure’ for a single event.

In our example of Aa x aa 1/2 Aa: 1/2 aa, then x = 1/2 and y = 1/2.

Copyright 2000©, Ted Helms

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