Normal Distribution

As the number of individuals in each class increases and p=q=1/2, the binomial distribution approaches a normal distribution. When the number of individuals in an experiment is large enough, the normal distribution can be used to provide confidence intervals for the observed frequency of a category. The normal distribution is used as an approximation to the binomial distribution when pqn is greater than 25.

1.96 standard deviations includes 95% of the samples of size n. This means that if we repeat an experiment 100 times, in 95 of the experiments the true mean will be included in the confidence interval which is

C.I. = 1.96 (s.d.); s.d. is standard deviation of a proportion

The standard deviation of a proportion is

The confidence interval is a measure of the precision of an unknown parameter. If we repeated an experiment an infinite number of times, under all sorts of environmental conditions, this would be the true mean. The sample mean is based on finite data and the experiments are evaluated under a limited number of environmental conditions.

Example:

In a cross of Bb x Bb genotypes, 70 bb plants were observed out of 200 progeny observed. Find the sample relative frequency and confidence interval.

We can use the confidence interval approach

Our hypothesized value of u = 0.25 falls outside the confidence interval. We have 95% confidence that the true proportion of bb plants is between 0.28 and 0.42.

The relative frequency =  observed number in class 
The relative frequency = total number in experiment

The relative frequency = 70/200 = 0.35 =

The standard deviation of a proportion = .
We use the and to determine the s.e.

The confidence interval is approximated by using the normal distribution. Let a = 0.05. We estimate the standard error using p and q.
C.I. =   + 1.96(s.d.)
C.I. = X + 1.96(s.d.)
C.I. = 0.35 + 1.96(0.034)
C.I. = 0.35 + 0.07 = X + s.e.

This means that we have 95% confidence that the true proportion of bb plants is between 0.28 and 0.42. This interval does not include 0.25, so we do not have evidence to support Ho:3:1 ratio. Another way to look at the problem is using a "t" test.

tcal =	- u =	0.35-0.25 =	2.94
		0.034

The critical value for a = 0.05 is t_tab=1.96.

Because t_cal > t_tab, we reject Ho.

We can use the confidence interval approach.

Our hypothesised value of u = 0.25 falls outside the confidence interval. We have 95% confidence that the true proportion of bb plants is between 0.28 and 0.42.