# Thread: Derivation of Margin of Error term in a normal distribution

1. ## Derivation of Margin of Error term in a normal distribution

I'm trying to understand the derivation of the formula:
$E = z_{\frac{\alpha}{2}}\sqrt{\frac{\hat{p}\hat{q}}{n} }$

The definition given in the text is: When data from a simple random sample are used to estimate a population proportion $p$, the Margin of Error, denoted by $E$, is the maximum likely difference (with probability $1 - \alpha$, such as 0.95) between the observed sample proportion $\hat{p}$ and the true value of the population proportion $p$. The margin of error $E$ is found by multiplying the critical value and the standard deviation of sample proportions, as shown in the above formula.

So here's my question: Where did $\sqrt{\frac{\hat{p}\hat{q}}{n}}$ come from?

In an earlier chapter it was stated that the standard deviation $\sigma = \sqrt{npq}$. And it's obvious that $z_{\frac{\alpha}{2}}$ is the critical value. So, why isn't $E = z_{\frac{\alpha}{2}}\sqrt{npq}$?

There is no other explanation given of why $\sqrt{\frac{\hat{p}\hat{q}}{n}}$ is a "standard deviation of a sample proportion".

Any help would be appreciated.

2. ## Re: Derivation of Margin of Error term in a normal distribution

Hey B9766.

You might want to figure out how the variance and standard deviation are related to your margin of error.

If you find E[X{P-p_hat}^2] then you should get something close to the variance and when you take the square root you should get the standard deviation.

Remember - You are looking at the difference between the observed proportion P [which is a random variable] and the estimated proportion [p_hat] and finding its expected value after you square this difference to get the variance.

Have you covered variance before? Do you know what standard deviations look at?

3. ## Re: Derivation of Margin of Error term in a normal distribution

Thanks Chiro,

Yes, I think I have a good grasp on the concepts and calculations involved in variance and standard deviation. Variance is the easy one because it simply establishes the average difference between the mean and all sample values. So,

$\sigma^2 = \frac{\sum{(x - \mu)^2}}{N}$

Although I think it more contrived than anything, the standard deviation is a somewhat arbitrary measure along the x-axis, equal to the square root of the variance. But the term in question has nothing to do with either of these concepts.

As I see it, the equation $\sigma = \sqrt{npq}$ is an approximation of population standard deviation based on a normal distribution. It's clear that the larger n becomes, the closer $\sigma$ comes to the true population parameter - basically an example of the Law of Large Numbers.

It's not clear to me why pq causes this approximation. For any proportion, p, the product of that proportion and it's complement somehow creates an approximation of the variance, as in:

$\sigma^2 = (n)(p)(1-p) = npq$

But even if you can explain that, it's entirely unclear to me how $\sigma^2 = \frac{\hat{p}\hat{q}}{n}$ as used in the Margin of Error calculation. The implication is that $\frac{\sum{(x - \mu)^2}}{N} = \frac{\hat{p}\hat{q}}{n}$ How can this be?