Solved – Cochran’s sample size formula p variable meaning and intuition

finite-populationsample-sizesampling

I want to calculate a sample size for a large population of about 50 million. I came across Cochran's formula and the finite population correction.
In short, Cochran's formula is the following:

$$
n_\infty = \frac{z^2 p(1-p)}{e^2}
$$
I have found multiple resources that describe p as a sample proportion or as estimated proportion of an attribute that is present in the population. Most of the time, p is considered by default 0.5.

1) What is p and how can I understand it intuitively with an example?

2) Is p found exclusively via experiments? If not, what is its formula?

3) Most people use p = 0.5 because it gives the maximum sample size. Is this true?

Best Answer

1) What is p and how can I understand it intuitively with an example?
You answered your own question here. "p as a sample proportion or as estimated proportion of an attribute that is present in the population" If example if you want to estimate the percentage of women attending the University, that would be equal to p and 1-p is the percentage of men attending.

2) Is p found exclusively via experiments? If not, what is its formula?
Yes, p in context is a measured variable from the population.

3) Most people use p = 0.5 because it gives the maximum sample size. Is this true?
Yes is case assuming p=0.5 will provide the worst case assessment and thus require the largest sample size. Perform the calculation your self by comparing: 0.5*(1-0.5) versus 0.1*(1-0.1)

Related Solutions

Finite Population Correction Factor – Detailed Explanation and Use

The threshold is chosen such that it ensures convergence of the hypergeometric distribution ($\sqrt{\frac{N-n}{N-1}}$ is its SD), instead of a binomial distribution (for sampling with replacement), to a normal distribution (this is the Central Limit Theorem, see e.g., The Normal Curve, the Central Limit Theorem, and Markov's and Chebychev's Inequalities for Random Variables). In other words, when $n/N\leq 0.05$ (i.e., $n$ is not 'too large' compared to $N$), the FPC can safely be ignored; it is easy to see how the correction factor evolves with varying $n$ for a fixed $N$: with $N=10,000$, we have $\text{FPC}=.9995$ when $n=10$ while $\text{FPC}=.3162$ when $n=9,000$. When $N\to\infty$, the FPC approaches 1 and we are close to the situation of sampling with replacement (i.e., like with an infinite population).

To understand this results, a good starting point is to read some online tutorials on sampling theory where sampling is done without replacement (simple random sampling). This online tutorial on Nonparametric statistics has an illustration on computing the expectation and variance for a total.

You will notice that some authors use $N$ instead of $N-1$ in the denominator of the FPC; in fact, it depends on whether you work with the sample or population statistic: for the variance, it will be $N$ instead of $N-1$ if you are interested in $S^2$ rather than $\sigma^2$.

As for online references, I can suggest you

Solved – how to determine sample size

You can always take a larger sample.
The "general" rule of thumb is that you want at least 30 samples to be statistically reasonable, so yes a sample size of 18 is likely to be too small with a population of 800.
Your calculation is missing something as you should get a sample size around 89 for a population of 800 with 10% precision.

Yamane (1967) has a simplified formula for calculating sample size. A 95% confidence level and P = .05 are assumed.

$n=N/(1+N(e^2))$

n=800/(1+800(.10^2)) = 88.88

Best Answer

Related Solutions

Finite Population Correction Factor – Detailed Explanation and Use

Solved – how to determine sample size

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