Solved – Calculate sample size from margin of error and confidence level

sample-size

I need to calculate the sample size for the data I have. I am given a margin of error5% with a confidence level 95% .

Which formula can I use to get the sample size value if the total population size is X?
Sorry, I searched the internet but couldn't find a formula for this purpose.

https://www.statisticshowto.com/probability-and-statistics/find-sample-size/#Cochran

Any help is appreciated.

Best Answer

If you are estimating a binomial proportion near to $p\approx 1/2$ based on a sample of size $n,$ then a 95% CI is of the form $\hat p \pm 1.96\sqrt{\hat p(1-\hat p)/n}.$ for which the margin of error is $$M = 1.96\sqrt{\hat p(1-\hat p)/n} \approx 1.96\sqrt{(1/2)(1/2)/n}\\ = 1.96\sqrt{1/4n} \approx 1/\sqrt{n}.$$

So, if $M = 0.05 = 5\%.$ then $n \approx 1/M^2 = 1/(.05)^2 = 400.$

You do not give many details about the parameter being estimated or the kind of confidence being used. My computations above are based on estimating binomial success probability $p \approx 1/2,$ as in some election polling situations. This is the only elementary application I know about where your information is sufficient.

If this is not the application you have in mind, then you will need to give more information: for example, the approximate value of $p$ for a binomial CI or an estimate of population variance $\sigma^2$ if you are trying to estimate the mean $\mu$ of a normal population. Other applications of confidence intervals may require various additional information.

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Solved – Confidence intervals vs sample size

In addition to Peter's great answer, here are some answers to your specific questions:

Who to trust will depend also on who is doing the poll and what effort they put into getting a good quality poll. A bigger sample size is not better if the sample is not representative, taking a huge poll, but only in one, non-swing state would not give very good results.

There is a relationship between sample size and the width of the confidence interval, but other things also influence the width, such as how close the percentage is to 0, 1, or 0.5; what bias adjustments were used, how the sample was taken (clustering, stratification, etc.). The general rule is that the width of the confidence interval will be proportional to $\frac{1}{\sqrt{n}}$, so to halve the interval you need 4 times the sample size.
If you know enough about how the sample was collected and what formula was used to compute the interval then you could solve for the standard deviation (you also need to know the confidence level being used, usually 0.05). But the formula is different for stratified vs. cluster samples. Also most polls look at percentages, so would use the binomial distribution.
There are ways to combine the information, but you would generally need to know something about how the samples were collected, or be willing to make some form of assumptions about how the intervals were constructed. A Bayesian approach is one way.

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

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