# Sample size in a population of 300

I have to conduct a survey in the discipline of International relations. I have a list of 300 diplomats to interview for close ended, open ended questionnaire. I want to select randomly every fifth from the list and the number of sample comes to 60. Is it a correct same size from a population of 300? would be grateful for your answer please.

You would need at least 169 samples (with 95% confidence and with 5%- Margin of error) which is a most common requirement.

With your current sample size you are having 11.3% margin of error (with 95% confidence). Use this on-line sample size calculator to calculate sample. Also please see the table given below for your quick reference.

Hope this helps!

The prior answers regarding sample size are good ones. I want to address the issue of using n-th select to choose your sample. In a famous example from the 50s Paul Lazarsfeld, the great Columbia sociologist, did a huge study of The American Soldier where he selected every 10th name from a list provided by the Army. Using this criteria, what he got back was a sample of non-commissioned officers (NCOs) since the basic unit in the Army was the platoon containing 9 men plus an NCO as the 10th. Luckily, Lazarsfeld discovered the error before the study was done and was able to correct it in time.

The point is that in using n-th select, unless your list of diplomats is randomized, any structure, organization, sorted order, etc., of any kind that exists in the original list will result in a nonrandom sample.

• Grateful to all for help. Hope to do better in my survey. Thanks Commented Oct 15, 2015 at 16:07

It will depend on the bound (margin of error) that you want. Here's the general sample size formula for a proportion:

$$n = \frac{Np(1-p)}{(N-1)\frac{B^2}{4} + p(1-p)}$$ where p is a guess at the true proportion, B is the desired bound, N population size. If you don't know p, the worst case is $p = \frac{1}{2}$, so with your N = 300, we get the formula $$n = \frac{300}{(299)B^2 + 1}$$

For the case you stated (n = 60), the bound you would expect (at worst) is $$B = 2\sqrt{\frac{300-60}{300}\times\frac{1}{4\times60}} = 0.11$$ So a sample of size n = 60 would give a margin of error of around plus or minus 11 percent.

This assumes the worse case where the proportion is really around 1/2. If it is close to zero or one you will get a better bound.