# Computing number of people with the same name and date of birth based on sample data

A country has a population of 40 000 000 people.

I have a sample with data for 1 500 000 people from this country.

In this sample 1.9% of people have a "pair" (one or more) with the same first name, last name and date of birth (but who is not the same person).

Can I use this information to compute how many people in the whole population have a "pair" with the same name and date of birth?

• Assuming a random sample, it's 760000 people plus/minus 1.96*the standard error (for 95% confidence) Jun 7, 2018 at 3:21
• @Yannis How would you compute the standard error?
– whuber
Jun 7, 2018 at 13:21
• 1.9% sharing the same full name AND date of birth? Geez, people in this country have no imagination whatsoever. Mar 27, 2020 at 8:13

Your sample proportion of having a "pair" is $\hat{p} = 0.019$, which is the best estimate for the population proportion (assuming random sampling). So we set $\hat{p} = p$, and therefore the number of people in the population that have a "pair" is $40,000,000 \times 0.019 = 760,000$.
Now assuming independent observations (your sample is about 3.75% of the population, so OK) and a sufficient number of both "failures" and "successes" (both have more than 10 observations, so OK), then: $$\hat{p} \sim N(p, \frac{p(1-p)}{N})$$ The standard error is defined as the standard deviation of the sampling distribution of $\hat{p}$, so in this case: $$SE = \sqrt{\frac{p(1-p)}{N}} = \sqrt{\frac{0.019(1-0.019)}{1,500,000}} = 0.0001$$ Note that this is so small, because N is so large! Now, in order to construct a 95% confidence interval: $$\hat{p} \pm 1.96\times SE = 0.019 \pm 0.000196 = [0.018804, 0.019196]$$ Translating this back to the population, the "true" number of people in the population that have a "pair" is in the interval $[752160, 767840]$. (A lot more goes into the interpretation of the confidence interval, but I'm not going to elaborate here.)