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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?

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

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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.)

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    $\begingroup$ Ok, but. Population of first and last names is over so with bigger sample, number of people with pairs should be higher. $\endgroup$ Jun 10, 2018 at 20:26
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    $\begingroup$ Having a "pair" is not an independent property of the sampling units. Consequently the analysis in this answer is incorrect. Moreover, it doesn't even obtain approximately the right result. The difficulty is that the result can vary depending on what is assumed about the distribution of duplicate names in the population. For an example of one solution (which is embodied in government regulations even!) see WAC 434-379-010 at apps.leg.wa.gov/wac/default.aspx?cite=434-379-010. $\endgroup$
    – whuber
    Mar 31, 2019 at 15:30

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