I have no solid background in statistics so please bear with me. In general, I am trying to figure out methods to determine sample sizes required to establish a certain level of confidence for a population (e.g. out of 1000 computers, how many do I need to sample to be sure that the rest has the same properties with a certain level of confidence?).

There are online calculators for "sample sizes", often using the Cochran's (below) formula. But I am unsure whether that is valid for this purposes as basically no quality control book uses that, quoting ANSI sampling tables instead.

$$\text{Sample Size} = \frac{n}{1 + (n/\text{population})}$$ in which $n$ is equal to $Z * Z [P (1-P)/(D*D)]$ (using a 95% confidence and $5\%$ margin of error and $p = 0.5$) which gives me sample size $323$.

This formula tells me that for a population of 2000, I need to sample 323 (CL 95% and 5% margin of error). Is it valid for any sampling where I expect random distribution? Because my related question (Determine required sample size with unknown standard deviation) got quite complicated answers/comments for me so I suspect this is not gonna work..just do not know why.

  • $\begingroup$ You're seeing the analogy between the sampling schemes clearly enough. If a quality engineer were interested in calculating a confidence interval of a given length for the proportion of defectives in a batch he might use Cochran's formula to work out the required sample size, given the conditions were met that make it a decent approximation. But he'd more often be interested in specifying rules for accepting or rejecting the batch that guarantee batches with more than a given number of defects are rejected with no less than a given probability. What are you interested in? ... $\endgroup$ – Scortchi Jan 5 '16 at 16:35
  • $\begingroup$ ... "How many do I need to sample to be sure that the rest has the same properties with a certain level of confidence" isn't at all clear. BTW You should define Cochran's rule in your question: there are variants & not everyone will know it by name. It seems you've based the sample size calculation on the normal approximation to the binomial assuming a proportion of one-half - giving a pessimistic estimate of the variance - & then carried out a finite population correction. $\endgroup$ – Scortchi Jan 5 '16 at 16:36
  • $\begingroup$ @Scortchi Thanks. What I mean by that sentence is: What sample size is enough for me to expect that it represents the whole population. If those 323 devices out of 2000 work, can I say that there is 95% confidence that all work? $\endgroup$ – Pietross Jan 5 '16 at 16:40
  • $\begingroup$ No. The 95% confidence interval for the number of defective computers in the whole batch of 2000 would be $\{0, 1, \ldots, 16\}$ - using the hypergeometric distribution, as explained in the post I linked to from your previous question. (Suppose there were just one defective computer in the batch. The probability of getting no defectives in your sample of 323 would still be 84%. Only when you get up to 17 defective computers in the whole batch would the probability of getting no defctives in your sample fall below 5%.) $\endgroup$ – Scortchi Jan 5 '16 at 16:45
  • $\begingroup$ @Scortchi Sorry I do not follow. So what does this 323 (result of the formula) mean? I mean, it is the sample size with 95% confidence but what does it actually mean? I thought the whole point of this sample size is to estimate the whole population. $\endgroup$ – Pietross Jan 5 '16 at 16:55

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