# From confidence interval to standard deviation - what am I missing?

Given:

Crude incidence rate per 100,000: 164

population: 211,101

95% CI: 147-182

What is the standard deviation of the incidence rate :

My calc: Standard error:(182-147) / (2 x 1.96) = 8.93

Standard deviation = 8.93 x sqrt(211,101) = 4103

This can't be right, and I think it must be to do with the 100,000

Is this correct: Standard deviation = 8.93 x sqrt(211,101/100000) = 12.98

? Many thanks !

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Why can't your answer of 4103 be right? What makes you suspicious of it? (I obtain approximately 4050 using formulas for the Binomial distribution.) Note that you don't even need the CI to compute this: the incidence rate and count ($164$ and $10^5$, respectively) completely determine the SD. –  whuber May 31 '12 at 21:21
@whuber when I think about the SD I think it should be on the same scale as the mean (164) so 4103 isn't intuitive to me. I'm a beginner so I am probably making some basic error, so thank you for your time and I'm sorry if I'm wasting it, but I am very grateful. –  Joe King May 31 '12 at 21:34
Thinking about thoughtful and clearly stated questions is never a waste of time, Joe. –  whuber May 31 '12 at 22:03

### Analysis

The calculations in the question support the assumption that a binomial model is being used. This treats the incidents as if 211,101 slips of paper were drawn randomly from a hat having millions of slips (far more than the number observed). On each slip is drawn either a $1$ (an "incident") or a $0$. Evidently, a total of $164/10000 \times 211101 = 346$ incidents (ones) were observed.

This total is large enough to let us estimate that a fraction $164/10000$ of all the slips in the hat are marked with ones. This is an estimate $\hat{p}$ of the hat's expectation, $p$. Probability theory tells us that the variance of the total of the numbers observed on $N$ draws from this hat is approximately $p(1-p)N$. Its square root is the standard deviation for the total of $N$ draws. ("Approximately" weasels around a potential, but likely small, "finite population" adjustment when the slips are not drawn with replacement.)

As a proxy for the unknown value of $p$, we have the estimate $\hat{p}$. Using the information $\hat{p} = 164/100000$ and $N=100000$ we get an (estimated) standard deviation ($\hat{s}$) equal to $4046$. It seems high compared to the expected total of only $164$ incidents per $100000$, but it's correct: binomial distributions with rare outcomes are highly skewed.

### Interpretations

This standard deviation itself isn't usually interpreted, but it is useful for constructing confidence intervals and other quantities related to the chance outcomes. For instance, the standard error of the sample mean is obtained by dividing $\hat{s}$ by the root of $N$. Taking $N=211101$ gives a standard error of $4046 / \sqrt{211101}$ = $8.81$.

We see a close connection between these results with the calculations in the question itself, which deduced (from the confidence interval) that the standard error is $8.93$. If we were able to peer into the hat and total the values on all its tickets, we would anticipate the result to lie between $147/100000$ and $182/100000$ times the number of tickets. In deducing this, I have used a procedure (the 95% CI) that will fool me at most 5% of the time (due to the chance behavior of the 211,101 random draws).

Similarly, the standard error for $N$ = $100000$ draws equals $12.80$. This is what has been calculated at the end of the question. It means that based on what we have seen so far, we would anticipate the number of incidents observed in another $100000$ draws from this hat would differ from $164$ (due to chance alone), but only by $12.8$ or so. A difference much larger than this--say, less than $125$ or more than $205$--would be surprising. (This is a prediction interval. It is wider than the confidence interval because it needs to account not only for the element of chance in the 211,101 outcomes already observed, which makes us somewhat uncertain about the true value of $p$, but also for the element of chance in the 100,000 future outcomes.)

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This is fantastic. Are you a teacher ? –  Joe King May 31 '12 at 22:07
Joe, I believe that everybody who has answered a question on our site is, by virtue of that activity, a teacher. This can quickly be seen by taking a glance through the highest-voted answers of which there are many contributed by many people. –  whuber May 31 '12 at 22:11
thanks again. The link is informative. I hope to be a real mathematician one day. This site gives me a lot of inspiration. –  Joe King May 31 '12 at 22:16
Good work Huber! –  Michael Chernick May 31 '12 at 23:07
Yeah, this is great. –  Colin K Jun 1 '12 at 0:13

If you are talking about measured (ratio or interval) data, then the standard deviation of the data measures scatter, and the standard error of the mean quantifies how precisely that mean is known (from its sample size and SD). The two are very different. Converting from one to the other is straightforward using the equation the original poster used.

But these data are binomial. There are two outcomes (new case of disease or not), and the incidence rate is the proportion of people who get a case this year (or whatever unit of time they used). It is possible to compute a standard error of almost any value computed from a sample of data assumed to be drawn from (or representative of) a larger population. In this case, it makes perfect sense to compute the standard error of the proportion, which the OP did.

The OP and WHuber then computed the SD of the number of expected cases, which is either 4103 or 4046 (I didn't try to figure out why those two calculations aren't identical). That is the SD for the number of cases you'd expect to see in a group of 211,101 people. It is not the SD of the incidence rate.

There really is no standard deviation of the incidence rate. Or, more accurately, the standard deviation and standard error are really the same thing when the goal is to quantify the precision of a computed parameter like the rate. The standard error of a mean, for example, can be thought of as the standard deviation of the mean. Two terms for the same value. That value is very very different than the standard deviation of the data. Similarly the standard error of a proportion is the same as the standard deviation of a proportion. The difference is that the term standard deviation is rarely used in that context.

My answer. The standard deviation of the incidence rate is 8.93 per 100,000.

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Your clarifications are much appreciated, Harvey. I believe the question was inspired by noting that the number of cases in 211,101 was 364 and its estimated SD was 4103. The OP (quite thoughtfully) wondered how the SD of a non-negative quantity could be so extraordinarily greater than the value of that quantity. That is what needs explaining. Concerning the differences between 4103/4046 and 8.93/8.81, note that one value in each pair is computed from a CI but the CI methodology is not given: many are possible. There may also be a finite-population correction. The agreement is close enough. –  whuber Jun 1 '12 at 13:53
Thank you @Harvey Motulsky . Is it appropriate to use this standard deviation for a power/sample size calculation based on a 2 sample t test when designing a clinical trial for an intervention targeting the incidence rate ? –  Joe King Jun 7 '12 at 8:12
Joe, there is no need to base sample size calculations on a two sample t test. It is easy enough to find the equations that compute sample size to compare two proportions. –  Harvey Motulsky Jun 7 '12 at 15:01