Skip
# Shouldn't a point estimate be the midpoint of its confidence interval?

(adsbygoogle = window.adsbygoogle || []).push({});

(adsbygoogle = window.adsbygoogle || []).push({});

Post

# Shouldn't a point estimate be the midpoint of its confidence interval?

June 22, 2011 5:49 PM Subscribe

Shouldn't a point estimate be in the center of its confidence interval?

I'm working with some survey data gathered and condensed by another agency. For various measures, they have provided point estimates and confidence intervals. For many of them, the point estimate is not the midpoint of the confidence interval.

Did I not understand this part of statistics class? Isn't a confidence interval supposed to be symmetrical?

If it matters, the subject of the survey was what percentage of the population between certain ages has undergone certain medical procedures.

I'm working with some survey data gathered and condensed by another agency. For various measures, they have provided point estimates and confidence intervals. For many of them, the point estimate is not the midpoint of the confidence interval.

Did I not understand this part of statistics class? Isn't a confidence interval supposed to be symmetrical?

If it matters, the subject of the survey was what percentage of the population between certain ages has undergone certain medical procedures.

Can you maybe give us some examples to make sure that's what they're saying?

posted by mandymanwasregistered at 5:56 PM on June 22, 2011

posted by mandymanwasregistered at 5:56 PM on June 22, 2011

Short version: the data are not normally distributed. In perfectly normal data, the mean, median, and mode are all the same value. Your "point estimate" is one of these, or some specialized or proprietary combination thereof, whereas a confidence interval means something real: extrapolating out, 95% of observations would fall in this range. In real data, especially data about age, normal distribution cannot be assumed.

Think about it this way: aging doesn't progress linearly from negative to positive infinity, even if lifespans were infinite. There's no such thing as a negative age, for one. Plus, the difference between ages 8 and 16 is much greater than the difference from 40 to 48, in real terms.

posted by supercres at 6:06 PM on June 22, 2011 [1 favorite]

Think about it this way: aging doesn't progress linearly from negative to positive infinity, even if lifespans were infinite. There's no such thing as a negative age, for one. Plus, the difference between ages 8 and 16 is much greater than the difference from 40 to 48, in real terms.

posted by supercres at 6:06 PM on June 22, 2011 [1 favorite]

What supercres said.

Also, imagine that the medical procedure is one that most people get at age 24 (point estimate), but that they can get any time between 18 and 70 (confidence interval).

posted by Etrigan at 6:09 PM on June 22, 2011

Also, imagine that the medical procedure is one that most people get at age 24 (point estimate), but that they can get any time between 18 and 70 (confidence interval).

posted by Etrigan at 6:09 PM on June 22, 2011

Thanks for the answers so far. Here are a couple of the more skewed points:

96.3% (88.9%-98.8%)

95.6% (82.1%-99.1%)

FYI, the data were in Excel but in text format, and I copied the cells I needed and pasted them into a workbook for my own use, then keyed the numbers in as values on separate worksheets so I could graph them. I've checked my keying, so it's not that, and there's really no way I could have corrupted their text, I don't think.

I will have to put my brain into a higher gear to think about the possibility of a non-normal distribution applying here...thanks for the help so far.

posted by lakeroon at 6:15 PM on June 22, 2011

96.3% (88.9%-98.8%)

95.6% (82.1%-99.1%)

FYI, the data were in Excel but in text format, and I copied the cells I needed and pasted them into a workbook for my own use, then keyed the numbers in as values on separate worksheets so I could graph them. I've checked my keying, so it's not that, and there's really no way I could have corrupted their text, I don't think.

I will have to put my brain into a higher gear to think about the possibility of a non-normal distribution applying here...thanks for the help so far.

posted by lakeroon at 6:15 PM on June 22, 2011

Do they mention the standard error anywhere? Or their alpha level, (since one mustn't assume it's .05)? Maybe how they arrived at these calculations?

posted by mandymanwasregistered at 6:51 PM on June 22, 2011

posted by mandymanwasregistered at 6:51 PM on June 22, 2011

The Wikipedia article on skewness will help your conceptual understanding, especially this figure, comparing mean, median and mode for distributions with skewness of 0.25 and 1. (0 is unimodal, though not necessarily normal.) Non-parametric stats will also point you in the right direction.

I guess it comes down to what "point estimate" means. To me, it would be the single value that summarizes the distribution best-- in your case, a single number for when

As for why any age measure must be skewed... Well, think about a medical procedure like orthodontic braces. Say the average age to get braces is 12 years, with a standard deviation of 3 years. (I'm making up plausible enough numbers; the theory is the same.) If the data were normally distributed, the same number of people would get braces older than 21 as would under 3-- given the dearth of people marketing braces for babies in the way braces for adults is marketed, I'd say this test fails. This can be applied to almost any procedure-- literally any procedure if you think theoretically enough.

Do they say how "point estimate" or "confidence interval" are calculated? I'm basically just BSing on

(This is a handy skill when critiquing any quantitative paper. One of the first errors in science that springs to mind is assuming normal distribution when it's not warranted, or at least without testing for it.)

posted by supercres at 6:54 PM on June 22, 2011

I guess it comes down to what "point estimate" means. To me, it would be the single value that summarizes the distribution best-- in your case, a single number for when

*most people have the procedure*, as**Etrigan**put it. For the high skewness distribution, where would you put the point estimate? Probably somewhere between the median and the mode, right? A confidence interval in the traditional 95% sense is only symmetric around the median-- it's a percentile statistic, like the median.As for why any age measure must be skewed... Well, think about a medical procedure like orthodontic braces. Say the average age to get braces is 12 years, with a standard deviation of 3 years. (I'm making up plausible enough numbers; the theory is the same.) If the data were normally distributed, the same number of people would get braces older than 21 as would under 3-- given the dearth of people marketing braces for babies in the way braces for adults is marketed, I'd say this test fails. This can be applied to almost any procedure-- literally any procedure if you think theoretically enough.

Do they say how "point estimate" or "confidence interval" are calculated? I'm basically just BSing on

*how*this could be the case, not*why*it is, in your specific instance.(This is a handy skill when critiquing any quantitative paper. One of the first errors in science that springs to mind is assuming normal distribution when it's not warranted, or at least without testing for it.)

posted by supercres at 6:54 PM on June 22, 2011

Short answer: no, not always. Usually it's at least kind of near the middle but weird things can happen. (

posted by madcaptenor at 7:02 PM on June 22, 2011

posted by madcaptenor at 7:02 PM on June 22, 2011

To be more specific about the "non-normal" distribution responses relative to what you're looking at: proportions of respondents with a certain characteristic or answer follow a binomial distribution. The probability/proportion of a response is never normal, because it is bounded at the lower end by 0 and at the upper end by 1.

When analyzing this sort of data, to get both point estimates and confidence bands, one might use a logistic model (general linear model, family = binomial, link = logit). The logit transformation is log(p/(1-p)), and the inverse is exp(x)/(1+exp(x)). So when point estimates and standard errors are generated, they're generated on this transformed scale. What you'll find is that the standard errors actually

For illustration: if my probability is .986, logit(.986) = 4.254599

If my standard error produces a confidence interval of +/- 1.2 around that estimate, then those upper and lower limits are

high = inverse logit of (4.254599 + 1.2) = 0.9957416

low = inverse logit of (4.254599 - 1.2) = 0.9549807

If you didn't work on this transformed scale, then your confidence bands would produce impossible probabilities of less than 0 or greater than 1.

posted by shelbaroo at 8:29 PM on June 22, 2011

When analyzing this sort of data, to get both point estimates and confidence bands, one might use a logistic model (general linear model, family = binomial, link = logit). The logit transformation is log(p/(1-p)), and the inverse is exp(x)/(1+exp(x)). So when point estimates and standard errors are generated, they're generated on this transformed scale. What you'll find is that the standard errors actually

*are*applied symmetrically - just not on the scale you're thinking of!For illustration: if my probability is .986, logit(.986) = 4.254599

If my standard error produces a confidence interval of +/- 1.2 around that estimate, then those upper and lower limits are

high = inverse logit of (4.254599 + 1.2) = 0.9957416

low = inverse logit of (4.254599 - 1.2) = 0.9549807

If you didn't work on this transformed scale, then your confidence bands would produce impossible probabilities of less than 0 or greater than 1.

posted by shelbaroo at 8:29 PM on June 22, 2011

When you’re dealing with proportions, the confidence intervals won’t be symmetric (because at the extremes your bounded by 0 and 100%). The easiest way to understand why they won’t be symmetric is to look at the extreme cases.

The reason this problem comes up is that most people are taught the wald formula for confidence intervals of proportions. The wald formula gives incorrect confidence intervals at extreme cases (because it creates symmetric confidence intervals). The Wikipedia article on binomial confidence intervals gives different formulas for calculating binomial confidence intervals. The Agresti-Coull interval is the formula you should be using.

This site is useful for understanding how the different formulas for binomial confidence intervals work. To get a good understanding of how confidence intervals work, get the book Statistics with Confidence.

posted by Jasper Friendly Bear at 8:47 PM on June 22, 2011 [1 favorite]

**Example**: Suppose 1% of the individuals had surgery. It wouldn’t make sense to have a confidence interval that included values less than zero percent (e.g., 1% +/- 2%). The same thing at the other extreme. For proportions, it wouldn’t make sense to have values greater than 100% in the confidence interval.The reason this problem comes up is that most people are taught the wald formula for confidence intervals of proportions. The wald formula gives incorrect confidence intervals at extreme cases (because it creates symmetric confidence intervals). The Wikipedia article on binomial confidence intervals gives different formulas for calculating binomial confidence intervals. The Agresti-Coull interval is the formula you should be using.

This site is useful for understanding how the different formulas for binomial confidence intervals work. To get a good understanding of how confidence intervals work, get the book Statistics with Confidence.

posted by Jasper Friendly Bear at 8:47 PM on June 22, 2011 [1 favorite]

Usually this has to do with the cause of error or binding of the data set. For example, if you're determining distances by measuring a line of sight angle, and you have plus or minus 1 degree of error in your measurement, the error on the near side will be smaller than on the far side (i.e. the sin of 75° > sin 74°).

It can also do with a bounded data set. When can a baby first sit upright? On average it's 7 months, but there are all kinds of circumstances that might make it take longer or shorter. But no special case is going to make the "shorter" side of the distribution less than seven months long.

posted by Kid Charlemagne at 8:57 PM on June 22, 2011

It can also do with a bounded data set. When can a baby first sit upright? On average it's 7 months, but there are all kinds of circumstances that might make it take longer or shorter. But no special case is going to make the "shorter" side of the distribution less than seven months long.

posted by Kid Charlemagne at 8:57 PM on June 22, 2011

No, in fact it is only special cases and a particularly famous kind of CI which give symmetric intervals on a particular scale. Producing non-symmetric intervals is regarded as a feature of other kinds of CIs; symmetric intervals aren't invariant to re-parameterization. Note: those links discuss tests, but intervals can always be obtained by pivoting a test.

posted by a robot made out of meat at 7:28 AM on June 23, 2011

posted by a robot made out of meat at 7:28 AM on June 23, 2011

This thread is closed to new comments.

posted by proj at 5:54 PM on June 22, 2011