# Confidence interval for combined populations

I'd like to compare two approaches for calculating the confidence interval ($$CI$$) when estimating the mean for a sample of weighted values:

The first approach is based upon the standard error of the mean itself ($$\hat{x}$$):

$$CI_{(1-\alpha)} = \hat{x} \pm t_{(\alpha/2, df)} * s / \sqrt{n}$$

The second is based upon the standard error of a proportion:

$$CI_{(1-\alpha)} = \hat{p} \pm t_{(\alpha/2, df)} * \sqrt{\hat{p} * (1 - \hat{p}) / n}$$

The tricky part (for me) is how to combine the standard error of the weighted proportions into a population.

Consider an example: If we start with a random sample of 50 coins taken from a jar (without replacement):

qty denomination value ($$w$$)
15 pennies \$0.01 7 nickels \$0.05
8 dimes \$0.10 20 quarters \$0.25

Incorporating the value (or weights) of each coin's denomination into a weighted mean for all coins is trivial. The combined value of the 50-coin sample is \$6.25, so the weighted mean of a coin is \$0.1250. The standard deviation is $0.1069, so the confidence interval is \$[0.0946, 0.1554].

Alternatively, we can look at the sample as the aggregate of four populations (i.e., the four denominations of coins) and calculate $$CI$$s for those individual populations:

denomination proportion confidence interval
pennies 0.3000 [0.1698, 0.4302]
nickels 0.1600 [0.0558, 0.2642]
dimes 0.1400 [0.0414, 0.2386]
quarters 0.4000 [0.2608, 0.5392]

A brute-force way to pool these intervals would be to combine them in ways that maximized proportions of low- and high-value coins. E.g., to establish a lower bound, I could combine 43.02% pennies, 26.42% nickels, 4.48% dimes, and 26.08% quarters. For the upper bound, I could combine 16.98% pennies, 5.58% nickels, 23.52% dimes, and 53.92% quarters. The combinations for both the lower and upper bounds will add up to 100%.

I'm not sure that approach captures all available information, however.

What is the formula to pool the standard error for weighted proportions? Intuition suggests that pooling the standard error for multiple proportions introduces more uncertainty, but I'd like to confirm/refute that with the appropriate formula.

• Why take this approach at all? Why not directly analyze the values themselves in the usual way? – whuber Apr 10 at 13:40
• Thanks, @whuber. I agree that it'd be straightforward to calculate a confidence interval for the mean. However, coins have discrete values, so the sampling distribution of the mean is actually truncated normal. In extreme cases (i.e., when most of the coins are pennies or quarters), the bounds of the CI for the mean will return impossible values (i.e., lb < 0.01 or ub > 0.25). While I could likewise truncate the confidence interval, I'd prefer to use the sampled proportions of coins to avoid having to deal with the messiness of a truncated normal sampling distribution for the mean. – curious Apr 10 at 14:12
• The sampling distribution of the mean is never exactly Normal. In your case, it is exceptionally close to Normal and the usual Normal-theory methods give optimal results. – whuber Apr 10 at 16:53
• Thanks again, @whuber. I've reframed the question to (hopefully) focus less on the example and more on the concept that I'm trying to better understand. – curious Apr 10 at 22:00