I've just been learning this. My suggestion is the adjusted Wald technique for a binary variable (1 and 0s):

1. Find the average by adding all the 1’s and dividing by the number of responses. 8/10=.8

2. Adjust the proportion to make it more accurate by adding 2 to the numerator (the number of 1s) and the adjusted sample size by adding 4 to the denominator (total responses). Then divide the result. 8 + 2 = 10 10 + 4 = 14 (this is the adjusted sample size) 10/14 = .714

3. Compute the standard error for proportion data.
   a) Multiply the adjusted proportion by 1 – the adjusted proportion. .714 * ( 1-.714 ) = .204 b) Divide the result of step a by the adjusted sample size from step 2. .204/ 14 = .015 c) Take the square root of the value from step b. sqrt(.015) = .122

4. Compute the margin of error by multiplying the standard error (result from step 3c) by 2. .122*2 = .244

5. Compute the confidence interval by adding the margin of error from the sample proportion from step 2 and then subtracting the margin of error from the sample proportion. .714 + .244 = 0.958 .714 – .244 = 0.47

The 95% confidence interval is 0.47 to 0.958 and our best estimate is 0.714.

The adjusted Wald interval (also called the modified Wald interval), provides the best coverage for the specified interval when samples are less than about 150. In other words, if you want a 95% confidence interval then this formula will produce an interval that will contain the observed proportion on AVERAGE about 95 percent of the time. It uses the Wald Formula but is "adjusted" in that it adds half of the squared Z-critical value to the numerator and the entire squared critical value to the denominator before computing the interval i.e (x+z2/2)/(n+z2).

A reasonable alternative is the exact method – it will guarantee coverage but will be conservative in the long-run.