2
$\begingroup$

I have survey data where I want to calculate a derived variable, based on 2 estimated proportions. (e.g., "Did you visit the web-page?" If Yes, "Did you place an order?", and I want to estimate total orders placed.)

Let's say I have a known population of 10,000, randomly survey 1,000 people, and get 80% Yes to the first question;

Some R code of the example;

N=10000
n_1= 1000
p_1= 0.8
prop_SE<- function(p,n) {sqrt(p*(1-p) / n)} # function for SE of a proportion
p1_SE= prop_SE(p_1,n_1)

0.80 +/- 0.0126

20% of those Yes respondents to Q1, respond Yes to the follow-up question. I treat those 800 as the sample size for calculating SE for that proportion.

n_2= p_1*n_1
p_2= 0.2
p2_SE= prop_SE(p_2,n_2)

0.2 +/- 0.0141

Then I want to estimate 'total orders placed', which would be N * p_1 * p_2 = 1600.

This seems like a simple problem, but I haven't found an explanation of how to calculate the standard error for this estimate. Is it simply the sum of the 2 SEs? p1_SE * n_1 + p2_SE * n_2= 23.96? Why or why not?

$\endgroup$
1
$\begingroup$

Total orders placed is generated from the estimate p_1*p_2 (a proportion of a proportion). By the reasoning of the cited link, the standard error will be given by prop_SE(p_1*p_2, n_1*p_1).

But you want the standard error of N*(p_1*p_2). This will just be N*(standard error of p_1*p_2). i.e.: 129.6148

Here's why, roughly: For fixed N, for the proportion X/n, the formula you're looking for is:

SE(N * X/n) = sqrt(Var(N * X/n)/n)

= sqrt(N^2 * Var(X/n)/n) = N * sqrt(Var(X/n) /n) = N * (standard error of X/n).

$\endgroup$
  • $\begingroup$ Thanks. I want to understand this a little better, can you provide any reference for further reading on where your explanation formula comes from? $\endgroup$ – Dave M Jan 8 '16 at 19:14
  • $\begingroup$ The first part comes from the link you provided and your own proposed solution. The explanation is based on the property of variance that for r.v. X and constants a, b: Var(aX + b) = (a^2)*Var(X) which can be found in any intro stats books, I think. For the proof, 1st note E(aX + b) = a*E(X) + b. Then write out the definition of variance in terms of expected value, expand (the b's 'drop out') and factor. Then use the note. Then reapply the definition of variance...Apologies for the terrible formatting... $\endgroup$ – Jordan Collins Jan 8 '16 at 19:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.