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# Standard error of the difference between two proportions

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1. I need to estimate , where and .

Here's the question for reference.

I calculated the proportion, , from the first sample, which had [; 256 ;] of the variable I'm looking at, in a sample of . came from the fact that there are out of in the second sample.

The problem I'm facing is that if I do the transformation first (), I get , which obviously won't work with the SQRT in the SE formula.

We are also told that (this is the population proportion).

Where am I going wrong? I can provide more info if needed. Also, apologies if some of my terminology is a bit off.

Thanks!

To clarify what I have already done:

An earlier question asked me to estimate [; SE[P_0] ;], which I did by first calculating the proportion, [; P_0 = 256/1250 = 0.2048 ;], then putting this value into the SE formula as follows:

I used this formula as it was how I was taught to find the SE of a sample proportion.

I then applied similar intuition for the next question, by first calculating [; P_1 ;], then the difference between the proportions:

The problem I then face is that if I try to use the same SE formula from above, I get tripped up by the negative:

[; SE[P_0 - 2P_1] = SE[-0.0352] = \sqrt\frac{-0.0352(1--0.0352)}{1250} = \sqrt{-0.00002915...};]
2. (Original post by danlocke)
I need to estimate , where and .
You need to go via the formula for the addition of variance.

So, take your standard errors, turn them into variances, get the variance of the linear combination, then square root to get the standard error.
3. (Original post by Gregorius)
You need to go via the formula for the addition of variance.

So, take your standard errors, turn them into variances, get the variance of the linear combination, then square root to get the standard error.
Ah, right. How does this look?
4. (Original post by danlocke)
Ah, right. How does this look?
Right approach, but you've lost the sample sizes...
5. (Original post by Gregorius)
Right approach, but you've lost the sample sizes...
Ah, so I just needed to divide the original two formula for variance by n, right?
6. (Original post by danlocke)
Ah, so I just needed to divide the original two formula for variance by n, right?
Yes.
7. (Original post by Gregorius)
Yes.
Great, would you say 0.02163... looks correct then? If you missed my edit, here's my working.

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