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A research institute surveyed 1000 online shoppers in the United States and China. One question asked if the online shopper followed brands they purchased through social media. Here are the results

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proportions:
The USA:    513/(513+487)=0.513, China: 928/(72+928) =0.928

I calculated the odd ratio as follows:

$$OR=(0.513/(1-0.513))/(0.928/(1-0.928))=1.053/12.889=0.0817$$

Then I constructed a 95% confidence interval for it (0.1377 was given as SE for $b_1$). Could I calculate it from data in the table above?

$$CI=-2.5043±1.96*0.1377=(-2.7742,-2.2344) \\ CIOR=(ex p⁡(-2.7742),exp⁡(-2.2344)=(0.062,0.107)$$

How can I sate this interval?

I wrote:

We are 95% confident that the U.S. online shoppers are as low as 0.062 and as high as 0.107 as likely to follow brands they purchased through social media as Chinese online shopper.

How about:

We are 95% confident that the odd ratio of the U.S. online shoppers who follow brands they purchased through social media vs. their Chinese counterparts is as low as 0.062 and as high as 0.107

Are they correct? I'm even afraid of their English grammar and meaning!

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The standard error given is for the log odds ratio. You obtain it by the square root of the sum of the reciprocals of the table frequencies.

$$ se = \sqrt( 1 / 487 + 1 / 513 + 1 / 72 + 1 / 928) $$

which by my calculations does equal 0.1377 as you state.

People usually just state that the 95% confidence interval for the odds ratio is from 0.062 to 0.107 and leave it at that. It means that 95% of the time when you construct similar intervals the true value will lie between the limits (and 5% it will not).

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