6
$\begingroup$

In 1878, C. S. Peirce performed a calculation that (I think) would be better done using chi-squared testing — but Pearson hasn’t introduced that yet.

What exactly is Peirce doing here in the last sentence? Is it valid? almost valid? (When I did a chi-squared test, I got a different answer than he did.)

Peirce calculates the probability that a difference can be due to chance.

The first paragraph is straightforward enough. Where we would now say the probable error in the distribution of a sample is within +/- .6745 standard deviations, he has .477 * sqrt(2), and so on down his table.

In the next paragraph, he approximates p = 0.5 and then correctly calculates probable error on the two distributions.

But then what does he do in the last sentence?

(Once I figure that out, I’ll need to figure out if this was a common pre-Pearson approach.)

$\endgroup$
1
$\begingroup$

It looks like he is saying the probable error of the difference between the two proportions is no more than the sum of the two probable errors, so $$d_1+d_2=0.0008+0.0003=0.0011$$ and that $$10(d_1+d_2)=0.011\approx 0.0105=\text{the actual discrepancy.}$$

$\endgroup$

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.