Statistical test for difference between means

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Aviram Shemen 2020-12-24

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https://ww2.mathworks.cn/matlabcentral/answers/702042-statistical-test-for-difference-between-means

评论： Aviram Shemen 2020-12-29

采纳的回答： Walter Roberson

Hi,

I have two mean values and standard errors (se) of two unknown distributions.

I would like to check whether the difference between these two means is statistically significant.

mean1=1.2545, se1=0.0145

mean2=1.6913, se2=0.0172

Thanks,

Aviram.

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Walter Roberson 2020-12-24

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https://ww2.mathworks.cn/matlabcentral/answers/702042-statistical-test-for-difference-between-means#answer_583997

在 MATLAB Online 中打开

mean1=1.2545; se1=0.0145;
mean2=1.6913; se2=0.0172;
rho = 0.01;
s = 2*normcdf(-abs(mean1-mean2), 0, min(se1,se2))
s = 2.3405e-199
s < rho
ans = logical
   1

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Walter Roberson 2020-12-25

在 MATLAB Online 中打开

Suppose that mean1 <= mean2 . Then

normcdf(mean1, mean2, se2)

gives the cdf that N(mean2, se2) <= mean1 . In turn you can recenter that as

N(0, se2) <= mean1 - mean2

Keep in mind here the assumption that mean1 <= mean2 so mean1 - mean2 is non-positive. "Given standard deviation #2, what is the cumulative probability that mean1 is (mean2 - mean1) to the left of it? Equal mean would be 0 difference which would be cdf 0.5

Now, suppose instead that mean2 <= mean1 then the you could look at the upper-tail probability that you are (mean1 - mean2) to the right of the center given that standard deviation, but if you try to calculate the cdf to the upper bound then you are likely to get a value too close to 1 to be useful. So it is easier to reflect it down to the other side and ask for the cdf to be that far to the left. The distance to the center is abs(mean1-mean2) and you the distance left of center, so -abs(mean1-mean2), tested against a normal distribution with mean 0 and second standard deviation. And since you have quietly moved from two-tail distribution to one-tail distribution, you need to double the probability -- if the difference in means was 0 then the probability would have accumulated to 1.0 . You want the test to succeed when the probability < 0.05 or < 0.01, one chance in 20 or one chance in 100 that the means are at least that far away.

Thus you have 2*cdf(-abs(mean1-mean2), 0, se2)

Then you flip the two around to find out the probability that sample 2 is at least as far as way based on se1, and by symmetry it is obvious you have 2*cdf(-abs(mean1-mean2), 0, se1) for that.

You now want to choose the more extreme of those two. But the more extreme of the two will be for the case with the smaller standard deviation. And that brings us to

2*cdf(-abs(mean1-mean2), 0, min(se1,se2))

Higher value meaning it is more likely that the two means are the same, lower value meaning it is less likely, more significant that they are different... so you want to test < rho

Walter Roberson 2020-12-27

在 MATLAB Online 中打开

mean1=1.2545; se1=0.0145;
mean2=1.6913; se2=0.0172;
rho = 0.01;
minse = min(se1,se2);
s = 2*normcdf(-abs(mean1-mean2)/minse, 0, 1)
s = 2.3405e-199

A small modification to the above that perhaps might make it marginally clearer based upon common statistics: we effectively normalize back to mean 0 and standard deviation 1, and then it is just a cdf calculation

Aviram Shemen 2020-12-28