Testing 2 Means: Got 0 in T-Test - Is Something Wrong?

[vanitymdl]

Y: 51 32 30 74 42 35 39 33 55 61
Y-hat: 49 35 29 72 44 32 38 36 57 60

Null hypothesis: mean y-hat = mean y
Alternative hypothesis: not null hypothesis

So I got as a t test 0 and I was wondering if I did something wrong because I never has this case happened to me.

 

[Mark44]

Y: 51 32 30 74 42 35 39 33 55 61
Y-hat: 49 35 29 72 44 32 38 36 57 60

Null hypothesis: mean y-hat = mean y
Alternative hypothesis: not null hypothesis

So I got as a t test 0 and I was wondering if I did something wrong because I never has this case happened to me.

View attachment 73295

Assuming all the numbers you show were correct, any value of t such that -2.1 < t < 2.1 would be within the acceptance region (of the null hypothesis). It seems a bit coincidental that the two sample means are equal, but I'm getting the same numbers for the two means and the two sample SDs.

 

[statdad]

The standard deviations are not the same. A question: you call one data set yhat: is that really a second data set or are those predicted values of your original data, as from a regression?

 

[vanitymdl]

The standard deviations are not the same. A question: you call one data set yhat: is that really a second data set or are those predicted values of your original data, as from a regression?

They are predicted values from a regression. The thing is that the means are the same giving me a zero t-test. Can that happen? The reason I rejected the null hypothesis is because that is still less than the critical point

 

[statdad]

That will always happen: when you fit a regression line to a set of data you minimize the sum of the squares of the residuals, which is
<br /> \sum (y - \widehat y)^2<br />

A consequence of that fitting process is that the sum of the residuals is zero - that is,

<br /> \sum (y - \widehat y) = 0<br />

which means that

<br /> \sum y = \sum \widehat y<br />

so the means of the original y values and the predicted y values are always the same.

Edited to add: The comments here apply only when the fitted line includes an intercept. If the line is forced to pass through the origin then the residuals no longer sum to zero and the predicted values and actual values do not have the same mean