Composite Scores and Standardized Composite Scores t test

I believe it should be the same.

Short answer: z-score standardization is a linear transformation and as such won't change the ratio that's the basis of the T-test.

Long: The basic formula for the independent two-sample T-test is: $$ t = frac{bar{X}_{1} - bar{X}_{2}}{s_{p}timessqrt{frac{2}{n}}} $$

If you did the z-score standardization, but have not changed the data otherwise. It is obvious that $sqrt{frac{2}{n}}$ is unchanged. So we just need to make sure that the ratio between the numerator and the denominator is unchanged too. Let's start with the denominator. The pooled standard deviation $s_p$ is:

$$ s_{p} = sqrt{frac{1}{2}times(sigma_{x_1}^{2}+sigma_{x_2}^{2})} $$

Where $sigma$ is the variance of the group:

$$ sigma_{x_{1}}^{2} = frac{sum_{i=1}^{n}{(x_{i}-bar{X_{1}})^{2}}}{{n-1}} $$

It can be assumed again that $n$ haven't changed. How much the sum part have changed due to standardization? For that let's look at the z-score formula: $$ z = x-bar{x} times frac{n-1}{sum{x-bar{x}}} $$ That's a transformation that we apply to every element in our initial dataset. The critical parts are $x_i - bar{x}$ from here and $bar{X_1} - bar{X_2}$ from the t-stat formula, as $n$ is unchanged. What we need to make sure essentially - to prove that the t statistics is the same - that these expressions have the same ratio in the initial and the z-score case. This can be proven by showing that the ratio of the mean and the individual values (the relative distance from the mean) is unchanged after the z-score transformation. Essentially:

$$ frac{x_{i}}{bar{x}} = frac{z_{i}}{bar{z}} $$

and this equation holds (see proof below) - the z-score doesn't change the relative distance between the values and the mean, actually it shows the distance from the mean in $sigma$ units. Even if the actual values of the $sigma$s will change their relative position to each other won't. That's kind of the point of the standardization - keep the distances, but lose the original level.

So back to the original t-statistic: $$ t = frac{bar{X}_{1} - bar{X}_{2}}{s_{p}timessqrt{frac{2}{n}}} $$ As individual values keep a relative distance from the mean, $bar{X_1} - bar{X_2}$ will be different from $bar{Z_1} - bar{Z_2}$, but as we've changed the pooled standard deviation (because of $x_i - bar{X_1}$) with the same scale once we move into calculating relative measures we end up with the same results.

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Proof: $$ frac{x_{i}}{frac{sum{x_{i}}}{n}} = frac{ frac{x_{i}-bar{x}}{sigma} }{frac{sum{frac{x_{i}-bar{x}}{sigma}}}{n}} $$ $$ sum{frac{x_i-bar{x}}{sigma}} times frac{x_i}{sum{x_i}} = frac{x_i-bar{x}}{sigma} $$ $$ frac{x_{i}}{sigma}timessum{x_i-frac{x_i}{sum{x_i}}}timesfrac{1}{sum{x_i}} = frac{x_{i}}{sigma}(1-frac{1}{sum{x_i}}) $$ $$ frac{sum{x_i-frac{x_i}{sum{x_i}}}}{x_i} = 1 - frac{1}{sum{x_i}} $$ where simplifying the LHS leaves us with $$ 1 - frac{1}{sum{x_i}} = 1 - frac{1}{sum{x_i}} $$ Thus proving that: $$ frac{x_{i}}{bar{x}} = frac{z_{i}}{bar{z}} $$