# Composite Scores and Standardized Composite Scores t test

I have a set of survey data related to 20 survey questions. Each of these questions represent a variable (Q1, Q2,...Q20). I created a new variable QCom which measures the response of the survey, and is given by a composite score obtained as the sum of the scores of responses from Q1 to Q20. I then perform a t-test for QCom to check for evidence of a difference in mean due to another variable (Sex -> Male or Female). I obtained the test statistics from SPSS's independent sample t test.

Subsequently, I created another variable called StCom (equal to standardized score of QCom). Again, I repeated the t-test as the above using StCom this time. The t test statistics that I obtained from SPSS was exactly the same as the first test using QCom.

In this case, I am not sure if this is normal or the z-scores transformation is incorrect. Can someone help to enlighten me why the composite scores and composite z-scores t test results are exactly the same?

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: $$t = \frac{\bar{X}_{1} - \bar{X}_{2}}{s_{p}\times\sqrt{\frac{2}{n}}}$$

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: $$t = \frac{\bar{X}_{1} - \bar{X}_{2}}{s_{p}\times\sqrt{\frac{2}{n}}}$$ 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.

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