How can I choose between homoscedastic and heteroscedastic I want to calculate the p-value between subgroups of my samples. For that, I am using the T.TEST function of Excel. But I do not understand the last parameter, type:

*

*Paired

*Two-sample equal variance (homoscedastic)

*Two-sample unequal variance (heteroscedastic)

In my case, I cannot use paired (not the same size). But how can I determine if my variance is equal or not? Can I just calculate the variance of my sample and compare them? If, yes, what is the threshold of equality? Or is it a general assumption, like the tails: gaussian or not?
Also, can I use the t-test if the size of my data are unbalanced? Some of my subgroups have about the same size, but some others are 85%-15%.
[EDIT] To add more context. I have a questionnaire that people answered with Likert-scale questions (from 1 to 5).
I split the data into different subgroups to make some analyses: gender, country of origin (Japanese or not)...
I calculated the t-test p-value for each variable (questions) for each subgroup.

I guess, the underlying assumption is that both sample groups come from normal distributions. In this case, the sample variance from either group should follow $\chi^2$ distribution, whilst the actual  variances should follow inverse $\chi^2$. You could construct a test that extracts variance from the first group and then tests how well the data from the second group follows it (i.e. is the second sample variance likely, given your guess based on the first one). Not sure whether there is a name for this procedure. Also, it is a good idea to know how much difference you want to be sensitive to. For example, do you care about 0.1% difference in variance?
I am sure there must be a software package that does this anyway, but since this is not something I need to do often, i prefer more manual route :)

ADDENDUM (keeping the previous text for comments to make sense).
How can we test whether two normal distributions have the same or similar variance, given two groups of samples. I get the feeling this is more detail than what you are after, but since there are now other answers, it makes sense to suggest a viable route to go
Firstly, you need to have some sense of what is a significant difference. This is subject matter expertise. One way to start may be to estimate the sample variance of either group and take a fraction of that. For example:
\begin{align}
\{X_1,\,X_2,\dots X_N\},\quad X_{1\dots N}\sim N\left(\mu_1,\sigma_X^2\right) \\
\{Y_1,\,Y_2,\dots Y_M\},\quad Y_{1\dots M}\sim N\left(\mu_2,\sigma_Y^2\right)
\end{align}
Let sample variances be:
\begin{align}
\bar{S}^2_X=\frac{\sum_{i=1\dots N}\left(X_i-\bar{X}\right)^2}{N-1},\quad \bar{X}=\frac{\sum_{i=1\dots N}X_i}{N} \\
\bar{S}^2_Y=\frac{\sum_{i=1\dots M}\left(Y_i-\bar{Y}\right)^2}{M-1},\quad \bar{Y}=\frac{\sum_{i=1\dots M}Y_i}{M}
\end{align}
You may then ask whether $\left|\sigma^2_X-\sigma^2_Y\right|\le MIN\left(\bar{S}^2_X,\,\bar{S}^2_Y\right)/10$. Or something similar. This gives you your null hypothesis.
Next you could use the fact that, by Student's Theorem the ratio of sample variance and actual variance is chi-2 distributed
$$
\frac{\left(N-1\right)\bar{S}^2_X}{\sigma^2_X}\sim\chi^2\left(N-1\right)
$$
Then the distribution of the variance, given data, is given by scaled-inverse-chi2:
$$
\sigma^2_{X}\sim Scaled-Inv-\chi^2\left(N-1,\,\bar{S}^2_X\right)
$$
And similar for $\sigma^2_Y$. So now you could actually compute the probability $\left|\sigma^2_X-\sigma^2_Y\right|\le MIN\left(\bar{S}^2_X,\,\bar{S}^2_Y\right)/10$ and do a proper hypothesis test of whether variances are the same.
