A lot of brave people are trying to answer your question, but they have two little information.

Are you comparing two groups?

are you sure your data are right? (Some people still superstitiously use 999 as a code for missing data, for example)

What are the actual means and standard deviations, and do these seem right?

At the moment, your question is a bit like I weighted something and it was 120 kilos – is this normal? Is it too heavy?

I think that the values are not logic, but anyway they depend on the data !

Konstantina, if you report n, Mean and SD for each group, others will be able to compute the t- and p-values independently verify your result.

it is depends on your data. please show mean and standard deviation for each group. check the variability of each groups. if the variability is same for the groups then choose equal variance t-test statistic otherwise unequal variance.

Dear Konstantina,

In addition to what Bruce suggested, you can do one more thing. There are different online t test calculators, where if you put the data and check the type of t test you want to run, will give you the value. From there you can cross check the result.

Type in google "online t test calculator"

The one i use is www.graphpad.com

Best,

Praheli

Assuming you did two-sample t-test, df tells us that you had a total sample size of 339+2 = 341 subjects, and the t-statistic is the mean difference Δ divided by the standard error of the mean difference, SEΔ. This ratio is t = Δ/SEΔ = 56.1.

Assuming equal variances and sample sizes in both groups (n ≈ 170.5) allows us to check what combinations of mean difference and variances would give such a ratio:

From the standard formulas for the t-test we find that

SEΔ = s/sqrt(n) * sqrt(2)

where s is the standard deviation of the data (in either group). Plugging this into t = Δ/SEΔ = 56.1 and solving for s gives

s = Δ/t * sqrt(n/2) ≈ Δ/56.1 * sqrt(170.5/2) ≈ Δ * 0.165

Given a mean difference, we can check what the standard deviation of the data had to be to get the t-value of 56.1. For instance, if the mean difference would be 20 years, the standard deviation would have to be about 3.3 years. If the mean difference was just 5 years, the standard deviation would have to be just about 0.82 years.

What is plausible depends on how you selected the subjects. If these are sampled from a youth camp or from viewers of a you-tube channel about teen-beauty, or from residents of retirement homes, I would not expect a large standard deviation. But if the samples were taken from the "general population", where you could get quite young as well as quite old people, I'd expect a larger standard deviation.

A lot of brave people are trying to answer your question, but they have two little information.

Are you comparing two groups?

are you sure your data are right? (Some people still superstitiously use 999 as a code for missing data, for example)

What are the actual means and standard deviations, and do these seem right?

At the moment, your question is a bit like I weighted something and it was 120 kilos – is this normal? Is it too heavy?

Just as Prof Ronán Michael Conroy rightly noted. Your question is not very informative. T-test is used to compared parametric data between two groups. Between which groups are you comparing the mean ages? If you are not comparing anything, you could just present the average age as it is.