I should not make this calculation in any of these ways. The way it is formulated, the question does not seem to me to make much sense. The Student t distribution is always centered at zero (unless it is a noncentral Student t distribution, which does not seem to be the case). So, for your problem, you’ll always be calculating a probability that won’t be tied to estimating your sample’s average. This may not be apparent with a small average like this example, but increase the average value to 100, for example, and see what I’m talking about.
Sample mean has asymptotically normal distribution, with mean equal to µ and variance Σ 2/n, where µ is the population mean, Σ 2 is the population variance and n is the sample size. So it’s easy to see that we can use the normal distribution to calculate the probability of a random variable being a standard deviation below the mean and a standard deviation above.
set.seed(1234)
x <- rnorm(86, mean=.5, sd=.3) # amostra aleatoria
media <- mean(x) # estimador pontual da media
erro_padrao <- sd(x)/sqrt(length(x)) # estimador do erro padrao
media-erro_padrao # media - erro padrao
[1] 0.4683514
media+erro_padrao # media + erro padrao
[1] 0.5321069
pnorm(media-erro_padrao, mean=media, sd=erro_padrao, lower.tail=TRUE)
[1] 0.1586553
pnorm(media+erro_padrao, mean=media, sd=erro_padrao, lower.tail=FALSE)
[1] 0.1586553
The question isn’t very detailed, so I can’t be sure what your real reason for calculating these odds is. Maybe if there are more details about your real goal, the people in the forum will be able to help you a little more.
Complement after editing the question: for me, this problem still does not make sense. I may simply be having trouble understanding it, but I will try to explain it in items because I believe it cannot be solved this way.
Where did the data analyzed come from? Saying that a distribution has an average -0.49 and standard deviation 3.029041 does not mean much. Is it symmetric to the average, for example? Does it have many outliers? Does it have bell shape? From U?
Why use t distribution? Even if your data came from a sample, I would only use t if I had any suspicions about heavy tails in your distribution. In addition, the calculation of variable standardization is only defined for variables with approximately normal distribution. Even if your data has t distribution, the heavy tails of this distribution will influence this calculation because, well, your variable has t Student distribution and the standardization is not defined in this case.
The formula (x-Mean(x))/sd(x) only works if x has an approximately normal distribution due to Central Limit Theorem. This theorem is only defined for random variables with asymptotically normal distribution. So I solved this problem in the way I presented earlier: the sample mean has asymptotically normal distribution, regardless of the distribution of the random variables
Is it possible to do it the way you are doing it? Yes, but it won’t be right. This proposed standardization does not exist for a t. Thus, you will get something similar to a z value, but that has no real meaning. After all, what does (x-Mean(x))/sd(x) mean in t? What is the distribution of this transformation? I don’t know if it’s t. I only know the case where x is normal or the case where we use the sample average.
If your data is normal, use the normal accumulated data directly. And it is not necessary to even standardize the variable, because it is possible to calculate these probabilities directly. Unless, of course, you want to find these values in a table. Then you can make this transformation smoothly.
This mean and this standard deviation are of the distribution or of a sample size equal to 86?
– Marcus Nunes
Sample size equal to 86.
– Laorie