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# Introduction to t statistics

An introduction to why we use t statistics.

## Want to join the conversation?

• What is the difference between a statistic and a parameter? Please explain like you would to someone who barely knew anything about statistics.
• We use 'statistic' in order to approximately estimate 'parameter'.

Let's say we want to know what percent of all male population of USA (or another random country) do some jogging in the morning. This percent is called 'parameter'.
Can we really survey and analyze every male in the USA ?
Well, maybe we can, but it would be to costly to do so in terms of the time, money or human rights infringements of those who don't want to share what the do in the morning.

So, in practice we just randomly select some men from all over the country and count what percent of them run in the morning. This percent is called 'statistic', which approximately estimates 'parameter'.
• This explanation of the distinction seems really confusing. If the population is Bernoulli distributed then the population proportion and population mean are the same thing! And yet we can estimate one with a Z-stat but the other needs a T-stat?

Also, when Sal calculates confidence intervals for the sample mean he uses the sample variance, which is presumably Bessel corrected and therefore less biased. But when he calculates the intervals for the sample proportion there's no Bessel correction!

Again, the population proportion and population mean are the same for a Bernoulli distribution. And the sample proportion and sample mean are also the same. Yet, when calculating confidence intervals, why do we use Z-stats for one and T-stats for the other? Why do we use Bessel's correction for one, and not for the other?

Finally, why is there no mention of the sample size? I thought that small n is the determining factor for when to use T-stats instead of Z-stats.
• why when calculating p hat (sample proportion), we dont use t score?
• When calculating phat, we know sigma. However, now we don't, as mentioned in , so we use a thing called a t score.

EDIT:
Sorry for my original unclear answer. Looking at Edexcel S3 and S4 manuals I am pleased to confirm that JW and chris are correct. When n is large(>30 for IAL) the Student-t tends toward a normal. Also remember that the t- and z-statistics are basically the same thing (s is unbiased estimate of \sigma) and the difference is that in one case s (sample variance) is also an r.v. and in the other it's not because of extra data given. So which on to use ultimately depends on whether you want to make the approximation that s==\sigma (which is accurate when n>30).

PS this vid is an intro to t-score so presumably he wants to connect the z- and t-scores first.
• Why is the expression at 'not so good'? Where can I get to read the math behind calculating z and t?
• What is the difference between z and t that fixes the problem?
• At , Sal claims that using z* as part of making the confidence interval for a sample mean actually leads to an underestimate for the confidence interval. Why is that?
• The actual sampling distribution of means doesn't really follow a normal distribution (which is what z is based on). The sampling distribution of means has more "extreme" values than does the normal distribution, particularly when you use small samples to estimate the mean. This means more of samples will have means further from the population mean than they would if the sampling distribution was normal. So the confidence interval is narrower than should be, and the intervals don't contain the parameter the "correct" proportion of time. The t-distribution accounts for these "fatter tails".
• here I'm a bit stuck...
p is the proportion of something in the population.
p_hat is the proportion of the same parameter in the sample we take. So to speak our statistic.
So isn't p "just" the population_mean (of the something) / N?
And isn't p_hat the sample proportion: p_hat = sample_mean / n ?
All this by definition?

What am I missing?
Is the X_mean we are searching for the mean of the sampling distribution of the sample means? But wouldn't that be mu = p.... so back to the beginning of my question...
And if it is the real mean value, so not the proportion wouldn't it be just p_hat * n ?
So if we have a mean, but not the proportion, then why can't we just do mean / n to get the p_hat. And from here go the old way with p*(1-p)... ?
• the x_ mean is the sample mean from some random independent sample of a population, which Sal discusses in the beginning of the video
(1 vote)
• Why using sample standard deviation leads to underestimate?!
It's square root of sample variance right? And sample variance is divided by "n-1" rather than "n", so it seems to have larger value. why doesn't it lead to overestimate?
• It's not about the sample standard deviation (the standard error), it's about the shape of the sampling distribution (all the possible means for a particular sample size). This distribution is not a normal distribution, particularly if you have small samples. It actually follows what's called student's t-distribution. This distribution has "fatter tails" (ie, more values that are far from the mean) than the normal distribution, and this is what causes the underestimation.
• Where does sigma over square root of n come from? Why and how did we put it there?
• Interesting question! In this discussion, we use theoretical (or population) standard deviation and variance.

To derive this, we use the following properties:
1) The variance of a sum of independent random variables is the sum of their variances.
2) When a random variable is multiplied by a factor that doesn't depend on the random variable, the variance is multiplied by the square of this factor.
3) The standard deviation is the (non-negative) square root of the variance, and so the variance is the square of the standard deviation.

Let the random variables X_1, X_2, X_3, ... , X_n represent a random sample of n data values, each of which has standard deviation sigma >= 0 (and therefore variance sigma^2). If we assume the population is very large, then it's reasonable to call these n random variables independent.

The sample mean is (X_1 + X_2 + X_3 + ... + X_n)/n, and the standard error of the mean is the standard deviation of the sample mean. Therefore, the standard error of the mean is

standard deviation[(X_1 + X_2 + X_3 + ... + X_n)/n]
= sqrt{variance[(X_1 + X_2 + X_3 + ... + X_n)/n]}
= sqrt[variance(X_1 + X_2 + X_3 + ... + X_n)/(n^2)]
= sqrt{[variance(X_1) + variance(X_2) + variance(X_3) + ... + variance(X_n)]/(n^2)}
= sqrt[n sigma^2 / (n^2)]
= sqrt(sigma^2 / n)
= sigma/sqrt(n).

Have a blessed, wonderful day!
• Hey
can anyone explain what is the difference between True population Proportion and True Population mean.
...
I am bit confused