# Probability a sample mean will fall in a range

1. Nov 21, 2015

### toothpaste666

1. The problem statement, all variables and given/known data
A random sample of size n = 81 is taken from an infinite population with the mean μ = 128 and the standard deviation σ = 6.3. With what probability can we assert that the value we obtain for the sample mean X will fall between 126.6 and 129.4?

3. The attempt at a solution
z = (x-μ)/(σ/sqrt(n))
so we have
z = (126.6-128)/(6.3/9) = -2 and z = (129.4-128)/(6.3/9) = 2
so the probability it will fall in the range is
F(2) - F(-2) = .9772 - .0228 = .9544

is this correct?

2. Nov 21, 2015

### Orodruin

Staff Emeritus
This depends on the actual distribution in the population. You can only do what you did if this distribution is assumed to be Gaussian.

3. Nov 21, 2015

### toothpaste666

Gaussian means "normal" right? I am confused a bit about that. In my book they seem to use "z" for the test statistic and use "t" when the population is known to be normal. From what I can tell they are the same thing except that with z you use the standard normal table and with t you use a different table with a certain amount of degrees of freedom. I don't think I fully get it.

4. Nov 21, 2015

### Ray Vickson

I do not actually believe you; I think you are mis-reading your book (although, to be honest, I am making this judgement sight-unseen). Typically, for an independent random sample from an underlying normal (=Gaussian) distribution with mean $\mu$ and variance $\sigma^2$: (1) we use $z$ and normal tables when we KNOW the value of $\sigma$; but (2) use $t$ and t-tables when we do not know $\sigma$, but have estimated it from the sample data itself.

In case (2), we estimate
$$\text{estimator of }\: \sigma^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2$$
where the sample values are $x_1, x_2, \ldots, x_n$ and $\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$ is the sample mean. In that case the jargon is that there are $n-1$ "degrees of freedom".

In the limit as $n \to \infty$ the t-distribution with (n-1) degrees of freedom becomes the standard normal, so using $z$ is like having infinitely many degrees of freedom.

5. Nov 21, 2015

### toothpaste666

so for either of the two statistics to work, the distribution must be normal?

6. Nov 21, 2015

### Ray Vickson

Theoretically, yes, but for a large sample-size, using the "normal" results give a "reasonably accurate" approximation. This is based on the so-called Central Limit Theorem; see, eg.,
https://en.wikipedia.org/wiki/Central_limit_theorem
or http://davidmlane.com/hyperstat/A14043.html
or http://www.statisticalengineering.com/central_limit_theorem.htm .

For a "reasonable" non-normal underlying distribution, a sample size of n = 81 is likely large enough that normal-based estimates will be informative, if not absolutely accurate.

7. Nov 21, 2015

### toothpaste666

ahh ok what my book actually says is use z for samples of n>30 with σ known and if σ is not known replace σ with s and if the sample is n<30 And the population is normal use t. so since my sample is large enough, my solution to this problem should be close enough?

8. Nov 21, 2015