Probability a sample mean will fall in a range

[toothpaste666]

Homework Statement

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?

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?

 

[Orodruin]

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

 

[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.

 

[Ray Vickson]

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.

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.

 

[toothpaste666]

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

 

[Ray Vickson]

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

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.

 

[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?

 

[Ray Vickson]

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?

Asked and answered.

 

[toothpaste666]

thank you