Distribution of the Sample Mean

The statistic used to estimate the mean of a population, μ, is the sample mean, .


If X has a distribution with mean μ, and standard deviation σ, and is approximately normally distributed or n is large, then is approximately normally distributed with mean μ and standard error..

When σ Is Known


If the standard deviation, σ, is known, we can transform to an approximately standard normal variable, Z:




From the previous example, μ=20, and σ=5. Suppose we draw a sample of size n=16 from this population and want to know how likely we are to see a sample average greater than 22, that is P(> 22)?


So the probability that the sample mean will be >22 is the probability that Z is > 1.6 We use the Z table to determine this:

P( > 22) = P(Z > 1.6) = 0.0548.



Exercise: Suppose we were to select a sample of size 49 in the example above instead of n=16. How will this affect the standard error of the mean? How do you think this will affect the probability that the sample mean will be >22? Use the Z table to determine the probability.



When σ Is Unknown

If the standard deviation, σ, is unknown, we cannot transform to standard normal. However, we can estimate σ using the sample standard deviation, s, and transform to a variable with a similar distribution, the t distribution. There are actually many t distributions, indexed by degrees of freedom (df). As the degrees of freedom increase, the t distribution approaches the standard normal distribution.




If X is approximately normally distributed, then

has a t distribution with (n-1) degrees of freedom (df)


Using the t-table

Note: If n is large, then t is approximately normally distributed.




The z table gives detailed correspondences of P(Z>z) for values of z from 0 to 3, by .01 (0.00, 0.01, 0.02, 0.03,…2.99. 3.00). The (one-tailed) probabilities are inside the table, and the critical values of z are in the first column and top row.

The t-table is presented differently, with separate rows for each df, with columns representing the two-tailed probability, and with the critical value in the inside of the table.

The t-table also provides much less detail; all the information in the z-table is summarized in the last row of the t-table, indexed by df = ∞.

So, if we look at the last row for z=1.96 and follow up to the top row, we find that

 P(|Z| > 1.96) = 0.05


Exercise: What is the critical value associated with a two-tailed probability of 0.01?


Now, suppose that we want to know the probability that Z is more extreme than 2.00. The t-table gives us

P(|Z| > 1.96) = 0.05


P(|Z| > 2.326) = 0.02


So, all we can say is that P(|Z| > 2.00) is between 2% and 5%, probably closer to 5%! Using the z-table, we found that it was exactly 4.56%.


In the previous example we drew a sample of n=16 from a population with μ=20 and σ=5. We found that the probability that the sample mean is greater than 22 is P( > 22) = 0.0548. Suppose that is unknown and we need to use s to estimate it. We find that s = 4. Then we calculate t, which follows a t-distribution with df = (n-1) = 24.


From the tables we see that the two-tailed probability is between 0.01 and 0.05.

P(|T| > 1.711) = 0.05


P(|T| > 2.064) = 0.01


To obtain the one-tailed probability, divide the two-tailed probability by 2.


P(T > 1.711) = ½ P(|T| > 1.711) = ½(0.05) = 0.025


P(T > 2.064) = ½ P(|T| > 2.064) = ½(0.01) = 0.005

So the probability that the sample mean is greater than 22 is between 0.005 and 0.025 (or between 0.5% and 2.5%)


Exercise: . If μ=15, s=6, and n=16, what is the probability that >18 ?