Check Your Knowledge

Randomness and Sampling

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Which of the following is a true statement about the standard error of the mean?

a) It provides a measure of the spread of observations in the sample data

b) It provides a measure of the spread of observations in the population

c) It provides a measure of the precision of the sample mean as an estimate of the population mean.

d) It is dependent on sample size

For a discussion of the above see Standard Error of the Mean.

c and d are both true.


The United Network for Organ Sharing (UNOS) is a registry of all organ transplants done in the United States. For example, according to their Website, in 2015 there were 30,969 organ transplants of which 17,878 involved liver transplants (58%).  We can summarize the data by derived measures, for example that 58% of the transplants done in the United States in 2015 involved liver transplants. Here we have the entire population of individuals who have had a transplant during these years.

Often in health sciences research we do not have access to the whole population but instead have a sample of the individuals.  On any sample we can calculate out a statistic based on the data collected (e.g. the sample mean.) A easy to use online calculator for generating sample statistics for relatively small samples (along with brief definitions) is the "Descriptive Statistics Calculator".  If our sampling method is random then we expect the calculated sample statistics to vary depending on which individuals are selected from the larger population. Sampling variation is this variation that we see in a sample statistic from sample to sample.

The distibution of a measurement or variable is a specification of the probability or frequency with which a specific value or range of values occurs.

The standard error of the mean, also called the standard deviation of the mean, is a method used to estimate the standard deviation of a sampling distribution.


For more information on sampling variation check out the following short articles at the British Medical Journal:

Try It Yourself


Use the Central Limit Theorem to calculate the probability that a sample mean from a random sample of 25 individuals will be greater than 12 if in the population the variable has a normal distribution with mean of 2 and a standard deviation of 5?

Answer: We want to find Prob( Xbar > 12) where Xbar is the sample mean.

Prob( Xbar > 12) = Prob( (Xbar - mu) / std > (12 - 2) / 5) 

= Prob (Z > 2.0)

= 0.23 (from online probability calculator - see below)

Using the online app

  • Go online to

  • Select "Statistical distributions and interpreting P values"

  • Select "Calculate P value from z, t, F, r or chi-square"

  • Enter 2 in the box for Z

  • Select "Compute P"

  • Divide this by 2 to get the one-sided calculation for greater than calculations.


data one; p=1-probnorm(2); run;
proc print; run;