Analysis of Simple Random Samples
The key to analyzing data from simple random samples is computing the variability of the sample estimate.
When you know the variability of the sample estimate, you can construct confidence intervals; and you can test
hypotheses.
Notation
The following notation is helpful, when we talk about analyzing data from simple random samples.

Σ = Summation symbol, used to compute sums
over the sample. ( To illustrate its use, Σ
x_{i} = x_{1} + x_{2} + x_{3} + ... + x_{m1}
+ x_{m} )
The Variability of the Estimate
The
precision
of a
sample design
is directly related to the variability of the estimate.
Two common measures of variability are the
standard deviation (SD) of the estimate and the
standard error (SE) of the estimate. The tables below show how to
compute both measures, assuming that the sample method is
simple random sampling.
The first table shows how to compute variability for a mean score. Note
that the table shows four sample designs. In two of the designs, the true
population variance is known; and in two, it is estimated from sample data.
Also, in two of the designs, the researcher sampled with replacement; and in
two, without replacement.
Population variance

Replacement strategy 
Variability 
Known

With replacement 
SD = sqrt [ σ^{2} / n ] 
Known

Without replacement 
SD = sqrt { [ ( N  n ) / ( N  1 ) ] * σ^{2} / n } 
Estimated

With replacement 
SE = sqrt [ s^{2} / n ] 
Estimated

Without replacement 
SE = sqrt { [ ( N  n ) / ( N  1 ) ] * s^{2} / n } 
The next table shows how to compute variability for a proportion. Like
the previous table, this table shows four sample designs. In two of the
designs, the true population proportion is known; and in two, it is estimated
from sample data. Also, in two of the designs, the researcher sampled with replacement;
and in two, without replacement.
Population proportion

Replacement strategy 
Variability 
Known

With replacement 
SD = sqrt [ P * ( 1  P ) / n ] 
Known

Without replacement 
SD = sqrt { [ ( N  n ) / ( N  1 ) ] * P * ( 1  P ) / n } 
Estimated

With replacement 
SE = sqrt [ p * ( 1  p ) / ( n  1 ) ] 
Estimated

Without replacement 
SE = sqrt [ [ ( N  n ) / ( N  1 ) ] * p * ( 1  p ) / n ] 
Sample Problem
This section presents a sample problem that illustrates how to analyze survey
data when the sampling method is simple random sampling. (In a
subsequent lesson, we revisit this problem and see how simple
random sampling compares to other sampling methods.)
Sample Planning Wizard
The analysis of data collected via simple random sampling can be complex and
timeconsuming. Stat Trek's Sample Planning Wizard can help. The Wizard computes
survey precision, sample size requirements, costs, etc., as well as estimates
population parameters and tests hypotheses. It also creates a summary report that
lists key findings and documents analytical techniques. The Wizard
is free. You can find the Sample Planning Wizard in Stat Trek's
main menu under the Stat Tools tab. Or you can tap the button below.
Sample Planning Wizard
Problem 1
At the end of every school year, the state administers a reading test to a
simple random sample drawn without replacement from a population of 20,000
third graders. This year, the test was administered to 36 students selected via
simple random sampling. The test score from each sampled student is shown
below:
50, 55, 60, 62, 62, 65, 67, 67, 70, 70, 70, 70, 72, 72, 73, 73, 75, 75,
75, 78, 78, 78, 78, 80, 80, 80, 82, 82, 85, 85, 85, 88, 88, 90, 90, 90
Using sample data, estimate the mean reading achievement level in the
population. Find the margin
of error and the
confidence interval. Assume a 95%
confidence level.
Solution: Elsewhere on this website, we described
how to compute the confidence interval for a mean score. We
follow that process below.
Therefore, the 95% confidence interval is 71.75 to 78.25. And the margin
of error is equal to 3.25. That is, we are 95%
confident that the true population mean is in the range
defined by 75 + 3.25.