### Beyond AP Statistics

#### Probability Basics

#### Small Samples

#### Distributions

#### Power

### Beyond AP Statistics

#### Probability Basics

#### Small Samples

#### Distributions

#### Power

# Sample Size: Stratified Random Samples

The precision and cost of a stratified design are influenced by the way that sample elements are allocated to strata.

## How to Assign Sample to Strata

One approach is proportionate stratification. With proportionate stratification, the sample size of each stratum is proportionate to the population size of the stratum. Strata sample sizes are determined by the following equation :

n_{h} = ( N_{h} / N ) * n

where n_{h} is the sample size for stratum *h*, N_{h}
is the population size for stratum *h*, N is total population size, and n
is total sample size.

Another approach is disproportionate stratification, which can be a better choice (e.g., less cost, more precision) if sample elements are assigned correctly to strata. To take advantage of disproportionate stratification, researchers need to answer such questions as:

- Given a fixed budget, how should sample be allocated to get the most precision from a stratified sample?
- Given a fixed sample size, how should sample be allocated to get the most precision from a stratified sample?
- Given a fixed budget, what is the most precision that I can get from a stratified sample?
- Given a fixed sample size, what is the most precision that I can get from a stratified sample?
- What is the smallest sample size that will provide a given level of survey precision?
- What is the minimum cost to achieve a given level of survey precision?
- Given a particular sample allocation plan, what level of precision can I expect?
- And so on.

Although a consideration of all these questions is beyond the scope of this tutorial, the remainder of this lesson does address the first two questions. (To answer the other questions, as well as the first two questions, consider using the Sample Planning Wizard.)

## Sample Planning Wizard

Stat Trek's Sample Planning Wizard can help you find the right sample allocation plan for your stratified design. You specify your main goal - maximize precision, minimize cost, stay within budget, etc. Based on your goal, the Wizard prompts you for the necessary inputs and handles all computations automatically. It tells you the best sample size for each stratum. The Wizard creates a summary report that lists key findings, including the margin of error. And it describes analytical techniques. And 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## How to Maximize Precision, Given a Stratified Sample With a Fixed Budget

The ideal sample allocation plan would provide the most precision for the least
cost. **Optimal allocation** does just that. Based on optimal
allocation, the best sample size for stratum *h* would be:

n_{h} = n * [ ( N_{h} * σ_{h} ) / sqrt( c_{h} )
] / [ Σ ( N_{i} * σ_{i} ) /
sqrt( c_{i} ) ]

where n_{h} is the sample size for stratum *h*, n is total sample
size, N_{h} is the population size for stratum *h*, σ_{h} is
the standard deviation of stratum *h*, and c_{h} is the direct
cost to sample an individual element from stratum *h*. Note that c_{h}
does not include indirect costs, such as overhead costs.

The effect of the above equation is to sample more heavily from a stratum when

- The cost to sample an element from the stratum is low.
- The population size of the stratum is large.
- The variability within the stratum is large.

## How to Maximize Precision, Given a Stratified Sample With a Fixed Sample Size

Sometimes, researchers want to find the sample allocation plan that provides the
most precision, given a fixed sample size. The solution to this problem is a
special case of optimal allocation, called **Neyman allocation**.

The equation for Neyman allocation can be derived from the equation for optimal
allocation by assuming that the direct cost to sample an individual element is
equal across strata. Based on Neyman allocation, the best sample size for
stratum *h* would be:

n_{h} = n * ( N_{h} * σ_{h} )
/ [ Σ ( N_{i} * σ_{i} ) ]

where n_{h} is the sample size for stratum *h*, n is total sample
size, N_{h} is the population size for stratum *h*, and σ_{h}
is the standard deviation of stratum *h*.

## Sample Problem

This section presents a sample problem that illustrates how to maximize precision, given a fixed sample size and a stratified sample. (In a subsequent lesson, we re-visit this problem and see how stratified sampling compares to other sampling methods.)

**Problem 1**

At the end of every school year, the state administers a reading test to a sample of 36 third graders. The school system has 20,000 third graders, half boys and half girls. The results from last year's test are shown in the table below.

Stratum | Mean score | Standard deviation |
---|---|---|

Boys | 70 | 10.27 |

Girls | 80 | 6.66 |

This year, the researchers plan to use a stratified sample, with one stratum consisting of boys and the other, girls. Use the results from last year to answer the following questions?

- To maximize precision, how many sampled students should be boys and how many should be girls?
- What is the mean reading achievement level in the population?
- Compute the confidence interval.
- Find the margin of error

Assume a 95% confidence level.

*Solution:* The first step is to decide how to allocate sample in order to
maximize precision. Based on Neyman allocation, the best sample size for
stratum *h* is:

n_{h} = n * ( N_{h} * σ_{h} )
/ [ Σ ( N_{i} * σ_{i} ) ]

where n_{h} is the sample size for stratum *h*, n is total sample
size, N_{h} is the population size for stratum *h*, and σ_{h}
is the standard deviation of stratum *h*. By this equation, the number of
boys in the sample is:

n_{boys} = 36 * ( 10,000 * 10.27 ) / [ ( 10,000 * 10.27 ) + ( 10,000 *
6.67 ) ]

n_{boys} = 21.83

Therefore, to maximize precision, the total sample of 36 students should consist of 22 boys and (36 - 22) = 14 girls.

The remaining questions can be answered during the process of computing the confidence interval. Elsewhere on this website, we described how to compute a confidence interval. We employ that process below.

- Identify a sample statistic. For this problem, we use
the overall sample mean to estimate the population mean. To compute the overall
sample mean, we use the following equation (which was introduced
in a
previous lesson):
x = Σ ( N

_{h}/ N ) * x_{h}x = ( 10,000/20,000 ) * 70 + ( 10,000/20,000 ) * 80

x = 75

Therefore, based on data from the sample strata, we estimate that the mean reading achievement level in the population is equal to 75. - Select a confidence level. In this analysis, the confidence level is defined for us in the problem. We are working with a 95% confidence level.
- Find the margin of error. Elsewhere on this site, we show
how to compute the margin of error when the sampling
distribution is approximately normal. The key steps are
shown below.
- Find standard deviation or standard error. The equation to
compute the standard error was introduced in a
previous lesson. We use that equation here:
SE = (1 / N) * sqrt { Σ [ N

_{h}^{2}* ( 1 - n_{h}/N_{h}) * s_{h}^{2}/ n_{h}] }SE = (1 / 20,000) * sqrt { [ 10,000

^{2}* ( 1 - 22/10,000 ) * (10.27)^{2}/ 22 ] + [ 10,000^{2}* ( 1 - 14/10,000 ) * (6.66)^{2}/ 14 ] }SE = 1.41

Thus, the standard deviation of the sampling distribution (i.e., the standard error) is 1.41.

- Find critical value. The critical value is a factor used to
compute the margin of error. We express the critical
value as a
z-score.
To find the critical value, we take these steps.
- Compute alpha (α):
α = 1 - (confidence level / 100)

α = 1 - 99/100 = 0.01

- Find the critical probability (p*):
p* = 1 - α/2 = 1 - 0.05/2 = 0.975

- The critical value is the z-score having a cumulative probability equal to 0.975. From the Normal Distribution Calculator, we find that the critical value is 1.96.

- Compute alpha (α):
- Compute margin of error (ME):
ME = critical value * standard error

ME = 1.96 * 1.41 = 2.76

- Find standard deviation or standard error. The equation to
compute the standard error was introduced in a
previous lesson. We use that equation here:
- Specify the confidence interval. The range of the confidence
interval is defined by the
*sample statistic*__+__*margin of error*. And the uncertainty is denoted by the confidence level. Thus, with this sample design, we are 95% confident that the sample estimate of reading achievement is 75__+__2.76.

In summary, given a total sample size of 36 students, we can get the greatest precision from a stratified sample if we sample 22 boys and 14 girls. This results in a 95% confidence interval of 72.24 to 77.76. The margin of error is 2.76.

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