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Statistics Made Easy

Random Selection vs. Random Assignment

Random selection and random assignment  are two techniques in statistics that are commonly used, but are commonly confused.

Random selection  refers to the process of randomly selecting individuals from a population to be involved in a study.

Random assignment  refers to the process of randomly  assigning  the individuals in a study to either a treatment group or a control group.

You can think of random selection as the process you use to “get” the individuals in a study and you can think of random assignment as what you “do” with those individuals once they’re selected to be part of the study.

The Importance of Random Selection and Random Assignment

When a study uses  random selection , it selects individuals from a population using some random process. For example, if some population has 1,000 individuals then we might use a computer to randomly select 100 of those individuals from a database. This means that each individual is equally likely to be selected to be part of the study, which increases the chances that we will obtain a representative sample – a sample that has similar characteristics to the overall population.

By using a representative sample in our study, we’re able to generalize the findings of our study to the population. In statistical terms, this is referred to as having  external validity – it’s valid to externalize our findings to the overall population.

When a study uses  random assignment , it randomly assigns individuals to either a treatment group or a control group. For example, if we have 100 individuals in a study then we might use a random number generator to randomly assign 50 individuals to a control group and 50 individuals to a treatment group.

By using random assignment, we increase the chances that the two groups will have roughly similar characteristics, which means that any difference we observe between the two groups can be attributed to the treatment. This means the study has  internal validity  – it’s valid to attribute any differences between the groups to the treatment itself as opposed to differences between the individuals in the groups.

Examples of Random Selection and Random Assignment

It’s possible for a study to use both random selection and random assignment, or just one of these techniques, or neither technique. A strong study is one that uses both techniques.

The following examples show how a study could use both, one, or neither of these techniques, along with the effects of doing so.

Example 1: Using both Random Selection and Random Assignment

Study:  Researchers want to know whether a new diet leads to more weight loss than a standard diet in a certain community of 10,000 people. They recruit 100 individuals to be in the study by using a computer to randomly select 100 names from a database. Once they have the 100 individuals, they once again use a computer to randomly assign 50 of the individuals to a control group (e.g. stick with their standard diet) and 50 individuals to a treatment group (e.g. follow the new diet). They record the total weight loss of each individual after one month.

Results:  The researchers used random selection to obtain their sample and random assignment when putting individuals in either a treatment or control group. By doing so, they’re able to generalize the findings from the study to the overall population  and  they’re able to attribute any differences in average weight loss between the two groups to the new diet.

Example 2: Using only Random Selection

Study:  Researchers want to know whether a new diet leads to more weight loss than a standard diet in a certain community of 10,000 people. They recruit 100 individuals to be in the study by using a computer to randomly select 100 names from a database. However, they decide to assign individuals to groups based solely on gender. Females are assigned to the control group and males are assigned to the treatment group. They record the total weight loss of each individual after one month.

Results:  The researchers used random selection to obtain their sample, but they did not use random assignment when putting individuals in either a treatment or control group. Instead, they used a specific factor – gender – to decide which group to assign individuals to. By doing this, they’re able to generalize the findings from the study to the overall population but they are  not  able to attribute any differences in average weight loss between the two groups to the new diet. The internal validity of the study has been compromised because the difference in weight loss could actually just be due to gender, rather than the new diet.

Example 3: Using only Random Assignment

Study:  Researchers want to know whether a new diet leads to more weight loss than a standard diet in a certain community of 10,000 people. They recruit 100 males athletes to be in the study. Then, they use a computer program to randomly assign 50 of the male athletes to a control group and 50 to the treatment group. They record the total weight loss of each individual after one month.

Results:  The researchers did not use random selection to obtain their sample since they specifically chose 100 male athletes. Because of this, their sample is not representative of the overall population so their external validity is compromised – they will not be able to generalize the findings from the study to the overall population. However, they did use random assignment, which means they can attribute any difference in weight loss to the new diet.

Example 4: Using Neither Technique

Study:  Researchers want to know whether a new diet leads to more weight loss than a standard diet in a certain community of 10,000 people. They recruit 50 males athletes and 50 female athletes to be in the study. Then, they assign all of the female athletes to the control group and all of the male athletes to the treatment group. They record the total weight loss of each individual after one month.

Results:  The researchers did not use random selection to obtain their sample since they specifically chose 100 athletes. Because of this, their sample is not representative of the overall population so their external validity is compromised – they will not be able to generalize the findings from the study to the overall population. Also, they split individuals into groups based on gender rather than using random assignment, which means their internal validity is also compromised – differences in weight loss might be due to gender rather than the diet.

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The Random Selection Experiment Method

Kendra Cherry, MS, is a psychosocial rehabilitation specialist, psychology educator, and author of the "Everything Psychology Book."

Emily is a board-certified science editor who has worked with top digital publishing brands like Voices for Biodiversity, Study.com, GoodTherapy, Vox, and Verywell.

When researchers need to select a representative sample from a larger population, they often utilize a method known as random selection. In this selection process, each member of a group stands an equal chance of being chosen as a participant in the study.

Random Selection vs. Random Assignment

How does random selection differ from  random assignment ? Random selection refers to how the sample is drawn from the population as a whole, whereas random assignment refers to how the participants are then assigned to either the experimental or control groups.

It is possible to have both random selection and random assignment in an experiment.

Imagine that you use random selection to draw 500 people from a population to participate in your study. You then use random assignment to assign 250 of your participants to a control group (the group that does not receive the treatment or independent variable) and you assign 250 of the participants to the experimental group (the group that receives the treatment or independent variable).

Why do researchers utilize random selection? The purpose is to increase the generalizability of the results.

By drawing a random sample from a larger population, the goal is that the sample will be representative of the larger group and less likely to be subject to bias.

Factors Involved

Imagine a researcher is selecting people to participate in a study. To pick participants, they may choose people using a technique that is the statistical equivalent of a coin toss.

They may begin by using random selection to pick geographic regions from which to draw participants. They may then use the same selection process to pick cities, neighborhoods, households, age ranges, and individual participants.

Another important thing to remember is that larger sample sizes tend to be more representative. Even random selection can lead to a biased or limited sample if the sample size is small.

When the sample size is small, an unusual participant can have an undue influence over the sample as a whole. Using a larger sample size tends to dilute the effects of unusual participants and prevent them from skewing the results.

Lin L.  Bias caused by sampling error in meta-analysis with small sample sizes .  PLoS ONE . 2018;13(9):e0204056. doi:10.1371/journal.pone.0204056

Elmes DG, Kantowitz BH, Roediger HL.  Research Methods in Psychology. Belmont, CA: Wadsworth; 2012.

By Kendra Cherry, MSEd Kendra Cherry, MS, is a psychosocial rehabilitation specialist, psychology educator, and author of the "Everything Psychology Book."

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AP®︎/College Statistics

Course: ap®︎/college statistics   >   unit 6.

• Statistical significance of experiment

Random sampling vs. random assignment (scope of inference)

• Conclusions in observational studies versus experiments
• Finding errors in study conclusions
• (Choice A)   Just the residents involved in Hilary's study. A Just the residents involved in Hilary's study.
• (Choice B)   All residents in Hilary's town. B All residents in Hilary's town.
• (Choice C)   All residents in Hilary's country. C All residents in Hilary's country.
• (Choice A)   Yes A Yes
• (Choice B)   No B No
• (Choice A)   Just the residents in Hilary's study. A Just the residents in Hilary's study.

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15.1: Random Selection

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• Page ID 10844

• Paul Pfeiffer
• Rice University

Introduction

The usual treatments deal with a single random variable or a fixed, finite number of random variables, considered jointly. However, there are many common applications in which we select at random a member of a class of random variables and observe its value, or select a random number of random variables and obtain some function of those selected. This is formulated with the aid of a counting or selecting random variable \(N\), which is nonegative, integer valued. It may be independent of the class selected, or may be related in some sequential way to members of the class. We consider only the independent case. Many important problems require optional random variables, sometimes called Markov times . These involve more theory than we develop in this treatment.

Some common examples:

Total demand of \(N\) customers— \(N\) independent of the individual demands. Total service time for \(N\) units— \(N\) independent of the individual service times. Net gain in \(N\) plays of a game— \(N\) independent of the individual gains. Extreme values of \(N\) random variables— \(N\) independent of the individual values. Random sample of size \(N\)— \(N\) is usually determined by propereties of the sample observed. Decide when to play on the basis of past results— \(N\) dependent on past

A useful model—random sums

As a basic model, we consider the sum of a random number of members of an iid class. In order to have a concrete interpretation to help visualize the formal patterns, we think of the demand of a random number of customers. We suppose the number of customers N is independent of the individual demands. We formulate a model to be used for a variety of applications.

A basic sequence \(\{X_n: 0 \le n\}\) [Demand of \(n\) customers] An incremental sequence \(\{Y_n:0 \le n\}\) [Individual demands] These are related as follows:

\(X_n = \sum_{k = 0}^{n} Y_k\) for \(n \ge 0\) and \(X_n = 0\) for \(n < 0\) \(Y_n = X_n - X_{n - 1}\) for all \(n\)

A counting random variable \(N\). If \(N = n\) then \(n\) of the \(Y_k\) are added to give the compound demand \(D\) (the random sum)

\(D = \sum_{k = 0}^{N} Y_k = \sum_{k = 0}^{\infty} I_{[N = k]} X_k = \sum_{k = 0}^{\infty} I_{\{k\}} (N) X_k\)

Note . In some applications the counting random variable may take on the idealized value \(\infty\). For example, in a game that is played until some specified result occurs, this may never happen, so that no finite value can be assigned to \(N\). In such a case, it is necessary to decide what value \(X_{\infty}\) is to be assigned. For \(N\) independent of the \(Y_n\) (hence of the \(X_n\)), we rarely need to consider this possibility.

Independent selection from an iid incremental sequence

We assume throughout , unless specifically stated otherwise, that: \(X_0 = Y_0 = 0\) \(\{Y_k: 1 \le k\}\) is iid \(\{N, Y_k: 0 \le k\}\) is an independent class

We utilize repeatedly two important propositions: \(E[h(D)|N = n] = E[h(X_n)]\), \(n \ge 0\) \(M_D (s) = g_N [M_Y (s)]\). If the \(Y_n\) are nonnegative integer valued, then so is \(D\) and \(g_D (s) = g_N[g_Y (s)]\)

We utilize properties of generating functions, moment generating functions, and conditional expectation. \(E[I_{\{n\}} (N) h(D)] = E[h(D)|N = n] P(N = n)\) by definition of conditional expectation, given an event, Now, \(I_{\{n\}} (N) h(D) = I_{\{n\}} (N) h(X_n)\) and \(E[I_{\{n\}} (N) h(X_n)] = P(N = n) E[h(X_n)]\). Hence \(E[h(D) |N = n] P(N = n) = P(N = n) E[h(X_n)]\). Division by \(P(N = n)\) gives the desired result. By the law of total probability (CE1b), \(M_D(s)= E[e^{sD}] = E\{E[e^{sD} |N]\}\). By proposition 1 and the product rule for moment generating functions,

\(E[e^{sD}|N = n] = E[e^{sX_n}] = \prod_{k = 1}^{n} E[e^{sY_k}] = M_Y^n (s)\)

\(M_D(s) = \sum_{n = 0}^{\infty} M_Y^n (s) P(N = n) = g_N[M_Y (s)]\)

A parallel argument holds for \(g_D\)

— □

Remark . The result on \(M_D\) and \(g_D\) may be developed without use of conditional expectation.

in the integer-valued case.

\(M_D(s) = E[e^{sD}] = \sum_{k = 0}^{\infty} E[I_{\{N = n\}} e^{sX_n}] = \sum_{k = 0}^{\infty} P(N = n) E[e^{sX_n}]\)

\(= \sum_{k = 0}^{\infty} P(N = n) M_Y^n (s) = g_N [M_Y (s)]\)

Example \(\PageIndex{1}\) A service shop

Suppose the number \(N\) of jobs brought to a service shop in a day is Poisson (8). One fourth of these are items under warranty for which no charge is made. Others fall in one of two categories. One half of the arriving jobs are charged for one hour of shop time; the remaining one fourth are charged for two hours of shop time. Thus, the individual shop hour charges \(Y_k\) have the common distribution

\(Y =\) [0 1 2] with probabilities \(PY =\) [1/4 1/2 1/4]

Make the basic assumptions of our model. Determine \(P(D \le 4)\).

\(g_N(s) = e^{8(s - 1)} g_Y (s) = \dfrac{1}{4} (1 + 2s + s^2)\)

According to the formula developed above,

\(g_D (s) = g_N [g_Y (s)] = \text{exp} ((8/4) (1 + 2s + s^2) - 8) = e^{4s} e^{2s^2} e^{-6}\)

Expand the exponentials in power series about the origin, multiply out to get enough terms. The result of straightforward but somewhat tedious calculations is

\(g_D (s) = e^{-6} ( 1 + 4s + 10s^2 + \dfrac{56}{3} s^3 + \dfrac{86}{3} s^4 + \cdot\cdot\cdot)\)

Taking the coefficients of the generating function, we get

\(P(D \le 4) \approx e^{-6} (1 + 4 + 10 + \dfrac{56}{3} + \dfrac{86}{3}) = e^{-6} \dfrac{187}{3} \approx 0.1545\)

Example \(\PageIndex{2}\) A result on Bernoulli trials

Suppose the counting random variable \(N\) ~ binomial \((n, p)\) and \(Y_i = I_{E_i}\), with \(P(E_i) = p_0\). Then

\(g_N = (q + ps)^n\) and \(g_Y (s) = q_0 + p_0 s\)

By the basic result on random selection, we have

\(g_D (s) = g_N [g_Y(s)] = [q + p(q_0 + p_0 s)]^n = [(1 - pp_0) + pp_0 s]^n\)

so that \(D\) ~ binomial \((n, pp_0)\).

In the next section we establish useful m-procedures for determining the generating function g D and the moment generating function \(M_D\) for the compound demand for simple random variables, hence for determining the complete distribution. Obviously, these will not work for all problems. It may helpful, if not entirely sufficient, in such cases to be able to determine the mean value \(E[D]\) and variance \(\text{Var} [D]\). To this end, we establish the following expressions for the mean and variance.

Example \(\PageIndex{3}\) Mean and variance of the compound demand

\(E[D] = E[N]E[Y]\) and \(\text{Var} [D] = E[N] \text{Var} [Y] + \text{Var} [N] E^2 [Y]\)

\(E[D] = E[\sum_{n = 0}^{\infty} I_{\{N = n\}} X_n] = \sum_{n = 0}^{\infty} P(N = n) E[X_n]\)

\(= E[Y] \sum_{n = 0}^{\infty} n P(N = n) = E[Y] E[N]\)

\(E[D^2] = \sum_{n = 0}^{\infty} P(N = n) E[X_n^2] = \sum_{n = 0}^{\infty} P(N = n) \{\text{Var} [X_n] + E^2 [X_n]\}\)

\(= \sum_{n = 0}^{\infty} P(N = n) \{n \text{Var} [Y] = n^2 E^2 [Y]\} = E[N] \text{Var} [Y] + E[N^2] E^2[Y]\)

\(\text{Var} [D] = E[N] \text{Var} [Y] + E[N^2] E^2 [Y] - E[N]^2 E^2[Y] = E[N] \text{Var} [Y] + \text{Var} [N] E^2[Y]\)

Example \(\PageIndex{4}\) Mean and variance for Example 15.1.1

\(E[N] = \text{Var} [N] = 9\). By symmetry \(E[Y] = 1\). \(\text{Var} [Y] = 0.25(0 + 2 + 4) - 1 = 0.5\). Hence,

\(E[D] = 8 \cdot 1 = 8\), \(\text{Var} [D] = 8 \cdot 0.5 + 8 \cdot 1 = 12\)

Calculations for the compound demand

We have m-procedures for performing the calculations necessary to determine the distribution for a composite demand \(D\) when the counting random variable \(N\) and the individual demands \(Y_k\) are simple random variables with not too many values. In some cases, such as for a Poisson counting random variable, we are able to approximate by a simple random variable.

The procedure gend

If the \(Y_i\) are nonnegative, integer valued, then so is \(D\), and there is a generating function. We examine a strategy for computation which is implemented in the m-procedure gend . Suppose

\(g_N (s) = p_0 + p_1 s + p_2 s^2 + \cdot\cdot\cdot p_n s^n\)

\(g_Y (s) = \pi_0 + \pi_1 s + \pi_2 s^2 + \cdot\cdot\cdot \pi_m s^m\)

The coefficients of \(g_N\) and \(g_Y\) are the probabilities of the values of \(N\) and \(Y\), respectively. We enter these and calculate the coefficients for powers of \(g_Y\):

\(\begin{array} {lcr} {gN = [p_0\ p_1\ \cdot\cdot\cdot\ p_n]} & {1 \times (n + 1)} & {\text{Coefficients of } g_N} \\ {y = [\pi_0\ \pi_1\ \cdot\cdot\cdot\ \pi_n]} & {1 \times (m + 1)} & {\text{Coefficients of } g_Y} \\ {\ \ \ \ \ \cdot\cdot\cdot} & { } & { } \\ {y2 = \text{conv}(y,y)} & {1 \times (2m + 1)} & {\text{Coefficients of } g_Y^2} \\ {y3 = \text{conv}(y,y2)} & {1 \times (3m + 1)} & {\text{Coefficients of } g_Y^3} \\ {\ \ \ \ \ \cdot\cdot\cdot} & { } & { } \\ {yn = \text{conv}(y,y(n - 1))} & {1 \times (nm + 1)} & {\text{Coefficients of } g_Y^n}\end{array}\)

We wish to generate a matrix \(P\) whose rows contain the joint probabilities. The probabilities in the \(i\)th row consist of the coefficients for the appropriate power of \(g_Y\) multiplied by the probability \(N\) has that value. To achieve this, we need a matrix, each of whose \(n + 1\) rows has \(nm + 1\) elements, the length of \(yn\). We begin by “preallocating” zeros to the rows. That is, we set \(P = \text{zeros}(n + 1, n\ ^*\ m + 1)\). We then replace the appropriate elements of the successive rows. The replacement probabilities for the \(i\)th row are obtained by the convolution of \(g_Y\) and the power of \(g_Y\) for the previous row. When the matrix \(P\) is completed, we remove zero rows and columns, corresponding to missing values of \(N\) and \(D\) (i.e., values with zero probability). To orient the joint probabilities as on the plane, we rotate \(P\) ninety degrees counterclockwise. With the joint distribution, we may then calculate any desired quantities.

Example \(\PageIndex{5}\) A compound demand

The number of customers in a major appliance store is equally likely to be 1, 2, or 3. Each customer buys 0, 1, or 2 items with respective probabilities 0.5, 0.4, 0.1. Customers buy independently, regardless of the number of customers. First we determine the matrices representing \(g_N\) and \(g_Y\). The coefficients are the probabilities that each integer value is observed. Note that the zero coefficients for any missing powers must be included.

Example \(\PageIndex{6}\) A numerical example

\(g_N (s) = \dfrac{1}{5} (1 + s + s^2 + s^3 + s^4)\) \(g_Y (s) = 0.1 (5s + 3s^2 + 2s^3\)

Note that the zero power is missing from \(gY\). corresponding to the fact that \(P(Y = 0) = 0\).

Example \(\PageIndex{7}\) Number of successes for random number \(N\) of trials.

We are interested in the number of successes in \(N\) trials for a general counting random variable. This is a generalization of the Bernoulli case in Example 15.1.2 . Suppose, as in Example 15.1.2 , the number of customers in a major appliance store is equally likely to be 1, 2, or 3, and each buys at least one item with probability \(p = 0.6\). Determine the distribution for the number \(D\) of buying customers.

We use \(gN\), \(gY\), and gend.

The procedure gend is limited to simple \(N\) and \(Y_k\), with nonnegative integer values. Sometimes, a random variable with unbounded range may be approximated by a simple random variable. The solution in the following example utilizes such an approximation procedure for the counting random variable \(N\).

Example \(\PageIndex{8}\) Solution of the shop time Example 15.1.1

The number \(N\) of jobs brought to a service shop in a day is Poisson (8). The individual shop hour charges \(Y_k\) have the common distribution \(Y =\) [0 1 2] with probabilities \(PY =\) [1/4 1/2 1/4].

Under the basic assumptions of our model, determine \(P(D \le 4)\).

Since Poisson \(N\) is unbounded, we need to check for a sufficient number of terms in a simple approximation. Then we proceed as in the simple case.

The m-procedures mgd and jmgd

The next example shows a fundamental limitation of the gend procedure. The values for the individual demands are not limited to integers, and there are considerable gaps between the values. In this case, we need to implement the moment generating function \(M_D\) rather than the generating function \(g_D\).

In the generating function case, it is as easy to develop the joint distribution for \(\{N, D\}\) as to develop the marginal distribution for \(D\). For the moment generating function, the joint distribution requires considerably more computation. As a consequence, we find it convenient to have two m-procedures: mgd for the marginal distribution and jmgd for the joint distribution.

Instead of the convolution procedure used in gend to determine the distribution for the sums of the individual demands, the m-procedure mgd utilizes the m-function mgsum to obtain these distributions. The distributions for the various sums are concatenated into two row vectors, to which csort is applied to obtain the distribution for the compound demand. The procedure requires as input the generating function for \(N\) and the actual distribution, \(Y\) and \(PY\), for the individual demands. For \(gN\), it is necessary to treat the coefficients as in gend. However, the actual values and probabilities in the distribution for Y are put into a pair of row matrices. If \(Y\) is integer valued, there are no zeros in the probability matrix for missing values.

Example \(\PageIndex{9}\) Noninteger values

A service shop has three standard charges for a certain class of warranty services it performs: $10, $12.50, and $15. The number of jobs received in a normal work day can be considered a random variable \(N\) which takes on values 0, 1, 2, 3, 4 with equal probabilities 0.2. The job types for arrivals may be represented by an iid class \(\{Y_i: 1 \le i \le 4\}\), independent of the arrival process. The \(Y_i\) take on values 10, 12.5, 15 with respective probabilities 0.5, 0.3, 0.2. Let \(C\) be the total amount of services rendered in a day. Determine the distribution for \(C\).

We next recalculate Example 15.1.6 , above, using mgd rather than gend.

Example \(\PageIndex{10}\) Recalculation of Example 15.1.6

In Example 15.1.6, we have

\(g_N (s) = \dfrac{1}{5} (1 + s + s^2 + s^3 + s^4)\) \(g_Y (s) = 0.1 (5s + 3s^2 + 2s^3)\)

The means that the distribution for \(Y\) is \(Y =\) [1 2 3] and \(PY =\) 0.1 * [5 3 2].

We use the same expression for \(gN\) as in Example 15.1.6.

As expected, the results are the same as those obtained with gend.

If it is desired to obtain the joint distribution for \(\{N, D\}\), we use a modification of mgd called jmgd . The complications come in placing the probabilities in the \(P\) matrix in the desired positions. This requires some calculations to determine the appropriate size of the matrices used as well as a procedure to put each probability in the position corresponding to its \(D\) value. Actual operation is quite similar to the operation of mgd, and requires the same data format.

A principle use of the joint distribution is to demonstrate features of the model, such as \(E[D|N = n] = nE[Y]\), etc. This, of course, is utilized in obtaining the expressions for \(M_D (s)\) in terms of \(g_N (s)\) and \(M_Y (s)\). This result guides the development of the computational procedures, but these do not depend upon this result. However, it is usually helpful to demonstrate the validity of the assumptions in typical examples.

Remark . In general, if the use of gend is appropriate, it is faster and more efficient than mgd (or jmgd). And it will handle somewhat larger problems. But both m-procedures work quite well for problems of moderate size, and are convenient tools for solving various “compound demand” type problems.

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As previously mentioned, one of the characteristics of a true experiment is that researchers use a random process to decide which participants are tested under which conditions. Random assignation is a powerful research technique that addresses the assumption of pre-test equivalence – that the experimental and control group are equal in all respects before the administration of the independent variable (Palys & Atchison, 2014).

Random assignation is the primary way that researchers attempt to control extraneous variables across conditions. Random assignation is associated with experimental research methods. In its strictest sense, random assignment should meet two criteria.  One is that each participant has an equal chance of being assigned to each condition (e.g., a 50% chance of being assigned to each of two conditions). The second is that each participant is assigned to a condition independently of other participants. Thus, one way to assign participants to two conditions would be to flip a coin for each one. If the coin lands on the heads side, the participant is assigned to Condition A, and if it lands on the tails side, the participant is assigned to Condition B. For three conditions, one could use a computer to generate a random integer from 1 to 3 for each participant. If the integer is 1, the participant is assigned to Condition A; if it is 2, the participant is assigned to Condition B; and, if it is 3, the participant is assigned to Condition C. In practice, a full sequence of conditions—one for each participant expected to be in the experiment—is usually created ahead of time, and each new participant is assigned to the next condition in the sequence as he or she is tested.

However, one problem with coin flipping and other strict procedures for random assignment is that they are likely to result in unequal sample sizes in the different conditions. Unequal sample sizes are generally not a serious problem, and you should never throw away data you have already collected to achieve equal sample sizes. However, for a fixed number of participants, it is statistically most efficient to divide them into equal-sized groups. It is standard practice, therefore, to use a kind of modified random assignment that keeps the number of participants in each group as similar as possible.

One approach is block randomization. In block randomization, all the conditions occur once in the sequence before any of them is repeated. Then they all occur again before any of them is repeated again. Within each of these “blocks,” the conditions occur in a random order. Again, the sequence of conditions is usually generated before any participants are tested, and each new participant is assigned to the next condition in the sequence. When the procedure is computerized, the computer program often handles the random assignment, which is obviously much easier. You can also find programs online to help you randomize your random assignation. For example, the Research Randomizer website will generate block randomization sequences for any number of participants and conditions ( Research Randomizer ).

Random assignation is not guaranteed to control all extraneous variables across conditions. It is always possible that, just by chance, the participants in one condition might turn out to be substantially older, less tired, more motivated, or less depressed on average than the participants in another condition. However, there are some reasons that this may not be a major concern. One is that random assignment works better than one might expect, especially for large samples. Another is that the inferential statistics that researchers use to decide whether a difference between groups reflects a difference in the population take the “fallibility” of random assignment into account. Yet another reason is that even if random assignment does result in a confounding variable and therefore produces misleading results, this confound is likely to be detected when the experiment is replicated. The upshot is that random assignment to conditions—although not infallible in terms of controlling extraneous variables—is always considered a strength of a research design. Note: Do not confuse random assignation with random sampling. Random sampling is a method for selecting a sample from a population; we will talk about this in Chapter 7.

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Random selection is how you draw the sample of people for your study from a population. Random assignment is how you assign the sample that you draw to different groups or treatments in your study.

It is possible to have both random selection and assignment in a study. Let’s say you drew a random sample of 100 clients from a population list of 1000 current clients of your organization. That is random sampling. Now, let’s say you randomly assign 50 of these clients to get some new additional treatment and the other 50 to be controls. That’s random assignment.

It is also possible to have only one of these (random selection or random assignment) but not the other in a study. For instance, if you do not randomly draw the 100 cases from your list of 1000 but instead just take the first 100 on the list, you do not have random selection. But you could still randomly assign this nonrandom sample to treatment versus control. Or, you could randomly select 100 from your list of 1000 and then nonrandomly (haphazardly) assign them to treatment or control.

And, it’s possible to have neither random selection nor random assignment. In a typical nonequivalent groups design in education you might nonrandomly choose two 5th grade classes to be in your study. This is nonrandom selection. Then, you could arbitrarily assign one to get the new educational program and the other to be the control. This is nonrandom (or nonequivalent) assignment.

Random selection is related to sampling . Therefore it is most related to the external validity (or generalizability) of your results. After all, we would randomly sample so that our research participants better represent the larger group from which they’re drawn. Random assignment is most related to design . In fact, when we randomly assign participants to treatments we have, by definition, an experimental design . Therefore, random assignment is most related to internal validity . After all, we randomly assign in order to help assure that our treatment groups are similar to each other (i.e., equivalent) prior to the treatment.

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Counteracting Methodological Errors in Behavioral Research pp 39–54 Cite as

Random Assignment

• Gideon J. Mellenbergh 2  
• First Online: 17 May 2019

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A substantial part of behavioral research is aimed at the testing of substantive hypotheses. In general, a hypothesis testing study investigates the causal influence of an independent variable (IV) on a dependent variable (DV) . The discussion is restricted to IVs that can be manipulated by the researcher, such as, experimental (E- ) and control (C- ) conditions. Association between IV and DV does not imply that the IV has a causal influence on the DV . The association can be spurious because it is caused by an other variable (OV). OVs that cause spurious associations come from the (1) participant, (2) research situation, and (3) reactions of the participants to the research situation. If participants select their own (E- or C- ) condition or others select a condition for them, the assignment to conditions is usually biased (e.g., males prefer the E-condition and females the C-condition), and participant variables (e.g., participants’ sex) may cause a spurious association between the IV and DV . This selection bias is a systematic error of a design. It is counteracted by random assignment of participants to conditions. Random assignment guarantees that all participant variables are related to the IV by chance, and turns systematic error into random error. Random errors decrease the precision of parameter estimates. Random error variance is reduced by including auxiliary variables into the randomized design. A randomized block design includes an auxiliary variable to divide the participants into relatively homogeneous blocks, and randomly assigns participants to the conditions per block. A covariate is an auxiliary variable that is used in the statistical analysis of the data to reduce the error variance. Cluster randomization randomly assigns clusters (e.g., classes of students) to conditions, which yields specific problems. Random assignment should not be confused with random selection. Random assignment controls for selection bias , whereas random selection makes possible to generalize study results of a sample to the population.

• Cluster randomization
• Cross-over design
• Independent and dependent variables
• Random assignment and random selection
• Randomized block design

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Mellenbergh, G.J. (2019). Random Assignment. In: Counteracting Methodological Errors in Behavioral Research. Springer, Cham. https://doi.org/10.1007/978-3-030-12272-0_4

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