7.2.1 Simple Random Sampling

According to Bolfarine and Bussab (2005), simple random sampling (SRS) is the simplest and most important method for selecting a sample.

Consider a population or universe (U) with N elements:

$U=\left\{1,2,\dots ,N\right\}$

According to Bolfarine and Bussab (2005), planning and selecting the sample include the following steps:

1. (a) Using a random procedure (as, for example, through a table with random numbers or a gravity-pick machine), we must draw an element from population U with the same probability;
2. (b) We repeat the previous process until a sample with n observations is generated (the calculation of the size of a simple random sample will be studied in Section 7.4);
3. (c) When the value drawn is removed from U before of the next draw, we have the SRS without replacement process. In case drawing a unit more than once is allowed, we have the SRS with replacement process.

According to Bolfarine and Bussab (2005), from a practical point of view, an SRS without replacement is much more interesting, because it satisfies the intuitive principle that we do not gain more information in case the same unit appears more than once in the sample. On the other hand, an SRS with replacement has mathematical and statistical advantages, such as, the independence between the units drawn. Let’s now study each of them.

7.2.2 Systematic Sampling

According to Costa Neto (2002), when the elements of the population are sorted and periodically removed, we have a systematic sampling. Hence, for example, in a production line, we can remove an element at every 50 items produced.

As advantages of systematic sampling, in comparison to simple random sampling, we can mention that it is carried out in a much faster and cheaper way, besides being less susceptible to errors made by the interviewer during the survey. The main disadvantage is the possibility of having variation cycles, especially if these cycles coincide with the period when the elements are removed from the sample. For example, let’s suppose that at every 60 parts produced in a certain machine, one part is inspected; however, in this machine a certain flaw usually happens, so, at every 20 parts produced, one is defective.

Assuming that the elements of the population are sorted from 1 to N and that we already know the sample size (n), systematic sampling works as follows:

1. (a) We must determine the sampling interval (k), obtained by the quotient of the population size and the sample size:

   $k=\frac{N}{n}$

   

   
   This value must be rounded to the closest integer.
2. (b) In this phase, we introduce an element of randomness, choosing the starting unit. The first element chosen {X₁} can be an any element between 1 and k;
3. (c) After choosing the first element, after each k element, a new element is removed from the population. The process is repeated until it reaches the sample size (n):

   $X_1,X_1+k,X_1+2k,\dots ,X_1+\left(n-1\right)k$

   

7.2.3 Stratified Sampling

In this type of sampling, a heterogeneous population is stratified or divided into subpopulations or homogeneous strata, and, in each stratum, a sample is drawn. Hence, initially, we define the number of strata and, by doing that, we obtain the size of each stratum. For each stratum, we specify how many elements will be drawn from the subpopulation, and this can be a uniform or proportional allocation. According to Costa Neto (2002), uniform stratified sampling, from which we draw an equal number of elements in each stratum, is recommended when the strata are approximately the same size. In proportional stratified sampling, on the other hand, the number of elements in each stratum is proportional to the number of elements in the stratum.

According to Freund (2006), if the elements selected in each stratum are simple random samples, the global process (stratification followed by random sampling) is called (simple) stratified random sampling.

According to Freund (2006), stratified sampling works as follows:

1. (a) A population of size N is divided into k strata of sizes N₁, N₂, …, N_k;
2. (b) For each stratum, a random sample of size n_i (i = 1, 2, …, k) is selected, resulting in k subsamples of sizes n₁, n₂, …, n_k.

In uniform stratified sampling, we have:

$n_1=n_2=\dots =n_k$

where the sample size obtained from each stratum is:

$n_i=\frac{n}{k},\text{para}i=1,2,\dots ,k$

where n = n₁ + n₂ + … + n_k

In proportional stratified sampling, on the other hand, we have:

$\frac{n_1}{N_1}=\frac{n_2}{N_2}=\dots =\frac{n_k}{N_k}$

In proportional sampling, the sample size obtained from each stratum can be obtained according to the following expression:

$n_i=\frac{N_i}{N}\cdot n,\text{for}i=1,2,\dots ,k$

As examples of stratified sampling, we can mention the stratification of a city into neighborhoods, of a population by gender or age group, of customers by social class or of students by school.

The calculation of the size of a stratified sample will be studied in Section 7.4.3.

7.2.4 Cluster Sampling

In cluster sampling, the total population must be subdivided into groups of elementary units, called clusters. The sampling is done from the groups and not from the individuals in the population. Hence, we must randomly draw a sufficient number of clusters and the objects from these will form the sample. This type of sampling is called one-stage cluster sampling.

According to Bolfarine and Bussab (2005), one of the inconveniences of cluster sampling is the fact that elements in the same cluster tend to have similar characteristics. The authors show that the more similar the elements in the cluster are, the less efficient the procedure is. Each cluster must be a good representative of the population, that is, it must be heterogeneous, containing all kinds of participants. It is the opposite of stratified sampling.

According to Martins and Domingues (2011), cluster sampling is a simple random sampling in which the sample units are the clusters; however, it is less expensive.

When we draw elements in the clusters selected, we have a two-stage cluster sampling: in the first stage, we draw the clusters and, in the second, we draw the elements. The number of elements to be drawn depends on the variability in the cluster. The higher the variability, the more elements must be drawn. On the other hand, when the units in the cluster are very similar, it is not advisable nor necessary to draw all the elements, because they will bring the same kind of information (Bolfarine and Bussab, 2005).

Cluster sampling can be generalized to several stages.

The main advantages that justify the wide use of cluster sampling are: a) many populations are already grouped into natural or geographic subgroups, facilitating its application; b) it allows a substantial reduction in the costs to obtain the sample, without compromising its accuracy. In short, it is fast, cheap, and efficient. The only disadvantage is that clusters are rarely the same size, making it difficult to control the range of the sample. However, to overcome this problem, we have to use certain statistical techniques.

As examples of clusters, we can mention the production in a factory divided into assembly lines, company employees divided by area, students in a municipality divided by schools, or the population in a municipality divided into districts.

Consider the following notation for cluster sampling:

• N: population size;
• M: number of clusters into which the population was divided;
• N_i: cluster size i (i = 1, 2, ..., M);
• n: sample size;
• m: number of clusters drawn (m < M);
• n_i: cluster size i of the sample (i = 1, 2, ..., m), where n_i = N_i;
• b_i: cluster size i of the sample (i = 1, 2, ..., m), where b_i < n_i.

In short, one-stage cluster sampling adopts the following procedure:

1. (a) The population is divided into M clusters (C₁, …, C_M) with sizes that are not necessarily the same;
2. (b) According to a sample plan, usually SRS, we draw m clusters (m < M);
3. (c) All the elements of each cluster drawn constitute the global sample $\left(n_i=N_i\mathrm{and}\sum _{i=1}^m{n}_i=n\right)$.

The calculation of the number of clusters (m) will be studied in Section 7.4.4.

On the other hand, two-stage cluster sampling works as follows:

1. (a) The population is divided into M clusters (C₁, …, C_M) with sizes that are not necessarily the same;
2. (b) We must draw m clusters in the first stage, according to some kind of sample plan, usually SRS;
3. (c) From each cluster i drawn, of size n_i, we draw b_i elements in the second stage, according to the same or to another sample plan $\left(b_i<n_i\mathrm{and}n=\sum _{i=1}^m{b}_i\right)$.

7.3.1 Convenience Sampling

Convenience sampling is used when participation is voluntary or the sample elements are chosen due to convenience or simplicity, such as, friends, neighbors, or students. The advantage this method offers is that it allows researcher to obtain information in a quick and cheap way.

However, the sample process does not guarantee that the sample is representative of the population. It should only be employed in extreme situations and in special cases that justify its use.

7.3.2 Judgmental or Purposive Sampling

In judgmental or purposive sampling, the sample is chosen according to an expert’s opinion or previous judgment. It is a risky method due to possible mistakes made by the researcher in his prejudgment.

Using this type of sampling requires knowledge of the population and of the elements selected.

7.3.3 Quota Sampling

Quota sampling presents greater rigor when compared to other nonrandom samplings. For Martins and Domingues (2011), it is one of the most used sampling methods in market surveys and election polls.

Quota sampling is a variation of judgmental sampling. Initially, we set the quotas based on a certain criterion. Within the quotas, the selection of the sample items depends on the interviewer’s judgment.

Quota sampling can also be considered a nonprobability version of stratified sampling.

Quota sampling consists of three steps:

1. (a) We select the control variables or the population’s characteristics considered relevant for the study in question;
2. (b) We determine the percentage of the population (%) for each one of the relevant variable categories;
3. (c) We establish the size of the quotas (number of people to be interviewed that have the characteristics needed) for each interviewer, so that the sample can have the same proportions as the population.

The main advantages of quota sampling are the low costs, speed, and convenience or ease in which the interviewer can select elements. However, since the selection of elements is not random, there are no guarantees that the sample will be representative of the population. Hence, it is not possible to generalize the results of the survey to the population.

7.3.4 Geometric Propagation or Snowball Sampling

Geometric propagation or snowball sampling is widely used when the elements of the population are rare, difficult to access, or unknown.

In this method, we must identify one or more individuals from the target population, and these will identify the other individuals that are in the same population. The process is repeated until the objective proposed is achieved, that is, the point of saturation. The point of saturation is reached when the last respondents do not add new relevant information to the research, thus, repeating the content of previous interviews.

As advantages, we can mention: a) it allows the researcher to find the desired characteristic in the population; b) it is easy to apply, because the recruiting is done through referrals from other people who are in the population; c) low cost, because we need less planning and people; and d) it is efficient to enter populations that are difficult to access.

7.4.1 Size of a Simple Random Sample

This section discusses how to calculate the size of a simple random sample to estimate the mean (a quantitative variable) and the proportion (a binary variable) of finite and infinite populations, with a maximum estimation error B.

The estimation error (B) for the mean is the maximum difference that the researcher accepts between μ (population mean) and $\overline{X}$ (sample mean), that is, $B\ge \left|\mu -\overline{X}\right|$.

On the other hand, the estimation error (B) for the proportion is the maximum difference that the researcher accepts between p (proportion of the population) and $\hat{p}$ (proportion of the sample), that is, $B\ge \left|p-\hat{p}\right|$.

7.4.2 Size of the Systematic Sample

In systematic sampling, we use the same expressions as in simple random sampling (as studied in Section 7.4.1), according to the type of variable (quantitative or qualitative) and population (infinite or finite).

7.4.3 Size of the Stratified Sample

This section discusses how to calculate the size of a stratified sample to estimate the mean (a quantitative variable) and the proportion (a binary variable) of finite and infinite populations, with a maximum estimation error B.

The estimation error (B) for the mean is the maximum difference that the researcher accepts between μ (population mean) and $\overline{X}$ (sample mean), that is, $B\ge \left|\mu -\overline{X}\right|$.

On the other hand, the estimation error (B) for the proportion is the maximum difference that the researcher accepts between p (proportion of the population) and $\hat{p}$ (proportion of the sample), that is, $B\ge \left|p-\hat{p}\right|$.

Let’s use the following notation to calculate the size of the stratified sample, as follows:

• k: number of strata;
• N_i: size of stratum i, i = 1, 2,..., k;
• N = N₁ + N₂ + … + N_k (population size);
• W_i = N_i/N (weight or proportion of stratum i, with $\sum _{i=1}^k{W}_i=1$);
• μ_i: population mean of stratum i;
• σ_i²: population variance of stratum i;
• n_i: number of elements randomly selected from stratum i;
• n = n₁ + n₂ + … + n_k (sample size);
• ${\overline{X}}_i$: sample mean of stratum i;
• S_i²: sample variance of stratum i;
• p_i: proportion of elements that have the characteristic desired in stratum i;

• $q_i=1-p_i.$

  

7.4.4 Size of a Cluster Sample

This section discusses how to calculate the size of a one-stage and a two-stage cluster sample.

Let’s consider the following notation to calculate the size of a cluster sample:

• N: population size;
• M: number of clusters into which the population was divided;
• N_i: size of cluster i (i = 1, 2, ..., M);
• n: sample size;
• m: number of clusters drawn (m < M);
• n_i: size of cluster i from the sample drawn in the first stage (i = 1, 2, ..., m), where n_i = N_i;
• b_i: size of cluster i from the sample drawn in the second stage (i = 1, 2, ..., m), where b_i < n_i;
• $\overline{N}=N/M$ (average size of the population clusters);
• $\overline{n}=n/m$ (average size of the sample clusters);
• X_ij: j-th observation in cluster i;
• σ_dc²: population variance in the clusters;
• σ_ec²: population variance between clusters;
• σ_i²: population variance in cluster i;
• μ_i: population mean in cluster i;
• σ_c² = σ_dc² + σ_ec² (total population variance).

According to Bolfarine and Bussab (2005), the calculation of σ_dc² and σ_ec² is given by:

${\sigma }_{\mathit{dc}}^2=\frac{\sum _{i=1}^M\sum _{j=1}^{N_i}{\left(X_{\mathit{ij}}-{\mu }_i\right)}^2}{N}=\frac{1}{M}\cdot \sum _{i=1}^M\frac{N_i}{\overline{N}}\cdot {\sigma }_i^2$

${\sigma }_{\mathit{ec}}^2=\frac{1}{N}.\sum _{i=1}^M{N}_i.{\left({\mu }_i-\mu \right)}^2=\frac{1}{M}.\sum _{i=1}^M\frac{N_i}{\overline{N}}.{\left({\mu }_i-\mu \right)}^2$

Assuming that all the clusters are the same size, the previous expressions can be summarized as follows:

${\sigma }_{\mathit{dc}}^2=\frac{1}{M}.\sum _{i=1}^M{\sigma }_i^2$

${\sigma }_{\mathit{ec}}^2=\frac{1}{M}.\sum _{i=1}^M{\left({\mu }_i-\mu \right)}^2$

7.4.4.2 Size of a Two-Stage Cluster Sample

In this case, we assume that all the clusters are the same size. Based on Bolfarine and Bussab (2005), let’s consider the following linear cost function:

$C=c_1.n+c_2.b$

  (7.22)

where:

• c₁: observation cost of one unit from the first stage;
• c₂: observation cost of one unit from the second stage;
• n: sample size in the first stage;
• b: sample size in the second stage.

The optimal size for b that minimizes the linear cost function is given by:

$b^{\ast }=\frac{{\sigma }_{\mathit{dc}}}{{\sigma }_{\mathit{ec}}}.\sqrt{\frac{c_1}{c_2}}$

  (7.23)

Example 7.15

Calculating the Size of a Cluster Sample

Consider the members of a certain club in Sao Paulo (N = 4,500). We would like to estimate the average evaluation score (0 to 10) given by these members regarding the main features of the club. The population is divided into 10 groups of 450 elements each, based on their membership number. The estimate of the mean and of the population variance per group, based on previous surveys, can be seen in Table 7.E.8. Assuming that the cluster sampling is based on a single stage, determine the number of clusters that must be drawn, considering B = 2% and α = 1%.

Table 7.E.8

Mean and population variance per group

i	1	2	3	4	5	6	7	8	9	10
μi	7.4	6.6	8.1	7.0	6.7	7.3	8.1	7.5	6.2	6.9
σi2	22.5	36.7	29.6	33.1	40.8	51.7	39.7	30.6	40.5	42.7

Solution

From the data given to us, we have:

$N=4,500,M=10,\overline{N}=4,500/10=450,B=0.02,\text{and}{z}_{\alpha }=2.575.$

Since all the clusters are the same size, the calculation of σ_dc² and σ_ec² is given by:

${\sigma }_{\mathit{dc}}^2=\frac{1}{M}.\sum _{i=1}^M{\sigma }_i^2=\frac{22.5+36.7+\dots +42.7}{10}=36.79$

${\sigma }_{\mathit{ec}}^2=\frac{1}{M}.\sum _{i=1}^M{\left({\mu }_i-\mu \right)}^2=\frac{{\left(7.4-7.18\right)}^2+\dots +{\left(6.9-7.18\right)}^2}{10}=0.35$

Therefore, σ_c² = σ_dc² + σ_ec² = 36.79 + 0.35 = 37.14

The number of clusters to be drawn in one stage, for a finite population, is given by Expression (7.19):

$m=\frac{M.{\sigma }_c^2}{M.\frac{B^2.{\overline{N}}^2}{z_{\alpha }^2}+{\sigma }_c^2}=\frac{10\times 37.14}{10\times \frac{{0.02}^2\times {450}^2}{{2.575}^2}+37.14}=2.33\cong 3$

Therefore, the population of N = 4, 500 members is divided into M = 10 clusters with the same size (N_i = 450,   i = 1, ...10). From the total number of clusters, we must randomly draw m = 3 clusters. In one-stage cluster sampling, all the elements of each cluster drawn constitute the global sample (n = 450 × 3 = 1, 350).

From the sample selected, we can infer, with a 99% confidence level, that the difference between the sample mean and the real population mean will be 2%, at the most.

Table 7.1 shows a summary of the expressions used to calculate the sample size for the mean (a quantitative variable) and the proportion (a binary variable) of finite and infinite populations, with a maximum estimation error B, and for each type of random sampling (simple, systematic, stratified, and cluster).

Table 7.1

Expressions to Calculate the Size of Random Samples

Type of Random Sample	Estimating the Mean (Infinite Population)	Estimating the Mean (Finite Population)	Estimating the Proportion (Infinite Population)	Estimating the Proportion (Finite Population)
Simple	$n=\frac{{\sigma }^2}{B^2/z_{\alpha }^2}$	$n=\frac{N.{\sigma }^2}{\left(N-1\right).\frac{B^2}{z_{\alpha }^2}+{\sigma }^2}$	$n=\frac{p.q}{B^2/z_{\alpha }^2}$	$n=\frac{N.p.q}{\left(N-1\right).\frac{B^2}{z_{\alpha }^2}+p.q}$
Systematic	$n=\frac{{\sigma }^2}{B^2/z_{\alpha }^2}$	$n=\frac{N.{\sigma }^2}{\left(N-1\right).\frac{B^2}{z_{\alpha }^2}+{\sigma }^2}$	$n=\frac{p.q}{B^2/z_{\alpha }^2}$	$n=\frac{N.p.q}{\left(N-1\right).\frac{B^2}{z_{\alpha }^2}+p.q}$
Stratified	$n=\frac{\sum _{i=1}^k{W}_i.{\sigma }_i^2}{B^2/z_{\alpha }^2}$	$n=\frac{\sum _{i=1}^k{N}_i^2.{\sigma }_i^2/W_i}{N^2.\frac{B^2}{z_{\alpha }^2}+\sum _{i=1}^k{N}_i.{\sigma }_i^2}$	$n=\frac{\sum _{i=1}^k{W}_i.p_i.q_i}{B^2/z_{\alpha }^2}$	$n=\frac{\sum _{i=1}^k{N}_i^2.p_i.q_i/W_i}{N^2.\frac{B^2}{z_{\alpha }^2}+\sum _{i=1}^k{N}_i.p_i.q_i}$
One-stage Cluster	$m=\frac{{\sigma }_c^2}{B^2/z_{\alpha }^2}$	$m=\frac{M.{\sigma }_c^2}{M.\frac{B^2.{\overline{N}}^2}{z_{\alpha }^2}+{\sigma }_c^2}$	$m=\frac{1/M.\sum _{i=1}^M\frac{N_i}{\overline{N}}.p_i.q_i}{B^2/z_{\alpha }^2}$	$m=\frac{\sum _{i=1}^M\frac{N_i}{\overline{N}}.p_i.q_i}{M.\frac{B^2.{\overline{N}}^2}{z_{\alpha }^2}+1/M.\sum _{i=1}^M\frac{N_i}{\overline{N}}.p_i.q_i}$