Random Sampling and Statistics

• This text introduces basic statistical concepts of random sampling, parameters, statistics, sampling distributions, and the central limit theorem.
• The learning objectives are to:
  • Illustrate random sampling
  • Distinguish between a parameter and a statistic
  • Find possible values of a random variable
  • Recognize sampling distributions of statistics, particularly the sample mean.

Population vs. Sample

• A population refers to the entire set of individuals sharing a common interest.
• Populations are often too large to collect data from all members.
• A sample is a smaller subset taken from this population.
• Samples are chosen for the purpose of convenience.
• The goal of a population and sample is to select a representation that captures the essential information of a specific characteristic in the population.
• It also shows the relationship between the sample and the population.

Sampling

• Sampling is applied when selecting a group or sample to represent the entire population.
• It is a statistical analysis process where a predetermined number of observations are taken from a larger population.
• Sampling reduces the number of respondents needed to represent the entire population.

Random Sampling

• Random sampling gives each member of the larger population an equal chance of being selected as part of a sample.
• Subjects of the population get an equal opportunity for selection as a representative sample.
• Types of random sampling include:
  • Simple Random Sampling
  • Systematic Random Sampling
  • Stratified Random Sampling
  • Cluster Random Sampling

Simple Random Sampling

• It is a common method to select a sample where everyone in the population is randomly chosen and has an equal chance of being part of the sample.
• The selection process is random, introducing no bias in the selection of individuals.
• Slovin's Formula can be applied to calculate the needed sample size from a given population.

Slovin's Formula

• Used to calculate the appropiate sample size for a populations.
• The formula is: n = N / (1 + Ne^2)
• N is the population size.
• e is the margin of error.
• Example:
  • If N = 1000 and e = 15% (0.15), then n = 1000 / (1 + 1000 * 0.15^2) = 42.55 ≈ 43.

Systematic Random Sampling

• A structured method of selecting a sample.
• The population is ordered, and a random starting point is chosen.
• List all the elements in the population and then select every nth element from that list, this is reliable, accurate and useful for long population lists.
• To calculate the interval (k) for selecting sample participants, use the formula k = N / n.
• N is the population size.
• n is the sample size.
• Example:
  • With N = 3000 and n = 600, then k = 3000 / 600 = 5.
• Once the interval (k) is determined, use a table of random numbers to choose the starting point.
  • If the interval is 5, select a random number between 1 and 5 to identify the starting point on a list of 3000 subjects; from there, select every 5th name on the list until the desired sample size is met.

Stratified Random Sampling

• This technique divides a population into subgroups or strata based on specific characteristics or attributes of interest.
• Each subgroup represents a homogeneous subset of the population in respect to the chosen characteristic.
• A random sample is selected for each stratum, and samples from different strata are combined to form the final representative sample.
• Example:
  • A population of 2000 students contains of 800 from grade school, 700 from high school, and 500 from senior high school, if a sample size of 600 is needed it can be determined using Slovin's formula to apply an error margin of 3.412% ≈ 0.03412.
  • You would allocate the samples of each stratum to obtain a proportion each sample is devided by the total population (2000), obtaining a proportion of 0.3 (600 / 2000 = 0.3)

Cluster Random Sampling

• It is applied when sampling individual elements directly from a large and diverse population is impractical or difficult.
• Instead of sampling individuals, the population is divided into clusters or groups.
• A random selection of clusters is made, including all elements within the selected clusters in the sample.
• This method is useful when the population is geographically dispersed or when reaching individual elements is costly or time-consuming.
• Example of simple random sampling: selecting 50 students from a university's enrollment list through a random number generator for a satisfaction survey.
• Example of systematic random sampling: surveying every 10th customer entering a grocery store for feedback on their shopping experience.
• Example of Stratified random sampling: Dividing a city's population into income groups, then randomly selecting 100 individuals from each group for a study on spending habits.
• Example of cluster random sampling: Randomly selecting 5 neighborhoods in a city and surveying all households within those neighborhoods to study local healthcare preferences.

Parameter

• A parameter is a numerical summary or characteristic of a population.
• The parameter value is a fixed value that describes a specific feature of the entire population being studied.
• In statistical analysis, parameters serve as the true values that researchers aim to estimate based on the collected data.
• Examples of parameters include the population mean, population standard deviation, and population proportion.
• It is the measurement or quantity that describes the population.

Statistics

• A statistic is a numerical summary or characteristic of a sample.
• It is a computed value derived from a subset of the population, known as a sample.
• Common examples include the sample mean, sample standard deviation, and sample proportion.
• It is the measurement or quantity that describes the sample.

Example Parameter and Statistics Problem

• Given pre-calculus scores from a grade of 11 STEM periodical examination, scores are (52, 33, 36, 26, 32, 21, 38, 43, 35, 29), computing the population mean, variance, and standard deviation are parameter metrics
• Alternatively to calculate the statistic metric from the same sample data; 5 random numbers can be chosen (52, 38, 43, 26, 29) to calculate the sample mean using the formula: x = (Σx) / N
  • Where Σx is the summation of x (Sum of measures) and N = Number of elements in the sample
  • This results in x = 37.60
• Using that average we can calculate the sample variance using this formula: s^2 = (Σ(x – x)²)/ n-1
  • Here s^2 = sample variance, x = given data x = sample mean n = number of elements in the sample
  • This results in s^2 = 89.04
• The sample standard deviation can be found by taking the square root of the sample variance. Use the formula: s = √(Σ(x-x)²)/n
  • This results in s = 9.44

Sampling Distribution

• Define the sampling distribution of the sample mean for a normal population when the variance is known.
• Define the sampling distribution of the sample mean for a normal population when the variance is unknown.
• Find the mean and variance of the sampling distribution of the sample mean.
• A sampling distribution of sample means shows the frequencies of the means obtained from all possible random samples of a certain size taken from a population
• These sample means can be either lower or higher than the mean of the population.
• Steps to construct the sampling distribution of the mean:
  • Determine the number of sets of all possible random samples that can be drawn from the given population by using the formula, NCn, where N is the population size and n is the sample size.
  • List all the possible samples and compute the mean of each sample.
  • Construct the sampling distribution.
  • Construct a histogram of the sampling distribution of the means.

Sampling Distribution, Example 1

• A population consists of the numbers 3, 5, 9, 10, and 8.
• The example shows computing all possible sample sizes of 3 and making a bar chart:
  • Use the formula of combinations: nCn = N! / n! (N - n)!
  • So nCn = 5! / 3! (5-3)! --> 5 x 4 / 2 x 1 = 10
  • Giving you 10 possible sample sizes of 3 from the given data.
• List all the possible samples and compute the mean of each sample.
  • This gives you the following result
    • 3, 5, 9 (5.67)
    • 3, 5, 10 (6)
    • 3, 5, 8 (5.33)
    • 3, 9, 10 (7.33)
    • 3, 9, 8 (6.67)
    • 3, 10, 8 (7)
    • 5, 9, 10 (8)
    • 5, 9, 8 (7.33)
    • 5, 10, 8 (7.67)
    • 9, 10, 8 (9)
• Note: sample mean is to be placed in ascending order*
  • This then needs to be arranged into Frequency
    • 5.33 (1)
    • 5.67 (1)
    • 6 (1)
    • 6.67 (1)
    • 7 (1)
    • 7.33 (2)
    • 7.67 (1)
    • 8 (1)
    • 9 (1)
• Total samples: 10*

Step 4 Probability for Bar Chart Visualization

• 5.33 (1/10 = 0.10)
• 5.67 (1/10 = 0.10)
• 6 (1/10 = 0.10)
• 6.67 (1/10 = 0.10)
• 7 (1/10 = 0.10)
• 7.33 (2/10 = 0.20)
• 7.67 (1/10 = 0.10)
• 8 (1/10 = 0.10)
• 9 (1/10 = 0.10)
• Total of 1

Example 2, Misha Laurice's Grades

• Misha Laurice receives a grade of 82 and 83 in her three major subjects.
• Follow the first step by listing all the possible samples
  • 82, 82, 82
  • 82, 82, 83
  • 82, 83, 82
  • 82, 83, 83
  • 83, 82, 82
  • 83, 82, 83
  • 83, 83, 82
  • 83, 83, 83
• Which yields the sample mean below:
  • 82 (82)
  • 82, 82, 83 (82.33)
  • 82, 83, 82 (82.33)
  • 82, 83, 83 (82.67)
  • 83, 82, 82 (82.33)
  • 83, 82, 83 (82.67)
  • 83, 83, 82 (82.67)
  • 83, 83, 83 (83)
• Creating the Proability for visualization
  • Sample Mean (82): Frequency (1): Probability (1/8 = 0.125)
  • Sample Mean (82.33): Frequency (3): Probability (3/8 = 0.375)
  • Sample Mean (82.67): Frequency (3): Probability (3/8 = 0.375)
  • Sample Mean (83): Frequency (1): Probability (1/8 = 0.125)
  • Total Sample: 8 samples Therefore can ask further questions like:
• What is the probability that her mean grade is lower than 83?
  • Identified as 82, 82.33, and 82.67.
  • Simply sum the numbers
  • P(X < 83) = 9.125 + 0.375 + 0.375 P(X < 83) = 0.875 or 87.5% What is the probability that her mean grade is greater than 82.33?
  • We know that the larger numbers are 82.67 and 83
  • P(X > 82.33) = 0.375 +0.125
  • P(X > 82.33) = 0.5 or 50% What is the probability that his mean grade is 82.67?
• P(X = 82.67) = 0.375 or 37.5%

Properties of Sample Mean Distributions

• The population mean is the same as the sample mean.
  • μx = μ
  • Where μx = all data and P(x) = probability
  • So μx = Σ x * P(x)
• Sample mean variance can be calculated with the following formula:
  • σx = ∑(x – μ)^2 * P(x) or Σ[X^2 x P(X)] – μ^2
• In cases where finite population (without replacement)
  • σx = σ^2 / n * N-n / N-1
• If infinite population (with replacement)
  • σx = σ^2 / n
• Standard deviation of the sampling distribution of the sample mean:
  • σx = √(σ^2 / n) · √(N - n) / (N - 1) (finite population)
  • σx = σ /√n (infinite population)
• To construct the sampling distribution of sample means
  • Compute for the population mean.
  • Compute for the population variance.
  • Determine the number of possible samples.
  • List all possible samples and their corresponding means.
  • Construct the sampling distribution of the sample means.
  • Compute the mean of the sampling distribution of the sample means.
  • Compute the variance of the sampling distribution of the sample means.
  • Construct the histogram.
• Example:
  • The sample consists of (3, 5, 9, 10, and 8), list possible variations while computing their mean variance. Use a sample size of 3, then compute the mean and variance of the sample mean samples for the sampling distribution.

Example 3 - Compute for Population Mean

• μ = (ΣX) / N
• μ = (3 + 5 + 9 + 10 + 8) / 5
• μ= 7

Population Variance

• σ2= ∑(X – μ)^2 /N
• σ2 = (3-7)2 + (5-7)2 + (9-7)2 + (10-7)2 + (8-7)2/ 5
• σ2 = 6.8

Samples

• NCn= N! / n! (N – n)!
• = 5! / 3! (5-3)!
• = (5 x 4) / (2 x 1)
• = 10 (possible samples)
• Listing sample corresponding means to the list below
  • 3, 5, 9 (5.67)
  • 3, 5, 10 (6)
  • 3, 5, 8 (5.33)
  • 3, 9, 10 (7.33)
  • 3, 9, 8 (6.67)
  • 3, 10, 8 (7)
  • 5, 9, 10 (8)
  • 5, 9, 8 (7.33)
  • 5, 10, 8 (7.67)
  • 9, 10, 8 (9)

Step 5, Construct A Sample Distribution

Sample Mean Frequency

• 5.33 (1)
• 5.67 (1)
• 6 (1)
• 6.67 (1)
• 7 (1)
• 7.33 (2)
• 7.67 (1)
• 8 (1)
• 9 (1)
• Total: 10* Computing this the average for this formula,
• μx = 7
• This is equal to a sample mean that is considered a standard sampling distribution.
• You can also compute the variance of this for some of its sample means. o