Sampling Distribution of Differences

Understand the sampling distribution of differences between means—what it is, why it matters, and how to apply it in hypothesis testing (with examples). Perfect for students, data scientists, and analysts! Ever wondered how statisticians compare two groups (e.g., test scores, sales performance, or medical treatments)? The key lies in the sampling distribution of differences between means—a fundamental concept for hypothesis testing, confidence intervals, and A/B testing.

Sampling Distribution of Differences Between Means

The Sampling Distribution of Differences Between Means is the probability distribution of differences between two sample means (e.g., $Mean_A – Mean_B$) if you repeatedly sampled from two populations.

Let there are two populations of size $N_1$ and $N_2$ having means $\mu_1$ and $\mu_2$ with variances $\sigma_1^2$ and $\sigma_2^2$. We need to draw all possible samples of size $n_1$ from the first population and $n_2$ from the second population, with or without replacement.

Let $\overline{x}_1$ be the means/averages of samples of the first population and $\overline{x}_2$ be the means/averages of the samples of the second population. After this, we will determine all possible differences between means/averages denoted by
$$d =\overline{x}_1 – \overline{x}_2$$

We call the frequency distribution differences as frequency distribution, while the probability distribution of the differences is the sampling distribution of differences between means.

Notations for Sampling Distribution of Differences between Means

NotationShort Description
$\mu_1$Mean of the first population
$\mu_2$Mean of the second population
$\sigma_1^2$Variance of the first population
$\sigma_2^2$Variance of the second population
$\sigma_1$Standard deviation of the first population
$\sigma_2$Standard deviation of the second population
$\mu_{\overline{x}_1 – \overline{x}_2}$Mean of the sampling distribution of difference between means
$\sigma^2_{\overline{x}_1 – \overline{x}_2}$Variance of the sampling distribution of difference between means
$\sigma_{\overline{x}_1 – \overline{x}_2}$Standard deviation of the sampling distribution of difference between means

Some Formulas for Sampling with/without Replacement

Sr. No.Sampling with ReplacementSampling without Replacement
1.$\mu_{\overline{x}_1 -\overline{x}_2} = \mu_1-\mu_2$$\mu_{\overline{x}_1 -\overline{x}_2} = \mu_1-\mu_2$
2.$\sigma^2_{\overline{x}_1 -\overline{x}_2}=\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}$$\sigma^2_{\overline{x}_1 -\overline{x}_2}=\frac{\sigma_1^2}{n_1}\left(\frac{N-1-n_2}{N_1-1}\right) + \frac{\sigma_2^2}{n_2}\left(\frac{N_2-n_2}{N_2-1}\right)$
3.$\sigma_{\overline{x}_1 -\overline{x}_2}=\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$$\sigma_{\overline{x}_1 -\overline{x}_2}=\sqrt{\frac{\sigma_1^2}{n_1}\left(\frac{N-1-n_2}{N_1-1}\right) + \frac{\sigma_2^2}{n_2}\left(\frac{N_2-n_2}{N_2-1}\right)}$

Example

Let $\overline{x}$ represent the mean of a sample of size $n_1=2$ selected at random with replacement from a finite population consisting of values 5, 7, and 9. Similarly, let $\overline{x}_2$ represent the mean of a sample of size $n_2=2$ selected at random from another finite population consisting of values 4, 6, and 8. Form the sampling distribution of the random variable $\overline{x}_1 – \overline{x}_2$ and verify that

  • $\mu_{\overline{x}_1 – \overline{x}_2} = \mu_1 – \mu_2$
  • $\sigma^2_{\overline{x}_1 – \overline{x}_2} = \frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}

Solution

Population IPopulation II
5, 7, 9
$N_1=3$
$n_1=2$
4, 6, 8
$N_2=3$
$n_2=2$
Possible samples with Replacement are $N_1^{n_1}=3^2 =9$Possible samples with Replacement are
$N_2^{n_2} = 3^2 = 9$
Sampling Distribution of Differences Between Means

All Possible Samples

All possible differences between samples means from both of the population is ($d=\overline{x}_1 – \overline{x}_2$).

$d=\overline{x}_1 =-\overline{x}_2$455666778
55-4= 100-1-1-1-2-2-3
6211000-1-1-2
6211000-1-1-2
732211100-1
732211100-1
732211100-1
8433222110
8433222110
9544333221

The Sampling Distribution of Differences Between Means

$d=\overline{x}_1 – \overline{x}_2$$f$$P(d)$$d\cdot P(d)$$d^2$$d^2 \cdot P(d)$
-311/81$-3 * 1/81 = -3/81$99/81
-244/81-8/81416/81
-11010/81-10/81110/81
01616/810/8100/81
11919/8119/81119/81
21616/8131/81464/81
31010/8130/81990/81
444/8116/811664/81
511/815/8125125/81
Total8181/81=1 297/81=3.67

\begin{align*}
\mu_{\overline{x}_1 – \overline{x}_2} &= E(d) = \Sigma(d\cdot P(d)) = \frac{81}{81}=1\\
\sigma^2_{\overline{x}_1 – \overline{x}_2} &= E(d^2) – [E(d)]^2\\
&=\Sigma d^2 P(d) – \left[\Sigma (d\cdot P(d))\right]^2\\
&= 3.67 – 1^2 = 2.67
\end{align*}

Sampling Distribution of differences between means, mean and variance of both populations

Verification

  • $\mu_{\overline{x}_1 – \overline{x}_2} = \mu_1 – \mu_2 \Rightarrow 7-6 = 1$
  • $\sigma_{\overline{x}_1 – \overline{x}_2}^2 = \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} = \frac{2.66}{2} + \frac{2.66}{2}\Rightarrow 2.66$

Sampling in R Language

Sampling with Replacement

In sampling with replacement, the units drawn are returned to the population before drawing the next unit. This means the same individual can be chosen more than once in the sampling process. The sampling with replacement may provide valuable insights while maintaining flexibility in selecting samples from a given population.

Key Characteristics of Sampling with Replacement

The following are key characteristics of Sampling with Replacement:

  1. Independence: Each selection is independent, as the same item can be selected multiple times.
  2. Population Size: The effective population size remains the same for each draw since previously selected items are replaced.
  3. Use Cases: This method is commonly used in algorithms, simulations, and bootstrapping techniques in statistics, where it’s important to assess variability or make inferences from a sample.

Example of Sampling with Replacement

As an example of sampling with replacement, suppose, you have a bag containing three colored balls (red, blue, and green), and you sample with a replacement, if you draw a red ball, you put it back into the bag before the next draw. As a result, in subsequent draws, you could again draw a red ball.

Drawing All Possible Samples Using Sampling with Replacement

Question: Consider a population with elements A, B, C, and D. Draw all possible samples of size 2 with replacement from this population.

Solution: In this problem, $N=4$ and $n=2$.

Possible number of samples (with replacement) = $N^n = 4^2 = 16$.

The 16 samples of size 2 are

AAABACAD
BABBBCBD
CACBCCCD
DADBDCDD

Question: Draw all possible samples of size 3 with replacement from a population having elements 2, 4, and 6.

Solution:

Population size = $N=3$, Sample size = n = 3$

Number of possible samples are $N^n = 3^3 = 27$

There are two ways to list these samples.

First Method:

First divide possible samples (27) by the population size unit quotient 1 is returned. For example, $\frac{27}{3} = 9, \quad \frac{9}{3}, \quad \frac{9}{3}=1$.

We obtained three quotients: 9, 3, and 1. These are the number of repetitions of population units. First, write every unit 9 times, then 3 times, and lastly, write every unit 1 time.

Sampling with Replacement

Second Method:

First, make the samples of size 2, which are easy to draw.

2, 2
2, 4
2, 6
4, 2
4, 4
4, 6
6, 2
6, 4
6, 6

Repeat these samples three times. Since the required number of samples is 27, add every population unit at (the start or) at the end of these samples of size two.

2, 2, 22, 2, 42, 2, 6
2, 4, 22, 4, 42, 4, 6
2, 6, 22, 6, 42, 6, 6
4, 2, 24, 2, 44, 2, 6
4, 4, 24, 4, 44, 4, 6
4, 6, 24, 6, 44, 6, 6
6, 2, 26, 2, 46, 2, 6
6, 4, 26, 4, 46, 4, 6
6, 6, 26, 6, 46, 6, 6

From the table above, 2 is added in the last of the first nine samples, then 4 is added in the last of the next 9 samples and finally 6 is added in the last nine samples.

Real-Life Examples of Sampling with Replacement

The following are some real-life examples of sampling with replacement:

  1. Lottery Draws: In some types of lotteries, numbers can be drawn multiple times before the final selection. For example, if a lottery allows for the same number to be drawn again after being selected, this is akin to sampling with replacement.
  2. Quality Control in Manufacturing: In a factory, inspectors might draw samples of products to test for defects. After testing each item, they return it to the production line before drawing the next sample to maintain the same population size and ensure each product has a chance of being selected again.
  3. Genetic Studies: In genetics, researchers might take DNA samples from a population to study traits or disorders. By replacing each sample with the population (considering genetic diversity), they can analyze the data while allowing for the possibility of selecting the same individual multiple times.
  4. Surveys: When conducting surveys, researchers might randomly select participants from a population (like voters or consumers) and, after querying each individual, they can include them again in the pool for subsequent selections, especially in larger datasets where the same individuals might provide valuable insights if repeated.
  5. Educational Testing: In standardized testing, students might take multiple attempts at a test where scores from previous attempts can be considered again in analyses to assess trends in learning or improvement.
  6. Customer Behavior Analysis: Companies may analyze customer purchase patterns by repeatedly sampling transactions. For instance, if a customer makes multiple purchases, their transaction data might be included in each analysis to understand their buying behavior over time.

Sampling Quiz Questions

Simulation and Sampling in R

Sampling Distribution of Means

Suppose, we have a population of size $N$ having mean $\mu$ and variance $\sigma^2$. We draw all possible samples of size $n$ from this population with or without replacement. Then we compute the mean of each sample and denote it by $\overline{x}$. These means are classified into a frequency table which is called frequency distribution of means and the probability distribution of means is called the sampling distribution of means.

Sampling Distribution

A sampling distribution is defined as the probability distribution of the values of a sample statistic such as mean, standard deviation, proportions, or difference between means, etc., computed from all possible samples of size $n$ from a population. Some of the important sampling distributions are:

  • Sampling Distribution of Means
  • Sampling Distribution of the Difference Between Means
  • Sampling Distribution of the Proportions
  • Sampling Distribution of the Difference between Proportions
  • Sampling Distribution of Variances

Notations of Sampling Distribution of Means

The following notations are used for sampling distribution of means:

$\mu$: Population mean
$\sigma^2$: Population Variance
$\sigma$: Population Standard Deviation
$\mu_{\overline{X}}$: Mean of the Sampling Distribution of Means
$\sigma^2_{\overline{X}}$: Variance of Sampling Distribution of Means
$\sigma_{\overline{X}}$: Standard Deviation of the Sampling Distribution of Means

Formulas for Sampling Distribution of Means

The following following formulas for the computation of means, variance, and standard deviations can be used:

\begin{align*}
\mu_{\overline{X}} &= E(\overline{X}) = \Sigma (\overline{X}P(\overline{X})\\
\sigma^2_{\overline{X}} &= E(\overline{X}^2) – [E(\overline{X})]^2\\
\text{where}\\
E(\overline{X}^2) &= \Sigma \overline{X}^2P(\overline{X})\\
\sigma_{\overline{X}} &= \sqrt{E(\overline{X}^2) – [E(\overline{X})]^2}
\end{align*}

Numerical Example: Sampling Distribution of Means

A population of $(N=5)$ has values 2, 4, 6, 8, and 10. Draw all possible samples of size 2 from this population with and without replacement. Construct the sampling distribution of sample means. Find the mean, variance, and standard deviation of the population and verify the following:

Sr. No.Sampling with ReplacementSampling without Replacement
1)$\mu_{\overline{X}} = \mu$$\mu_{\overline{X}} = \mu$
2)$\sigma^2_{\overline{X}}=\frac{\sigma^2}{n}$$\sigma^2_{\overline{X}}=\frac{\sigma^2}{n}\frac{N-n}{N-1}$
3)$\sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}}$$\sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}} \sqrt{\frac{N-n}{N-1}}$

Solution

The solution to the above example is as follows:

Sampling with Replacement (Mean, Variance, and Standard Deviation)

The number of possible samples is: $N^n = 5^2 = 25.

Samples$\overline{X}$Samples$\overline{X}$Samples$\overline{X}$
2, 224, 1078, 88
2, 436, 248, 109
2, 646, 4510, 26
2, 856, 6610, 47
2, 1066, 8710, 68
4, 236, 10810, 89
4, 448, 2510, 1010
4, 658, 46
4, 868, 67

The sampling distribution of sample means will be

$\overline{X}$Freq$P(\overline{X}$$\overline{X}P(\overline{X})$$\overline{X}^2$$\overline{X}^2P(\overline{X}$
211/252/2544/25
322/256/25918/25
433/2512/251648/25
544/2520/2525100/25
655/2530/2536180/25
744/2528/2549196/25
833/2524/2564192/25
922/2518/2581162/25
10112510/25100100/25
Total25/25=1150/25 = 61000/25=40

\begin{align*}
\mu_{\overline{X}} &= E(\overline{X}) = \Sigma \left[\overline{X}P(\overline{X})\right] = \frac{150}{25}=6\\
\sigma^2_{\overline{X}} &= E(\overline{X}^2) – [E(\overline{X}]^2=\Sigma [\overline{X}^2P(\overline{X})] – [\Sigma [\overline{X}P(\overline{X})]]^2\\
&= 40 – 6^2 = 4\\
\sigma_{\overline{X}} &= \sqrt{4}=2
\end{align*}

Mean, Variance, and Standard Deviation for Population

The following are computations for population values.

$X$24681030
$X^2$4163664100220

\begin{align*}
\mu &= \frac{\Sigma}{N} = \frac{30}{5} = 6\\
\sigma^2 &= \frac{\Sigma X^2}{N} – \left(\frac{\Sigma X}{n} \right)^2\\
&=\frac{220}{5} – (6)^2 = 8\\
\sigma&= \sqrt{8} = 2.82
\end{align*}

Verifications:

  1. Mean: $\mu_{\overline{X}} = \mu \Rightarrow 6=6$
  2. Variance: $\sigma^2_{\overline{X}} = \frac{\sigma^2}{n} \Rightarrow 4=\frac{8}{2}$
  3. Standard Deviation: $\sigma_{\overline{X}}=\frac{\sigma}{\sqrt{n}} \Rightarrow 2=\frac{2.82}{\sqrt{2}}=2$

Sampling without Replacement

The possible samples for sampling without replacement are: $\binom{5}{2}=10$

Samples$\overline{x}$Samples$\overline{x}$
2, 434, 86
2, 644, 107
2, 856, 87
2, 1066, 108
4, 648, 109

The sampling distribution sample means for sampling without replacement is

$\overline{x}$Freq$P(\overline{x})$$\overline{x}P(\overline{x})$$\overline{x}^2$$\overline{x}^2P(\overline{x})$
311/103/1099/10
411/104/101616/10
522/1010/102550/10
622/1012/103672/10
722/1014/104998/10
811/108/106464/10
911/209/108181/10
Total10/10=160/10=6390/10 = 39

\begin{align*}
\mu_{\overline{X}} &= E(\overline{X}) = \Sigma \left[\overline{X}P(\overline{X})\right] = \frac{60}{10}=6\\
\sigma^2_{\overline{X}} &= E(\overline{X}^2) – [E(\overline{X}]^2=\Sigma [\overline{X}^2P(\overline{X})] – [\Sigma [\overline{X}P(\overline{X})]]^2\\
&= 39 – 6^2 = 3\\
\sigma_{\overline{X}} &= \sqrt{3}=1.73
\end{align*}

Verifications:

  1. Mean: $\mu_{\overline{X}} = \mu \Rightarrow 6=6$
  2. Variance: $\sigma^2_{\overline{X}} = \frac{\sigma^2}{n}\cdot \left(\frac{N-n}{N-1}\right) \Rightarrow 3=\frac{8}{2}\cdot\left(\frac{5-2}{5-1}\right)=3$
  3. Standard Deviation: $\sigma_{\overline{X}}=\frac{\sigma}{\sqrt{n}} \Rightarrow 1.73=\sqrt{3}$

Why is Sampling Distribution Important?

  • Inference: Sampling distribution of means allows users to make inferences about the population mean based on sample data.
  • Hypothesis Testing: It is crucial for hypothesis testing, where the researcher compares sample statistics to population parameters.
  • Confidence Intervals: It helps construct confidence intervals, which provide a range of values likely to contain the population mean.
Sampling Distribution of Means

Note that the sampling distribution of means provides a framework for understanding how sample means vary from sample to sample and how they relate to the population mean. This understanding is fundamental to statistical inference and decision-making.

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