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1588 LearnersLast updated on November 20, 2025

Inferential statistics is a branch of statistics that helps draw conclusions or make predictions about a whole population using a smaller, representative sample. It relies on probability theory, saving time and resources while ensuring reliable and meaningful results from limited data.
Inferential statistics helps us learn about a large group (the population) by examining only a small portion of it (a sample). By using different tests and analytical methods, we can make predictions or infer conclusions about the whole population from this smaller group.
To choose a sample that truly represents the population, we can use the different sampling techniques. Some widely used methods include simple random sampling, stratified sampling, cluster sampling, and systematic sampling.
Example: A school has 1,000 students, and the principal wants to know the average time students spend on homework every day. It is impossible to ask all 1,000 students individually, so the principal uses a sampling method.
Here’s how sampling works in this situation:
Simple Random Sampling: Select 50 students randomly from the whole school, just like picking names out of a box.
Stratified Sampling: Divide the school into groups by class (6, 7, 8, 9, 10, 11, and 12). Then choose a few students from each class so that every group is included.
Cluster Sampling: Choose a few entire classes, for example, 7A, 8C, and 9B, and survey all the students in those selected classes.
Systematic Sampling: Use a complete list of all students and pick every 20th name (20th, 40th, 60th, etc.).
After collecting homework-time data from the chosen sample, the principal uses inferential statistics to estimate the average homework time for all 1,000 students.
There are mainly two types that researchers use to draw conclusions from small samples. They are called hypothesis testing and regression analysis. Let us take a closer look at these two types of inferential statistics.
| Inferential Statistics | |
| Hypothesis Testing | Regression Analysis |
| Z test | Linear Regression |
| F test | Nominal Regression |
| Anova test | Logistics Regression |
| Wilcoxon Signed Rank test | Ordinal Regression |
| Mann-whitney U test | |
Hypothesis Testing: Inferential statistics includes testing hypotheses and drawing conclusions about a population using sample data. It involves creating a null hypothesis and an alternative hypothesis before conducting a statistical test.
A hypothesis distribution can be left-tailed, right-tailed, or two-tailed. The conclusions are made by using the test statistic’s value, the critical value, and the confidence intervals. Some important hypothesis tests used in inferential statistics are:
Null hypothesis: H0: μ = μ0
Alternative hypothesis: H1: μ > μ0
Test statistic: Z Test = (x̄ – μ) / (σ / √n)
Here, x̄ = Sample mean
μ = Population mean
σ = Standard deviation of the population
n = Sample size
One thing we should keep in mind is that we can reject the null hypothesis if the z statistic is greater than the z critical value.
Null hypothesis: H0: μ = μ0
Alternative hypothesis: H1: μ > μ0
Test statistic: t = x̄ – μs / √n
Here, x̄ = Sample mean
μ = Population mean
s = Standard deviation of the population
n = Sample size
If the 't' statistic is greater than the 't' critical value, then the null hypothesis can be rejected.
Null hypothesis: H0: σ21 = σ22
Alternative hypothesis: H1: σ21 > σ22
Test statistic: f = σ21 / σ22
Here, σ21 = Variance of the first population
σ22 = Variance of the second population
If the f test statistic is greater than the crucial value, we can reject the null hypothesis.
For example:
A school wants to know if the average math score of Class A differs from the school average. They take a sample of 25 students and use a T-test. The T-test helps decide if Class A really performs differently or if the difference is due to random chance.
Regression Analysis: Regression analysis expresses the relationship between two variables. It determines how one variable responds to another. Simple linear, multiple linear, nominal, logistic, and ordinal regression are a few of the numerous regression models that can be applied.
A commonly used regression form in inferential statistics is linear regression. Linear regression analyzes how the dependent variable reacts to a unit change in the independent variable. Some crucial formulas for regression analysis in inferential statistics are as follows:
With α and β as regression coefficients, the straight-line equation is expressed as y = α + βx
β = ∑(xi − x̄) (yi − ȳ) / ∑(xi − x̄)2
β = rxy σy / σx
α = ȳ − βx̄
Here, x̄ = The mean of the independent variable
σx = Standard deviation of the first data set
ȳ = The mean of the dependent variable
σy = Standard deviation of the second data set
Here, you have to mention hypothesis testing and regression analysis.
Inferential statistics and descriptive statistics are key branches of statistics. Here are the main differences between them:
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Inferential Statistics
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Descriptive Statistics
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It employs diverse analysis techniques to generate inferences about a population based on sample data
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It is used to quantify the characteristics of the data
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Facilitates population-based conclusions
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Deals with summarizing the characteristics of a dataset
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Analytical tools used are regression analysis and hypothesis testing
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Methods used are measures of dispersion and central tendency
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Makes conclusions about an unknown population
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Expresses the features of a known population or sample
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Some of the measures used are linear regression, t-test, and z-test
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Variance, range, mean, and median are some of the measures used
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Pollsters conduct a survey of voters before an election in order to determine which candidate is likely to win. It is impossible to collect data from every single voter. Instead, they collect data from a sample group. To determine the overall voter preference, the pollsters use inferential statistics and sample analysis instead of asking millions of people. Inferential statistics helps us to:
Inferential statistics gives you the means to interpret data beyond the information you currently have and make well-informed conclusions regarding the information you lack.
Inferential statistics provide powerful tools for testing ideas, estimating values for entire populations, and making predictions based on sample data. Here are the main techniques explained in an easy-to-understand way:
Confidence Intervals
A confidence interval gives a range of values that probably contains the true population value. It helps us to understand how accurate our estimate is.
The formula for a confidence interval for the mean is:
\(CI = \bar{x} \pm Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}\)
Where:
x = the average of the sample
Zα/2 = a value from the Z-table (like 1.96 for 95% confidence)
σ = population standard deviation
n = sample size
Hypothesis Testing
Hypothesis testing is a method for testing whether a claim about a population is true.
It has two parts:
We collect data and calculate a test statistic. For a Z-test, we use:
\(Z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}\)
Where:
x = sample mean
μ0= the mean stated in the null hypothesis
σ = population standard deviation
n = sample size
Then we compare the Z value with a critical value or find the p-value:
p-value = 2⋅P (Z>∣zobs ∣). If the p-value is less than 0.05, we say the result is significant and reject the null hypothesis.
Central Limit Theorem (CLT)
The Central Limit Theorem says that when we take large enough samples, the average of those samples usually becomes distributed (bell-shaped), even if the original data was not normal.
Mathematically:
\(\bar{X} \sim N\left(\mu,\ \frac{\sigma}{\sqrt{n}}\right)\)
Where:
μ = true population mean
σ = population standard deviation
n = sample size
This is useful because many statistical tests require normal data. The CLT allows us to use these tests even with skewed or uneven data, such as income or shopping patterns.
When doing hypothesis testing, two common types of mistakes can occur:
Type I Error (False Positive)
This error happens when we reject the null hypothesis even though it is actually true. In simple words, we believe that there is an effect or difference when none exists. The probability of making this mistake is called α (alpha), also known as the significance level.
Type II Error (False Negative)
This error occurs when we fail to reject the null hypothesis even though it is actually false. This means we overlook a real effect or difference. The probability of this error is called β (beta). The power of a test is 1 − β, which indicates how well the test detects true effects.
The main aim in statistics is to minimize both types of errors. This can be done by selecting the appropriate sample size and setting an appropriate significance level.
Inferential statistics is a branch of statistics that allows us to make predictions, decisions, or generalizations about an entire population using data gathered from a smaller sample. It helps in understanding patterns, testing hypotheses, and drawing meaningful conclusions from limited information.
To make inferences about a population from sample data, we can use a technique called inferential statistics. However, using the technique improperly might result in incorrect decision-making, wrong interpretations, and false conclusions. Here are some of the common mistakes and useful solutions to avoid these typical mistakes.
Inferential statistics is a useful tool that helps us make inferences or conclusions based on sample data. Also, it helps us to make predictions and check the accuracy of hypotheses. However, the real-life applications of inferential statistics are limitless.
A nutritionist wants to check if a new diet plan reduces weight. A sample of 40 people follow the diet, and their average weight loss is 8 kg with a standard deviation of 4 kg. Previously, the average weight loss without the diet was 5 kg. Can we conclude that the new diet is effective at α = 0.05?
The new diet plan is effective in increasing weight loss.
Here, sample mean (x̄) = 8 kg
Population mean (μ) = 5 kg
Standard deviation of the population (s) = 4 kg
Sample size (n) = 40
Significance level (α) = 0.05
The null hypothesis is that the new diet does not lead to weight loss.
The alternative hypothesis is that the new diet increases weight loss.
So, the test statistic: \(t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \)
Now, we can substitute the values:
\(t = \frac{8 - 5}{4 / \sqrt{40}} \)
Calculating the denominator, we get,
\(\frac{4}{\sqrt{40}} = \frac{4}{6.32} \approx 0.633\)
So, \(\frac{8 - 5}{0.633} = \frac{3}{0.633} \approx 4.74\)
Next, we have to find the degrees of freedom (df) = n − 1 = 40 − 1 = 39
From the 't' table, we can find the critical t-value for α = 0.05, which is 1.685
Since, 4.75 > 1.685, we can reject the null hypothesis (H0).
The new diet plan is effective in increasing weight loss.
A company claims its average customer satisfaction rating is 8.5 out of 10. A random sample of 50 customers gives an average rating of 4.5, with a standard deviation of 0.9. Test the claim at α = 0.05.
The actual customer satisfaction rating is significantly lower than the claimed 8.5 out of 10.
Here, sample mean (x̄) = 4.5
Population mean (μ) = 8.5
The standard deviation of the population (s) = 0.9
Sample size (n) = 50
Significance level (α) = 0.05
The test statistic: \(t = \frac{\bar{x} - \mu}{s / \sqrt{n}}\)
Now we can substitute the values:
\(t = \frac{4.5 - 8.5}{0.9 / \sqrt{50}} \)
\(\frac{0.9}{\sqrt{50}} = \frac{0.9}{7.07} = 0.1273\)
So, \(t = \frac{4.5 - 8.5}{0.1273}\)
\(\frac{-4}{0.1273} = -31.40\)
Next, we have to find the degrees of freedom (df) = n − 1 = 50 − 1 = 49
From the 't' table, we can find the critical t-value for α = 0.05, which is ±2.009
Here, |t| = −31.40 and critical t-value = ±2.009.
Since, |t| > critical t-value, we reject the null hypothesis.
A professor believes that students who attend extra classes score higher than 65 marks on average. A sample of 45 students has an average score of 80, with a standard deviation of 8 marks. Can we conclude that attending extra classes improves performance at α = 0.05?
Attending extra classes significantly improves student performance.
Here, sample mean (x̄) = 80
Population mean (μ) = 65
Standard deviation of the population (s) = 8
Sample size (n) = 45
Significance level (α) = 0.05
The test statistic: \(t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \)
Now we can substitute the values:
\(t = \frac{80 - 65}{8 / \sqrt{45}} \)
\(\frac{8}{\sqrt{45}} = \frac{8}{6.71} = 1.19\)
So, \(t = \frac{80 - 65}{1.19} = \frac{15}{1.19} = 12.61\)
t = 12.61
Next, we have to find the degrees of freedom \((df) = n − 1 = 45 − 1 = 44\)
From the 't' table, we can find the critical t-value for α = 0.05, which is 1.680
Since 12.61 > 1.680, we can reject the null hypothesis. This means that attending extra classes significantly improves the performance of students.
After a new sales training is given to employees, the average sale goes up to $200 (a sample of 30 employees was examined) with a standard deviation of $15. Before the training, the average sale was $150. Check if the training helped at α = 0.05.
The training significantly increased the average sales.
Here, sample mean (x̄) = 200
Population mean (μ) = 150
Standard deviation of the population (\(\sigma \)) = 15
Sample size (n) = 30
Significance level (α) = 0.05
The test statistic: \(t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \)
Now we can substitute the values:
\(t = \frac{200 - 150}{15 / \sqrt{30}} \)
\(\frac{15}{\sqrt{30}} = \frac{15}{5.477} = 2.74 \)
So, \(\frac{50}{2.74} = 18.25\)
Next, we have to find the degrees of freedom \((df) = n − 1 = 30 − 1 = 29\)
From the 't' table, we can find the critical t-value for α = 0.05 and df = 29, which is 1.699
Since 18.25 > 1.699, we can reject the null hypothesis.
This means, the sales training significantly increased the average sales.
A test was conducted with the variance = 110 and n = 10. Certain changes were made in the test and it was again conducted with variance = 80 and n = 4. At a 0.05 significance level, was there any improvement in the test results?
The changes made in the test did not affect a significant improvement.
Here, we can apply the f test for variance comparison.
Test statistic: f = σ21 / σ22
σ21 = 110 (variance before change)
σ22 = 80 (variance after change)
Now we can substitute the values:
\(f = \frac{110}{80} = 1.375\)
Next, we can determine the degrees of freedom \((df_1) = n - 1 = 10 - 1 = 9\)
The degrees of freedom \((df_2) = n - 1 = 4 - 1 = 3\)
By using the 'f' table, we find that the critical t-value for α = 0.05, df1 = 9, and df2 = 3 is 9.28
Since f = 1.375 is much smaller than the critical value 9.28, we fail to reject the null hypothesis. Thus, the changes did not lead to significant improvement.
Jaipreet Kour Wazir is a data wizard with over 5 years of expertise in simplifying complex data concepts. From crunching numbers to crafting insightful visualizations, she turns raw data into compelling stories. Her journey from analytics to education ref
: She compares datasets to puzzle games—the more you play with them, the clearer the picture becomes!






