Last updated on June 18th, 2025
In statistics, inferential statistics is a branch that deals with making conclusions, predictions, or judgments about a population from a sample of data. It uses probability theory and draws conclusions about an entire population based on a smaller, representative sample. Employing this method saves us from studying each and every individual. It assesses the reliability of results and saves time and resources. In this topic, we will delve deeper into the concepts and properties of inferential statistics.
Inferential statistics uses different analytical tools to analyze data and make inferences about a whole population based on a smaller sample. It helps to draw conclusions about a population. In inferential statistics, a value from a sample data, like the sample mean used to estimate the entire population, such as the population mean.
The sample must reflect the entire population. Inferential statistics uses probability theory and statistical models to estimate population parameters and test hypotheses based on sample data.
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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.
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:
Z Test: Z test is used when data has a normal distribution and a sample size of at least 30. It is applied if the sample mean matches the population mean when the population variance is known. The right-tailed hypothesis can be tested using the following setup:
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.
T Test: A t test is used when the data has a student t distribution and the sample size is less than 30. When the population variance is unknown, the sample and population mean are compared. The inferential statistics hypothesis test is as follows:
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.
F Test: An F-test helps compare two samples or populations to see if there are any prominent differences. A right-tailed F-test can be set up as follows:
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.
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.
The most 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 u have to mention hypothesis testing and regression analysis
Inferential statistics and descriptive statistics are key branches of statistics.
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 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.
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.
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A nutritionist wants to check if a new diet plan reduces weight. A sample of 40 people follows 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 affect weight loss.
The alternative hypothesis is that the new diet increases weight loss.
So, the test statistic: t = x̄ – μs / √n
Now, we can substitute the values:
t = 8 - 54 / √40
Calculating the denominator we get,
4 / √40 = 4 / 6.32 ≈ 0.633
So, 8 - 5 / 0.633 = 3 / 0.633 ≈ 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 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 = x̄ – μ / s / √n
Now we can substitute the values:
t = 4.5 - 8.5 / 0.9 / √50
0.9 / √50 = 0.9 / 7.07 = 0.1273
So, t = 4.5 - 8.5 / 0.1273
-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 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 = x̄ – μ / s / √n
Now we can substitute the values:
t = 80 - 65 / 8 / √45
8 / √45 = 8 / 6.71 = 1.19
So, t = 80 - 65 / 1.19 = 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 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 sales training significantly increased the average sales.
Here, Sample mean (x̄) = 200
Population mean (μ) = 150
Standard deviation of the population (s) = 15
Sample size (n) = 30
Significance level (α) = 0.05
The test statistic: t = x̄ – μ / s / √n
Now we can substitute the values:
t = 200 - 150 / 15 / √30
15 / √30 = 15 / 5.477 = 2.74
So, 200 - 150 / 2.74 = 50 / 2.74
t = 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 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 = 110 / 80 = 1.375
Next, we can determine the degrees of freedom (df1) = n - 1 = 10 - 1 = 9
The degrees of freedom (df2) = 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 can reject the null hypothesis.
This means the changes made in the test did not affect a significant improvement.
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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!