hypothesis testing ordinal data

311

Statistics for Table of poured by lady

Sample Size = 8

OPTION EXACT in SAS indicates that we are doing exact tests which consider ALL tables with the exact same margins as the observed table. This option will work for any \(I \times J\)   table. OPTION FISHER, more specifically performs Fisher's exact test which is an exact test only for a \(2 \times 2\) table in SAS.

For R, see TeaLady.R where you can see we used the fisher.test() function to perform Fisher's exact test for the \(2 \times 2\) table in question.

The same could be done using chisq.test() with option, simulate.p.value=TRUE . By reading the help file on fisher. test() function, you will see that certain options in this function only work for \(2 \times 2\) tables. For the output, see TeaLady.out

The basic idea behind exact tests is that they enumerate all possible tables that have the same margins, e.g., row sums and column sums. Then to compute the relevant statistics, e.g., \(X^2\), \(G^2\), odds-ratios, we look for all tables where the values are more extreme than the one we have observed. The key here is that in the set of tables with the same margins, once we know the value in one cell, we know the rest of the cells. Therefore, to find a probability of observing a table, we need to find the probability of only one cell in the table (rather than the probabilities of four cells). Typically we use the value of cell (1,1).

Under the null hypothesis of independence, more specifically when odds-ratio \(\theta = 1\), the probability distribution of that one cell \(n_{11}\) is hypergeometric, as discussed in the Tea lady example.

  Stop and Think!

Extension of fisher's test.

For problems where the number of possible tables is too large, Monte Carlo methods are used to approximate "exact" statistics (e.g., option MC in SAS FREQ EXACT and in R under chisq.test() you need to specify simulate.p.value = TRUE and indicate how many runs you want MC simulation to do; for more see the help files). This extension, and thus these options in SAS and R, of the Fisher's exact test for a \(2 \times 2\) table, in effect, takes samples from a large number of possibilities in order to simulate the exact test.

This test is "exact" because no large-sample approximations are used. The \(p\)-value is valid regardless of the sample size. Asymptotic results may be unreliable when the distribution of the data is sparse, or skewed. Exact computations are based on the statistical theory of exact conditional inference for contingency tables.

Fisher's exact test is definitely appropriate when the row totals and column totals are both fixed by design. Some have argued that it may also be used when only one set of margins is truly fixed. This idea arises because the marginal totals \(n_{1+}, n_{+1}\) provide little information about the odds ratio \(\theta\).

Exact non-null inference for \(\theta\)

When \(\theta = 1\), this distribution is hypergeometric, which we used in Fisher's exact test. More generally, Fisher (1935) gave this distribution for any value of \(\theta\). Using this distribution, it is easy to compute Fisher's exact \(p\)-value for testing the null hypothesis \(H_0:\theta=\theta^*\) for any \(\theta^*\). The set of all values \(\theta^*\) that cannot be rejected at the \(\alpha=.05\) level test forms an exact 95% confidence region for \(\theta\).

Let's look at a part of the SAS output a bit closer, we get the same CIs in the R ouput. First, notice the sample estimate of the odds ratio equal to 9, which we can compute from the cross-product ratio as we have discussed earlier. Note also that fisher.test() in R for \(2 \times 2\) tables will give so-called "conditional estimate" of the odds-ratio so the value will be different (in this case, approximately 6.408).

Notice the difference between the exact and asymptotic CIs for the odds ratio for the two-sided alternative (e.g., \(\theta\ne1\)). The exact version is larger. Recalling the interpretation of the odds ratio, what do these CIs tell us about true unknown odds-ratio? This is a simple example of how inference may vary if you have small samples or sparseness.

Bias correction for estimating \(\theta\)

Earlier, we learned that the natural estimate of \(\theta\) is

\(\hat{\theta}=\dfrac{n_{11}n_{22}}{n_{12}n_{21}}\)

and that \(\log\hat{\theta}\) is approximately normally distributed about \(\log \theta\) with estimated variance

\(\hat{V}(\log\hat{\theta})=\dfrac{1}{n_{11}}+\dfrac{1}{n_{12}}+\dfrac{1}{n_{21}}+\dfrac{1}{n_{22}}\)

Advanced note: this is the estimated variance of the limiting distribution, not an estimate of the variance of \(\log\hat{\theta}\) itself. Because there is a nonzero probability that the numerator or the denominator of \(\hat{\theta}\) may be zero, the moments of \(\hat{\theta}\) and \(\log\hat{\theta}\) do not actually exist. If the estimate is modified by adding \(1/2\) to each \(n_{ij}\), we have

\(\tilde{\theta}=\dfrac{(n_{11}+0.5)(n_{22}+0.5)}{(n_{12}+0.5)(n_{21}+0.5)}\)

with estimated variance

\(\hat{V}(\log\tilde{\theta})=\sum\limits_{i,j} \dfrac{1}{(n_{ij}+0.5)}\)

In smaller samples, \(\log\tilde{\theta}\) may be slightly less biased than \(\log\hat{\theta}\).

4.6 - Lesson 4 Summary

In this lesson, we extended the idea of association in two-way contingency tables to accommodate ordinal data and linear trends. The key concepts are that ordinal data has a particular nature that we can summarize and potentially take advantage of when carrying out a significance test. We also introduced an alternative "exact" test for \(2\times2\) tables that can be used when the observed counts are too small for the usual chi-square approximation to apply.

Coming up next, we consider new ways to measure associations if more than two variables are in the picture. Specifically, we'll see how one relationship can be confounded with another so that interpretations will change, depending on whether we condition on certain variables in advance.

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Hypothesis Testing | A Step-by-Step Guide with Easy Examples

Published on November 8, 2019 by Rebecca Bevans . Revised on June 22, 2023.

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics . It is most often used by scientists to test specific predictions, called hypotheses, that arise from theories.

There are 5 main steps in hypothesis testing:

• State your research hypothesis as a null hypothesis and alternate hypothesis (H o ) and (H a  or H 1 ).
• Collect data in a way designed to test the hypothesis.
• Perform an appropriate statistical test .
• Decide whether to reject or fail to reject your null hypothesis.
• Present the findings in your results and discussion section.

Though the specific details might vary, the procedure you will use when testing a hypothesis will always follow some version of these steps.

Table of contents

Step 1: state your null and alternate hypothesis, step 2: collect data, step 3: perform a statistical test, step 4: decide whether to reject or fail to reject your null hypothesis, step 5: present your findings, other interesting articles, frequently asked questions about hypothesis testing.

After developing your initial research hypothesis (the prediction that you want to investigate), it is important to restate it as a null (H o ) and alternate (H a ) hypothesis so that you can test it mathematically.

The alternate hypothesis is usually your initial hypothesis that predicts a relationship between variables. The null hypothesis is a prediction of no relationship between the variables you are interested in.

• H 0 : Men are, on average, not taller than women. H a : Men are, on average, taller than women.

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For a statistical test to be valid , it is important to perform sampling and collect data in a way that is designed to test your hypothesis. If your data are not representative, then you cannot make statistical inferences about the population you are interested in.

There are a variety of statistical tests available, but they are all based on the comparison of within-group variance (how spread out the data is within a category) versus between-group variance (how different the categories are from one another).

If the between-group variance is large enough that there is little or no overlap between groups, then your statistical test will reflect that by showing a low p -value . This means it is unlikely that the differences between these groups came about by chance.

Alternatively, if there is high within-group variance and low between-group variance, then your statistical test will reflect that with a high p -value. This means it is likely that any difference you measure between groups is due to chance.

Your choice of statistical test will be based on the type of variables and the level of measurement of your collected data .

• an estimate of the difference in average height between the two groups.
• a p -value showing how likely you are to see this difference if the null hypothesis of no difference is true.

Based on the outcome of your statistical test, you will have to decide whether to reject or fail to reject your null hypothesis.

In most cases you will use the p -value generated by your statistical test to guide your decision. And in most cases, your predetermined level of significance for rejecting the null hypothesis will be 0.05 – that is, when there is a less than 5% chance that you would see these results if the null hypothesis were true.

In some cases, researchers choose a more conservative level of significance, such as 0.01 (1%). This minimizes the risk of incorrectly rejecting the null hypothesis ( Type I error ).

The results of hypothesis testing will be presented in the results and discussion sections of your research paper , dissertation or thesis .

In the results section you should give a brief summary of the data and a summary of the results of your statistical test (for example, the estimated difference between group means and associated p -value). In the discussion , you can discuss whether your initial hypothesis was supported by your results or not.

In the formal language of hypothesis testing, we talk about rejecting or failing to reject the null hypothesis. You will probably be asked to do this in your statistics assignments.

However, when presenting research results in academic papers we rarely talk this way. Instead, we go back to our alternate hypothesis (in this case, the hypothesis that men are on average taller than women) and state whether the result of our test did or did not support the alternate hypothesis.

If your null hypothesis was rejected, this result is interpreted as “supported the alternate hypothesis.”

These are superficial differences; you can see that they mean the same thing.

You might notice that we don’t say that we reject or fail to reject the alternate hypothesis . This is because hypothesis testing is not designed to prove or disprove anything. It is only designed to test whether a pattern we measure could have arisen spuriously, or by chance.

If we reject the null hypothesis based on our research (i.e., we find that it is unlikely that the pattern arose by chance), then we can say our test lends support to our hypothesis . But if the pattern does not pass our decision rule, meaning that it could have arisen by chance, then we say the test is inconsistent with our hypothesis .

If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

• Normal distribution
• Descriptive statistics
• Measures of central tendency
• Correlation coefficient

Methodology

• Cluster sampling
• Stratified sampling
• Types of interviews
• Cohort study
• Thematic analysis

Research bias

• Implicit bias
• Cognitive bias
• Survivorship bias
• Availability heuristic
• Nonresponse bias
• Regression to the mean

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics. It is used by scientists to test specific predictions, called hypotheses , by calculating how likely it is that a pattern or relationship between variables could have arisen by chance.

A hypothesis states your predictions about what your research will find. It is a tentative answer to your research question that has not yet been tested. For some research projects, you might have to write several hypotheses that address different aspects of your research question.

A hypothesis is not just a guess — it should be based on existing theories and knowledge. It also has to be testable, which means you can support or refute it through scientific research methods (such as experiments, observations and statistical analysis of data).

Null and alternative hypotheses are used in statistical hypothesis testing . The null hypothesis of a test always predicts no effect or no relationship between variables, while the alternative hypothesis states your research prediction of an effect or relationship.

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• Indian J Crit Care Med
• v.25(Suppl 2); 2021 May

An Introduction to Statistics: Choosing the Correct Statistical Test

Priya ranganathan.

1 Department of Anaesthesiology, Critical Care and Pain, Tata Memorial Centre, Homi Bhabha National Institute, Mumbai, Maharashtra, India

The choice of statistical test used for analysis of data from a research study is crucial in interpreting the results of the study. This article gives an overview of the various factors that determine the selection of a statistical test and lists some statistical testsused in common practice.

How to cite this article: Ranganathan P. An Introduction to Statistics: Choosing the Correct Statistical Test. Indian J Crit Care Med 2021;25(Suppl 2):S184–S186.

In a previous article in this series, we looked at different types of data and ways to summarise them. 1 At the end of the research study, statistical analyses are performed to test the hypothesis and either prove or disprove it. The choice of statistical test needs to be carefully performed since the use of incorrect tests could lead to misleading conclusions. Some key questions help us to decide the type of statistical test to be used for analysis of study data. 2

W hat is the R esearch H ypothesis ?

Sometimes, a study may just describe the characteristics of the sample, e.g., a prevalence study. Here, the statistical analysis involves only descriptive statistics . For example, Sridharan et al. aimed to analyze the clinical profile, species distribution, and susceptibility pattern of patients with invasive candidiasis. 3 They used descriptive statistics to express the characteristics of their study sample, including mean (and standard deviation) for normally distributed data, median (with interquartile range) for skewed data, and percentages for categorical data.

Studies may be conducted to test a hypothesis and derive inferences from the sample results to the population. This is known as inferential statistics . The goal of inferential statistics may be to assess differences between groups (comparison), establish an association between two variables (correlation), predict one variable from another (regression), or look for agreement between measurements (agreement). Studies may also look at time to a particular event, analyzed using survival analysis.

A re the C omparisons M atched (P aired ) or U nmatched (U npaired )?

Observations made on the same individual (before–after or comparing two sides of the body) are usually matched or paired . Comparisons made between individuals are usually unpaired or unmatched . Data are considered paired if the values in one set of data are likely to be influenced by the other set (as can happen in before and after readings from the same individual). Examples of paired data include serial measurements of procalcitonin in critically ill patients or comparison of pain relief during sequential administration of different analgesics in a patient with osteoarthritis.

W hat are the T ype of D ata B eing M easured ?

The test chosen to analyze data will depend on whether the data are categorical (and whether nominal or ordinal) or numerical (and whether skewed or normally distributed). Tests used to analyze normally distributed data are known as parametric tests and have a nonparametric counterpart that is used for data, which is distribution-free. 4 Parametric tests assume that the sample data are normally distributed and have the same characteristics as the population; nonparametric tests make no such assumptions. Parametric tests are more powerful and have a greater ability to pick up differences between groups (where they exist); in contrast, nonparametric tests are less efficient at identifying significant differences. Time-to-event data requires a special type of analysis, known as survival analysis.

H ow M any M easurements are B eing C ompared ?

The choice of the test differs depending on whether two or more than two measurements are being compared. This includes more than two groups (unmatched data) or more than two measurements in a group (matched data).

T ests for C omparison

( Table 1 lists the tests commonly used for comparing unpaired data, depending on the number of groups and type of data. As an example, Megahed and colleagues evaluated the role of early bronchoscopy in mechanically ventilated patients with aspiration pneumonitis. 5 Patients were randomized to receive either early bronchoscopy or conventional treatment. Between groups, comparisons were made using the unpaired t test for normally distributed continuous variables, the Mann–Whitney U -test for non-normal continuous variables, and the chi-square test for categorical variables. Chowhan et al. compared the efficacy of left ventricular outflow tract velocity time integral (LVOTVTI) and carotid artery velocity time integral (CAVTI) as predictors of fluid responsiveness in patients with sepsis and septic shock. 6 Patients were divided into three groups— sepsis, septic shock, and controls. Since there were three groups, comparisons of numerical variables were done using analysis of variance (for normally distributed data) or Kruskal–Wallis test (for skewed data).

Tests for comparison of unpaired data

A common error is to use multiple unpaired t -tests for comparing more than two groups; i.e., for a study with three treatment groups A, B, and C, it would be incorrect to run unpaired t -tests for group A vs B, B vs C, and C vs A. The correct technique of analysis is to run ANOVA and use post hoc tests (if ANOVA yields a significant result) to determine which group is different from the others.

( Table 2 lists the tests commonly used for comparing paired data, depending on the number of groups and type of data. As discussed above, it would be incorrect to use multiple paired t -tests to compare more than two measurements within a group. In the study by Chowhan, each parameter (LVOTVTI and CAVTI) was measured in the supine position and following passive leg raise. These represented paired readings from the same individual and comparison of prereading and postreading was performed using the paired t -test. 6 Verma et al. evaluated the role of physiotherapy on oxygen requirements and physiological parameters in patients with COVID-19. 7 Each patient had pretreatment and post-treatment data for heart rate and oxygen supplementation recorded on day 1 and day 14. Since data did not follow a normal distribution, they used Wilcoxon's matched pair test to compare the prevalues and postvalues of heart rate (numerical variable). McNemar's test was used to compare the presupplemental and postsupplemental oxygen status expressed as dichotomous data in terms of yes/no. In the study by Megahed, patients had various parameters such as sepsis-related organ failure assessment score, lung injury score, and clinical pulmonary infection score (CPIS) measured at baseline, on day 3 and day 7. 5 Within groups, comparisons were made using repeated measures ANOVA for normally distributed data and Friedman's test for skewed data.

Tests for comparison of paired data

T ests for A ssociation between V ariables

( Table 3 lists the tests used to determine the association between variables. Correlation determines the strength of the relationship between two variables; regression allows the prediction of one variable from another. Tyagi examined the correlation between ETCO 2 and PaCO 2 in patients with chronic obstructive pulmonary disease with acute exacerbation, who were mechanically ventilated. 8 Since these were normally distributed variables, the linear correlation between ETCO 2 and PaCO 2 was determined by Pearson's correlation coefficient. Parajuli et al. compared the acute physiology and chronic health evaluation II (APACHE II) and acute physiology and chronic health evaluation IV (APACHE IV) scores to predict intensive care unit mortality, both of which were ordinal data. Correlation between APACHE II and APACHE IV score was tested using Spearman's coefficient. 9 A study by Roshan et al. identified risk factors for the development of aspiration pneumonia following rapid sequence intubation. 10 Since the outcome was categorical binary data (aspiration pneumonia— yes/no), they performed a bivariate analysis to derive unadjusted odds ratios, followed by a multivariable logistic regression analysis to calculate adjusted odds ratios for risk factors associated with aspiration pneumonia.

Tests for assessing the association between variables

T ests for A greement between M easurements

( Table 4 outlines the tests used for assessing agreement between measurements. Gunalan evaluated concordance between the National Healthcare Safety Network surveillance criteria and CPIS for the diagnosis of ventilator-associated pneumonia. 11 Since both the scores are examples of ordinal data, Kappa statistics were calculated to assess the concordance between the two methods. In the previously quoted study by Tyagi, the agreement between ETCO 2 and PaCO 2 (both numerical variables) was represented using the Bland–Altman method. 8

Tests for assessing agreement between measurements

T ests for T ime-to -E vent D ata (S urvival A nalysis )

Time-to-event data represent a unique type of data where some participants have not experienced the outcome of interest at the time of analysis. Such participants are considered to be “censored” but are allowed to contribute to the analysis for the period of their follow-up. A detailed discussion on the analysis of time-to-event data is beyond the scope of this article. For analyzing time-to-event data, we use survival analysis (with the Kaplan–Meier method) and compare groups using the log-rank test. The risk of experiencing the event is expressed as a hazard ratio. Cox proportional hazards regression model is used to identify risk factors that are significantly associated with the event.

Hasanzadeh evaluated the impact of zinc supplementation on the development of ventilator-associated pneumonia (VAP) in adult mechanically ventilated trauma patients. 12 Survival analysis (Kaplan–Meier technique) was used to calculate the median time to development of VAP after ICU admission. The Cox proportional hazards regression model was used to calculate hazard ratios to identify factors significantly associated with the development of VAP.

The choice of statistical test used to analyze research data depends on the study hypothesis, the type of data, the number of measurements, and whether the data are paired or unpaired. Reviews of articles published in medical specialties such as family medicine, cytopathology, and pain have found several errors related to the use of descriptive and inferential statistics. 12 – 15 The statistical technique needs to be carefully chosen and specified in the protocol prior to commencement of the study, to ensure that the conclusions of the study are valid. This article has outlined the principles for selecting a statistical test, along with a list of tests used commonly. Researchers should seek help from statisticians while writing the research study protocol, to formulate the plan for statistical analysis.

Priya Ranganathan https://orcid.org/0000-0003-1004-5264

Source of support: Nil

Conflict of interest: None

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Statistics By Jim

Making statistics intuitive

Comparing Hypothesis Tests for Continuous, Binary, and Count Data

By Jim Frost 46 Comments

In a previous blog post, I introduced the basic concepts of hypothesis testing and explained the need for performing these tests. In this post, I’ll build on that and compare various types of hypothesis tests that you can use with different types of data, explore some of the options, and explain how to interpret the results. Along the way, I’ll point out important planning considerations, related analyses, and pitfalls to avoid.

A hypothesis test uses sample data to assess two mutually exclusive theories about the properties of a population . Hypothesis tests allow you to use a manageable-sized sample from the process to draw inferences about the entire population.

I’ll cover common hypothesis tests for three types of variables —continuous, binary, and count data. Recognizing the different types of data is crucial because the type of data determines the hypothesis tests you can perform and, critically, the nature of the conclusions that you can draw. If you collect the wrong data, you might not be able to get the answers that you need.

Related posts : Qualitative vs. Quantitative Data , Guide to Data Types and How to Graph Them , Discrete vs. Continuous , and Nominal, Ordinal, Interval, and Ratio Scales

Hypothesis Tests for Continuous Data

Continuous data can take on any numeric value, and it can be meaningfully divided into smaller increments, including fractional and decimal values. There are an infinite number of possible values between any two values. You often measure a continuous variable on a scale. For example, when you measure height, weight, and temperature, you have continuous data . With continuous variables, you can use hypothesis tests to assess the mean, median, and standard deviation.

When you collect continuous data, you usually get more bang for your data buck compared to discrete data. The two key advantages of continuous data are that you can:

• Draw conclusions with a smaller sample size.
• Use a wider variety of analyses, which allows you to learn more.

I’ll cover two of the more common hypothesis tests that you can use with continuous data—t-tests to assess means and variance tests to evaluate dispersion around the mean. Both of these tests come in one-sample and two-sample versions. One-sample tests allow you to compare your sample estimate to a target value. The two-sample tests let you compare the samples to each other. I’ll cover examples of both types.

There is also a group of tests that assess the median rather than the mean. These are known as nonparametric tests and practitioners use them less frequently. However, consider using a nonparametric test if your data are highly skewed and the median better represents the actual center of your data than the mean.

Related posts : Nonparametric vs. Parametric Tests and Determining Which Measure of Central Tendency is Best for Your Data

Graphing the data for the example scenario

Suppose we have two production methods, and our goal is to determine which one produces a stronger product. To evaluate the two methods, we draw a random sample of 30 products from each production line and measure the strength of each unit. Before performing any analyses, it’s always a good idea to graph the data because it provides an excellent overview. Here is the CSV data file in case you want to follow along: Continuous_Data_Examples .

These histograms suggest that Method 2 produces a higher mean strength while Method 1 produces more consistent strength scores. The higher mean strength is good for our product, but the greater variability might produce more defects.

Graphs provide a good picture, but they do not test the data statistically. The differences in the graphs might be caused by random sample error rather than an actual difference between production methods. If the observed differences are due to random error, it would not be surprising if another sample showed different patterns. It can be a costly mistake to base decisions on “results” that vary with each sample. Hypothesis tests factor in random error to improve our chances of making correct decisions.

Keep this graph in mind when we look at binary data because they illustrate how much more information continuous data convey.

Related posts : Using Histograms to Understand Your Data and How Hypothesis Tests Work: Significance Levels and P-values

Two-sample t-test to compare means

The first thing we want to determine is whether one of the methods produces stronger products. We’ll use a two-sample t-test to determine whether the population means are different. The hypotheses for our 2-sample t-test are:

• Null hypothesis: The mean strengths for the two populations are equal.
• Alternative hypothesis : The mean strengths for the two populations are different.

A p-value less than the significance level indicates that you can reject the null hypothesis. In other words, the sample provides sufficient evidence to conclude that the population means are different. Below is the output for the analysis.

The p-value (0.034) is less than 0.05. From the output, we can see that the difference between the mean of Method 2 (98.39) and Method 1 (95.39) is statistically significant. We can conclude that Method 2 produces a stronger product on average.

That sounds great, and it appears that we should use Method 2 to manufacture a stronger product. However, there are other considerations. The t-test tells us that Method 2’s mean strength is greater than Method 1, but it says nothing about the variability of strength values. For that, we need to use another test.

Related posts : How T-Tests Work and How to Interpret P-values Correctly and Step-by-Step Instructions for How to Do t-Tests in Excel .

2-Variances test to compare variability

A production method that has excessive variability creates too many defects. Consequently, we will also assess the standard deviations of both methods. To determine whether either method produces greater variability in the product’s strength, we’ll use the 2 Variances test. The hypotheses for our 2 Variances test are:

• Null hypothesis: The standard deviations for the populations are equal.
• Alternative hypothesis: The standard deviations for the populations are different.

A p-value less than the significance level indicates that you can reject the null hypothesis. In other words, the sample provides sufficient evidence for concluding that the population standard deviations are different. The 2-Variances output for our product is below.

Both of the p-values are less than 0.05. The output indicates that the variability of Method 1 is significantly less than Method 2. We can conclude that Method 1 produces a more consistent product.

Related post : Measures of Variability

What we learned and did not learn with the hypothesis tests

The hypothesis test results confirm the patterns in the graphs. Method 2 produces stronger products on average while Method 1 produces a more consistent product. The statistically significant test results indicate that these results are likely to represent actual differences between the production methods rather than sampling error.

Our example also illustrates how you can assess different properties using continuous data, which can point towards different decisions. We might want the stronger products of Method 2 but the greater consistency of Method 1. To navigate this dilemma, we’ll need to use our process knowledge.

Finally, it’s crucial to note that the tests produce estimates of population parameters—the population means (μ) and the population standard deviations (σ). While these parameters can help us make decisions, they tell us little about where individual values are likely to fall. In certain circumstances, knowing the proportion of values that fall within specified intervals is crucial.

For the examples, the products must fall within spec limits. Even when the mean falls within the spec limit, it’s possible that too many individual items will fall outside the spec limits if the variability is too high.

Other types of analyses

To better understand the distribution of individual values rather than the population parameters, use the following analyses:

Tolerance intervals : A tolerance interval is a range that likely contains a specific proportion of a population. For our example, we might want to know the range where 99% of the population falls for each production method. We can compare the tolerance interval to our requirements to determine whether there is too much variability.

Capability analysis : This type of analysis uses sample data to determine how effectively a process produces output with characteristics that fall within the spec limits. These tools incorporate both the mean and spread of your data to estimate the proportion of defects.

Related post : Confidence Intervals vs. Prediction Intervals vs. Tolerance Intervals

Proportion Hypothesis Tests for Binary Data

Let’s switch gears and move away from continuous data. Suppose we take another random sample of our product from each of the production lines. However, instead of measuring a characteristic, inspectors evaluate each product and either accept or reject it.

Binary data can have only two values. If you can place an observation into only two categories, you have a  binary variable . For example, pass/fail and accept/reject data are binary. Quality improvement practitioners often use binary data to record defective units.

Binary data are useful for calculating proportions or percentages, such as the proportion of defective products in a sample. You simply take the number of defective products and divide by the sample size. Hypothesis tests that assess proportions require binary data and allow you to use sample data to make inferences about the proportions of populations.

2 Proportions test to compare two samples

For our first example, we will make a decision based on the proportions of defective parts. Our goal is to determine whether the two methods produce different proportions of defective parts.

To make this determination, we’ll use the 2 Proportions test. For this test, the hypotheses are as follows:

• Null hypothesis: The proportions of defective parts for the two populations are equal.
• Alternative hypothesis: The proportions of defective parts for the two populations are different.

A p-value less than the significance level indicates that you can reject the null hypothesis. In this case, the sample provides sufficient evidence for concluding that the population proportions are different. The 2 Proportions output for our product is below.

Both p-values are less than 0.05. The output indicates that the difference between the proportion of defective parts for Method 1 (~0.062) and Method 2 (~0.146) is statistically significant. We can conclude that Method 1 produces defective parts less frequently.

1 Proportion test example: comparison to a target

The 1 Proportion test is also handy because you can compare a sample to a target value. Suppose you receive parts from a supplier who guarantees that less than 3% of all parts they produce are defective. You can use the 1 Proportion test to assess this claim.

First, collect a random sample of parts and determine how many are defective. Then, use the 1 Proportion test to compare your sample estimate to the target proportion of 0.03. Because we are interested in detecting only whether the population proportion is greater than 0.03, we’ll use a one-sided test. One-sided tests have greater power to detect differences in one direction, but no ability to detect differences in the other direction. Our one-sided 1 Proportion test has the following hypotheses:

• Null hypothesis: The proportion of defective parts for the population equals 0.03 or less.
• Alternative hypothesis: The proportion of defective parts for the population is greater than 0.03.

For this test, a significant p-value indicates that the supplier is in trouble! The sample provides sufficient evidence to conclude that the proportion of all parts from the supplier’s process is greater than 0.03 despite their assertions to the contrary.

Comparing continuous data to binary data

Think back to the graphs for the continuous data. At a glance, you can see both the central location and spread of the data. If we added spec limits, we could see how many data points are close and far away from them. Is the process centered between the spec limits? Continuous data provide a lot of insight into our processes.

Now, compare that to the binary data that we used in the 2 Proportions test. All we learn from that data is the proportion of defects for Method 1 (0.062) and Method 2 (0.146). There is no distribution to analyze, no indication of how close the items are to the specs, and no indication of how they failed the inspection. We only know the two proportions.

Additionally, the samples sizes are much larger for the binary data than the continuous data (130 vs. 30). When the difference between proportions is smaller, the required sample sizes can become quite large. Had we used a sample size of 30 like before, we almost certainly would not have detected this difference.

In general, binary data provide less information than an equivalent amount of continuous data. If you can collect continuous data, it’s the better route to take!

Related post : Estimating a Good Sample Size for Your Study Using Power Analysis

Poisson Hypothesis Tests for Count Data

Count data can have only non-negative integers (e.g., 0, 1, 2, etc.). In statistics , we often model count data using the Poisson distribution. Poisson data are a count of the presence of a characteristic, result, or activity over a constant amount of time, area, or other length of observation. For example, you can use count data to record the number of defects per item or defective units per batch. With Poisson data, you can assess a rate of occurrence.

For this scenario, we’ll assume that we receive shipments of parts from two different suppliers. Each supplier sends the parts in the same sized batch. We need to determine whether one supplier produces fewer defects per batch than the other supplier.

To perform this analysis, we’ll randomly sample batches of parts from both suppliers. The inspectors examine all parts in each batch and record the count of defective parts. We’ll randomly sample 30 batches from each supplier. Here is the CSV data file for this example: Count_Data_Example .

Performing the Two-Sample Poisson Rate Test

We’ll use the 2-Sample Poisson Rate test. For this test, the hypotheses are as follows:

• Null hypothesis: The rates of defective parts for the two populations are equal.
• Alternative hypothesis: The rates of defective parts for the two populations are different.

A p-value less than the significance level indicates that you can reject the null hypothesis because the sample provides sufficient evidence to conclude that the population rates are different. The 2-Sample Poisson Rate output for our product is below.

Both p-values are less than 0.05. The output indicates that the difference between the rate of defects per batch for Supplier 1 (3.56667) and Supplier 2 (5.36667) is statistically significant. We can conclude that Supplier 1 produces defects at a lower rate than Supplier 2.

Hypothesis tests are a great tool that allow you to take relatively small samples and draw conclusions about entire populations. There is a selection of tests available, and different options within the tests, which make them useful for a wide variety of situations.

To see an alternative approach to these traditional hypothesis testing methods, learn about bootstrapping in statistics !

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Reader Interactions

May 10, 2024 at 11:06 am

Ah that explains why I couldn’t find an R function that returned the same output! Thank you very much for your reply – I can stop looking now!

May 9, 2024 at 11:24 am

Thank you for this article, I’ve learned a lot from reading it.

Your Two-Sample Poisson Rate Test example is very similar in structure to my data so I am trying to follow the same approach. The results pictured look like output from an R function – but I have been unable to find one that outputs results in this way. If these were indeed created by an R library/function, would you mind sharing which one you used, please?

Kind regards, Ash

May 9, 2024 at 4:49 pm

Sorry, I’m not using R. The results are from Minitab statistical software.

September 29, 2023 at 3:21 pm

Hello guys, what is the outcome varaible in independent sample t test? binary or not? Because it compares the means of two independent populations as which is greater or lower

September 29, 2023 at 5:47 pm

The outcome variable for an independent sample t-test is continuous because you’re using it to calculate the means for two groups.

The binary variable is the categorical factor in the design. The binary variable defines the two groups for which you’re calculating the means. For example, your binary variable could be gender (male/female), experimental group (control/treatment), or material type (A/B). But the outcome variable is continuous so you can calculate the means for both groups and compare them. Click the link to learn more about the test.

March 22, 2023 at 3:32 pm

The document can’t be found, is the link still working?

March 22, 2023 at 3:35 pm

You’ll need to specify which link you’re talking about so I can check it. All links should be working.

October 9, 2021 at 3:28 am

Greetings!!! Very intuitive explanation. Liked the way you have explained with sufficient examples.

Jim based on Inferential Statistics, could you include an article on A/B Testing Methodology incorporating from basics like —Data Collection Process, Dataset Splitting Procedures & Duration for carrying out such experiments.

Also if you could incorporate illustrations from different industries viz. Healthcare, Manufacturing, Logistics, Quality, Ecommerce, Marketing, Advertisement Domains, this would indeed be useful.

Nowadays A/B Testing & Multivariate Testing is being incorporated & implemented in a robust manner across Data Science domain. Article or Write-up regarding this would immensely be useful.

Looking forward to a favourable and positive response.

August 21, 2021 at 1:22 pm

The poisson test example has N of 30. I am wondering the appropriate distribution if the sample is lower than 30. Is it a t statistic or chi-square

June 16, 2021 at 12:02 pm

Hi, great post! I have an expected and observed data set and want to do additional testing to see if they differ signficantly from each other. Furthermore, the specific entries that contribute to the most weight in that significance or places that should have special attention. I did chi-square goodness of fit, but want to go further. Just to add, this is count data.

June 19, 2021 at 4:17 pm

I’m not 100% sure what you want to do to go further. Because it’s count data, you could model it with the Poisson or Negative Binomial distribution. If you have other relevant variables, you can fit a Poisson or Negative Binomial regression model to explore relationships in your data. I talk a bit about those types of models in my post about choosing the correct type of regression model . You can also perform hypothesis tests designed for that type of data. The chi-squared test you performed seems like a good one to see how the expected and observed differs!

April 29, 2021 at 11:44 pm

How do you do a independence test in Stata for a categorical variable with 6 levels and a binary variable.

April 30, 2021 at 12:23 am

I’m not a Stata user, but it sounds like you need to perform a chi-square test of independence .

December 13, 2020 at 1:00 pm

Hi Jim – thank you for this great site! I have a situation where there is a reference standard (tells me if there is truly fat in a mass) and I have 2 different methods of detecting if there is (or is not) fat in the mass. My null hypothesis is that there is no difference in detection. I have a set of masses where I know if there is fat in the masses and used the 2 methods to detect whether they were able to detect the fat. Is the 2 proportions test the most appropriate for this question? Thank you so much!

December 3, 2020 at 8:31 am

Thank you Jim for the wonderful post. It was clearly written and I enjoyed reading through it. I have an additional query. I wanted to compare the variances of two methods of measurement applied at each observation point of a field survey. The variables from both methods have binary data type. How can I do the statistical test. Thank you in advance for your help.

December 3, 2020 at 3:02 pm

With binary data, you can’t compare variances. You can compare proportions using a proportions test. I discuss these tests in the binary section of this post. To read an example of a 2-sample proportions test, read my post about flu shot effectiveness . In it, I use 2-proportions tests to evaluate real flu study data. Or read my post about Mythbusters test about whether yawns are contagious , where I use a 2-proportions test. That way you can see what these tests can do. I cover them, and many other tests, in much more detail in my Hypothesis Testing book !

I hope this helps!

August 5, 2020 at 9:36 pm

Hi Jim. Just wanted to follow up and see if you’ve had a chance to review this question yet?

August 6, 2020 at 1:09 am

Hi Jack, thanks for the reminder! I just replied!

August 3, 2020 at 4:45 am

Hi Jim , a green belt has a project on flu vaccinations , with 5 data points, % vaccination rates per year averaging about 36% of staff numbers. Her project was to increase vaccination rates this year , and has accumulated a lot of data points to measure vaccination rates in different office areas as a percent of total staff numbers which have almost doubled. Should she use 2 sample t test to measure difference in means between before and after data ( continuous) or should she use 2 sample test for proportions (attribute). There is small sample size for before data and large sample size for after data

August 5, 2020 at 12:46 am

I see two possibilities. The choice depends on whether she measured the same sites in the before and after. If they’re different sites before and after, she has independent groups and can use the regular 2-sample proportions test.

If they’re the same sites before and after, she can use either the test of marginal homogeneity or McNemar’s test. I have not used these tests myself and cannot provide more information. However, if she used the same sites before and after, she has proportions data for dependent groups (same groups) and should not use the regular 2-sample proportions test. These two tests can handle proportions for dependent groups.

July 19, 2020 at 3:18 am

Hi Jim. Would I be able to use the 2 Proportion Test for comparing 2 proportions from different time periods? Example scenario: I run a satisfaction survey on a MOOC site during Q1 to a random sample of visitors and find that 80% of them were satisfied with their experience. The following quarter I run the same survey and find that 75% were satisfied. Is the 5 percentage point drop statistically significant or just due to random noise?

Sorry about the delay in replying! Sometimes comments slip through the cracks!

Yes, you can do as you suggest assuming the respondents are different in the two quarters and assuming that the data are binary (satisfied/not satisfied). The 2 proportions test is designed for independent groups and binary data.

I hope that helps even belatedly!

May 19, 2020 at 9:58 pm

Thanks…Let me see that document

May 19, 2020 at 7:04 pm

I would like to ask about 2 sample poisson rate. How do you calculate the 95% CI and test for difference ? Your answer is really appreciated. Thank you so much for giving this tutorial.

May 19, 2020 at 9:17 pm

This document describes the calculations. I hope it helps!

March 6, 2020 at 4:52 am

Thank you so much for your kind support. Esteemed regards.

March 5, 2020 at 9:23 pm

Thanks for your helpful comments. Basically I have developed my research model based on competing theories. For example, I have one IV (Exploitation) and Two DV’s (Incremental innovation and Radical Innovation). Each variable in the model has own indicators. Some researchers claims that exploitation support only incremental innovations. On the other hand there are also studies that claims that in-depth exploitation also support radical innovation. However, these researchers claim that exploitation support radical innovations in a limited capacity as compared to incremental innovation. ON the basis of these competing theories I developed my hypothesis as: Exploitation significantly and positively influences both incremental and radical innovation, however exploitation influence incremental innovation more than radical innovation.

Thank you very much for your quick response. Its really helpful.

March 6, 2020 at 1:46 am

Hi Shabahat,

Thanks you for the additional information. I messed up one thing in my previous reply to you. Because you only have the one IV, you don’t need to worry about standardizing that IV for comparability. However, assuming the DVs use different units of measurement, that’ll make comparisons problematic. Consequently, you probably need to standardize the two dependent variables instead. You’d learn how a one unit change in the IV relates to the standard deviation of each DV. That puts the DVs on the same scale. See how other researchers have handled this situation in your field to be sure that’s an accepted approach.

Additionally, if you find a significant relationship between exploitation and radical innovation, then you’d have good evidence to support that claim you’re making.

March 5, 2020 at 2:37 am

Hi Jim, Its really amazing. However I have a query regarding my analysis. I have one independent variable (Continuous) and Two dependent variables (Continuous). In the linear regression, the Independent variable significantly explains both dependent variables. Problem: Now i want to compare the effect of my Independent variable on both dependent variables. How can I compare?. If the effect is different, how can I test whether the effect difference is statistically significant or not in SPSS.

March 5, 2020 at 4:01 pm

The fact that you’re talking about different DVs complicates things because they’re presumably measuring different things and using different units, which makes comparison difficult. The standardized coefficients can help you get around that but it changes the interpretation of the results.

Assuming that the two DV variables measure different characteristics, you might try standardizing your IVs and fitting the model using the standardized values. This process produces standardized coefficients, which use the same units and allows you to compare–although it does change the interpretation of the coefficients. I write about this in my post about assessing the importance of your predictors . You can also look at the CIs for the standardized coefficients and see if they overlap. If they don’t overlap, you know the difference between the standardized coefficients is statistically significant.

January 20, 2019 at 12:40 pm

Great Post! Can we test proportions from a continuos variable with unknown distribution using the poisson distribution using a cut-off value for good and bad samples and couting them?

November 14, 2018 at 4:05 pm

Hi Jim, I really enjoy reading your posts and they have cleared many stat concepts!

I had a question about the chi square probability distribution. Although it is a non-parametric test, why does it fall into a continuous probability distribution and why can we use the chi square distribution for categorical data if it’s a continuous probability distribution?

November 14, 2018 at 10:53 pm

Hi Sanjana,

That’s a great question! I’m glad you’re thinking about the types of variables and distributions, and how they’re used together.

You’re correct on both counts. Chi-squared test of independent is nonparameteric because it doesn’t assume a particular data distribution. Additionally, analysts use it to test the independence of categorical variables . There are other ways to use this distribution as well.

Now, onto why we use chi-square (a distribution for continuous data) with categorical variables! Yes, it involves categorical variables, but the analysis assesses the observed and expected counts of these variables. For each cell, the analysis takes the squared difference between the observed count and the expected count and then divides that by the expected count. These values are summed acrossed all cells to produce the chi-square value. This process produces a continuous variable that is based on the differences between the observed and expected counts of the categorical variables. When the value of this variable is large enough, we know that the difference between the observed counts and the expected counts is large enough to be unlikely due to chance. And, that’s why we use a continuous distribution to analyze categorical variables.

May 27, 2018 at 12:58 am

This is very helpful! Thank you!

May 26, 2018 at 11:38 pm

Hi Jim, Great post. I was wondering, do you know of any references that discuss the difference in sample size between binary and continuous data? I am looking for a reference to cite in a journal article. Thanks, Amanda.

May 26, 2018 at 11:55 pm

The article I cite below discusses the different sample sizes in terms of observations per model term in order to avoid overfitting your model. I also cover these ideas in my post about how to avoid overfitting regression models . For regression models, this provides a good context for sample size requirements.

Babyak, MA., What You See May Not Be What You Get: A Brief, Nontechnical Introduction to Overfitting in Regression-Type Models, Psychosomatic Medicine 66:411-421 (2004).

February 3, 2018 at 7:04 am

I am totally new to statistics,

Following a small sample from my dataset.

Views PosEmo NegEmo 1650077 2.63 1.27 753826 2.39 0.47 926647 1.71 1.02

Views = Dependent continous Variable PosEmo = Independent Continous Variable NegEmo = Independent Continous Variable

My query : 1. How to run Hypothesis testing on same, Im pretty confused what to use , what to do , I am using SPSS modeler and SPSS statistics tool. 2.I think Multiple Regression is Ok for this . Let me know how to use it in SPSS modeler or stats tool.

Regards Sarika

February 5, 2018 at 1:30 am

Hi Sarika, yes, it sounds like you can use multiple regression for those data. The hypothesis test in this case would be the p-values for the regression coefficients . Click that link to learn more about that. In your stats software, choose multiple linear regression and then specify the dependent variable and the two independent variables. Fit the model and then check the statistical output and the residual plots to see if you have a good model. Be sure to check out my regression tutorial too. That covers many aspects of regression analysis.

November 30, 2017 at 8:45 pm

Thanks for your sharing!

In the binary case (or proportion case), is there any comparison between “two proportion test” and “Chi-square” test? Is there any guideline to choose which test to use?

November 30, 2017 at 9:46 pm

You’re welcome! Regarding your question, a two proportion test requires one categorical variable with two levels. For example, the variable could be “test result” and the two levels are “pass” and “fail.”

A chi-square test of independence requires at least two categorical variables. Those variables can have two or more levels. You can read an example of the chi-square test of independence that I’ve written about. The example is based on the original Star Trek TV series and determines whether the uniform color affects the fatality rate. That analysis has two categorical variables–fatalities and uniform color. Fatalities has two levels that indicate whether a crewmember survived or died. Uniform color has three levels–gold, blue, and red.

As you can see, the data requirements for the two tests are different.

I hope this helps! Jim

November 30, 2017 at 2:18 am

November 29, 2017 at 10:41 pm

Great post. Thanks for sharing your expertise.

November 29, 2017 at 11:38 pm

Thank you! I’m glad it was helpful.

November 29, 2017 at 9:02 pm

Very nice article. Could you explain more on hypothesis testing on median?

November 29, 2017 at 11:39 pm

Thank you! For more information about testing the median, click the link in the article for where I compare parametric vs nonparametric analyses.

November 29, 2017 at 6:51 pm

Please let me know when one can use Probit Analysis. May I know the Procedure in SPSS.

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December 15

Ordinal Data – What Is It, And How Do You Analyse It?

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Ordinal data is one of only 4 types of data in statistics.

Do you know what they all are and what you can do with them?

If you want to know everything there is to know about Ordinal data - definitions, examples, analysis and statistics - then you're in the right place.

When you're done here, you'll also want to read this post's sister articles on quantitative data and qualitative data , Nominal data , Interval data and Ratio data .

For now, though, here is our guide to Ordinal data  and how to deal with them...

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This post forms part of a series on the 4 types of data in statistics.

For more detail, choose from the options below:

4 Types of Data in Statistics: Introduction

Nominal data, ordinal data, interval data, all 4 types of data compared, statistical hypothesis testing, what is ordinal data.

If you want a simple definition of Nominal data, it would be this:

Ordinal Data Definition

Ordinal data is a type of categorical data in which the values follow a natural order

Ordinal data is the statistical data type that has the following characteristics:

Ordinal Data are observed, not measured, are ordered but non-equidistant and have no meaningful zero

Their categories can be ordered (1st, 2nd, 3rd, etc. - hence the name 'ordinal'), but there is no consistency in the relative distances between adjacent categories.

As with Nominal data, Ordinal data can have 2 categories, and we also call these dichotomous data. The only difference between Nominal dichotomous data and Ordinal dichotomous data is that Ordinal dichotomous data have an order, whereas Nominal dichotomous data do not.

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Ordinal Data Examples

Examples of Ordinal data include:

• Health (healthy, sick)
• Opinion (agree, mostly agree, neutral, mostly disagree, disagree)
• Tumour Grade (1, 2, 3)
• Tumour Stage (I, IIA, IIB, IIIA, IIIB, etc.)
• Time of day (morning, noon, night)

In the examples of Ordinal data above, there is an obvious order to the categories. Are any of the examples of Ordinal data above examples of dichotomous data?

What Can You Calculate With Ordinal Variables?

When Ordinal data are used in analysis, they are called Ordinal Variables, so that's what we'll call them from here.

With Ordinal variables you can group the data by assessing whether they are the same or different.

As Ordinal variables are ordered, they can be sorted by making simple comparisons between the categories, such as Greater/Less than , M ore / Less , Higher/Lower , etc..

You can't do any mathematical operations with Ordinal variables, though, because they aren't numerical data.

For example, you can group people according to their socioeconomic status, but you can also sort them as well. Upper class is higher on the scale than Middle class, which in turn is higher than Lower class.

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What Descriptive Statistics Can You Do With Ordinal Data?

With Ordinal variables you can calculate precisely the same things as you can with Nominal variables, with a couple of extra things (see if you can spot them). That is:

• Frequencies - count how many you have in each category
• Proportions - determine how often something happens by dividing the frequency by the total number of events
• Percentages - transform the proportions to percentages by multiplying by 100
• Central point - since there is an order to the data you can rank them and compute the  median (or mode, but not the mean) to find the central value.
• Summary statistics - as the data are ordered, you can use percentiles and the inter-quartile range to summarise your data

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What data visualisations can you do with ordinal variables.

As with Nominal variables, since the only descriptive statistics you can do with Ordinal data are frequencies, proportions and percentages, the only ways to visualise these are with pie charts and bar charts.

Ordinal Data Also Make Great Dummy Variables!

By now you might be wondering whether you can create dummy variables with Ordinal data. Well, you can!

It can be slightly different with Ordinal data, though, because their categories are ordered. Instead of choosing a category, you would usually use a cut-off value and base your dummy variable around that.

Let's look at an example. Say you have ages categorised as 20's, 30's, 40's, 50's and 60's, and you suspect that your data are different for those 50 and above. Here, you would code all ages below 50 as 0 and all ages 50+ as 1. You now have a dummy variable, and you can analyse it using all the same techniques as you would a dummy variable created using Nominal data.

Ordinal variables are very common in surveys and questionnaires, and the Likert scale is one example of collecting Ordinal data.

You can use Ordinal variables in non-parametric hypothesis tests that rely on ranking, such as the Mann-Whitney U-test or the Wilcoxon Matched-Pairs test.

Ordinal Data - What Is It, And How Do You Analyse It? Everything You Need To Know (And More) @chi2innovations #dataanalytics #datatypes #statistics

What Statistics Can You Do With Ordinal Variables?

Ordinal variables can be used in precisely the same statistical hypothesis tests as Nominal variables. Actually, that's not quite true - there are some additional tests you can use with Ordinal variables, but they are more advanced and are beyond the scope of this article. If you want to know more, you can look up the Chi-Squared For Trend Test, Ordinal Logistic Regression and Nominal Logistic Regression.

For now, just know that for the pairwise tests, you can treat Ordinal variables and Nominal variables exactly the same.

Ordinal vs Nominal Data - What is the Difference?

Ordinal data and Nominal data are both qualitative data, and the difference between them is that Nominal data can only be classified - arranged into classes or categories - whereas Ordinal data can be classified and ordered.

One of the assumptions of Ordinal data is that although the categories are ordered, they do not have equal intervals.

It is interesting to note that in practice some Ordinal data are treated as though they do have equal intervals because the statistical tests that can be used on these data are much more powerful than those typically used on Ordinal data. This is OK as long as your data collection methods ensure that the equidistant rule isn't bent too much.

Tumour Grade [1, 2, 3] is a classic example of this in healthcare.

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Ordinal Data: Summary

The basics of statistics, like data collection , data cleaning and data integrity aren't sexy, and as a result are often neglected, and that is also the case with data types.

In my experience, few people that have to do statistics as part of their research know and understand the statistical data types, and as a result struggle to get to grips with what they can and can't do with their data.

That's a shame, because as you've seen, if you know the 4 types of data in statistics you know:

• How to handle them correctly
• What you can calculate with them
• Which descriptive statistics and visualisations are appropriate
• Which statistical hypothesis tests you can use

In short, data types are a roadmap to doing your entire study properly.

They really are that important!

Hopefully, by now you have a good understanding of what Ordinal data are, and what you can do with them.

Ordinal Data are observed, not measured. They are ordered but non-equidistant and have no meaningful zero. Categories are named, and once data are grouped you can sort them.

Ordinal data are types of Qualitative data (also known as categorical data), and you cannot perform any mathematical operations on Ordinal data.

Now that you know everything there is to know about Ordinal data, you might also like to read this post's sister articles on quantitative data and qualitative data , Nominal data , Interval data and Ratio data .

In the final posts we'll compare each of the 4 types of data and I'll also show you how to choose the correct statistical hypothesis test .

Do you have any questions about Ordinal data? Is there something that I've missed out?

Let me know in the comments below - your feedback will help me to improve the post and make learning about data and statistics easier for everybody!

data tips, ordinal data, qualitative data

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Have you ever looked at your data and wondered how and where to get started?

If you don't know the difference between Quantitative and Qualitative data, or between Ratio, Interval, Ordinal, and Nominal data, then you're in the right place.

Here is our guide to statistical data types and how to deal with them.

Five Ways to Analyze Ordinal Variables (Some Better than Others)

by Karen Grace-Martin   6 Comments

But that doesn’t mean that you’re stuck with few options.  There are more than you’d think.

Some are better than others, but it depends on the situation and research questions.

Here are five options when your dependent variable is ordinal.

1. Analyze ordinal variables as if they’re nominal

Ordinal variables are fundamentally categorical. One simple option is to ignore the order in the variable’s categories and treat it as nominal . There are many options for analyzing categorical variables that have no order.

This can make a lot of sense for some variables. For example, when there are few categories and the order isn’t central to the research question.

The biggest advantage to this approach is you won’t violate any assumptions .

2. Analyze ordinal variables as if they’re numeric

Because the ordering of the categories often is central to the research question, many data analysts do the opposite: ignore the fact that the ordinal variable really isn’t numerical and treat the numerals that designate each category as actual numbers.

This approach requires the assumption that the distance between each set of subsequent categories is equal. And that can be very difficult to justify.

So think long and hard about whether you’re able to justify this assumption.

3. Non-parametric tests

Some good news: there are other options.

Many non-parametric descriptive statistics are based on ranking numerical values. Ranks are themselves ordinal–they tell you information about the order, but no distance between values.

Just like other ordinal variables.

So while we think of these tests as useful for numerical data that are non-normal or have outliers, they work for ordinal variables as well, especially when there are more than just a few ordered categories.

Common rank-based non-parametric tests include Kruskal-Wallis , Spearman correlation , Wilcoxon-Mann-Whitney, and Friedman.

Each test has a specific test statistic based on those ranks, depending on whether the test is comparing groups or measuring an association.

The limitation of these tests, though, is they’re pretty basic.  Sure, you can compare groups one-way ANOVA style or measure a correlation, but you can’t go beyond that.  You can’t, for example, include interactions among two independent variables or include covariates.

You need a real model to do that.

4. Ordinal logistic & probit regression

There aren’t many tests that are set up just for ordinal variables, but there are a few.  One of the most commonly used is ordinal models for logistic (or probit ) regression.

There are a few different ways of specifying the logit link function so that it preserves the ordering in the dependent variable. The most commonly available in software is the cumulative link function, which allows you to measure the effect of predictors on the odds of moving into any next-highest-ordered category.

These models are complex, have their own assumptions, and can take some practice to interpret. But they are also sometimes exactly what you need.

They are a very good tool to have in your statistical toolbox.

5. Rank transformations

Another model-based approach combines the advantages of ordinal logistic regression and the simplicity of rank-based non-parametrics.

The basic idea is a rank transformation: transform each ordinal outcome score into the rank of that score and run your regression, two-way ANOVA, or other model on those ranks.

The thing to remember though, is that all results need to be interpreted in terms of the ranks.  Just as a log transformation on a dependent variable puts all the means and coefficients on a log(DV) scale, the rank transformation puts everything on a rank scale. Your interpretations are going to be about mean ranks, not means.

Reader Interactions

June 26, 2024 at 7:39 am

How ordinal data can be analyzed by mean?

June 26, 2024 at 9:08 am

Well, you have to make some really big assumptions to do it.

December 4, 2022 at 5:32 am

What’s the inferential statistical techniques for all the level of measurement in statistics

October 21, 2022 at 8:12 pm

For #5, it has been shown that interaction effects under simple rank transformations explode Type I error rates. Put simply, you can’t just rank-transform your DV and run, say, a two-way ANOVA and safely interpret the interaction effect. An increasingly common fix is to use the ALIGNED rank transform (ART), which aligns the response for each main effect and interaction separately, before ranking. This procedure preserves correct Type I error rates for all effects, including interactions. We created an R package called ARTool that performs the ART procedure and gives ANOVA table results. We also have a Windows executable for aligning-and-ranking data. You can find out more on our project page here: http://depts.washington.edu/acelab/proj/art/

October 26, 2022 at 9:58 am

Thanks, Jacob. I’ll take a look. I hadn’t heard of aligned rank transformation.

October 14, 2020 at 9:17 am

1. am running ordinal logistic regression in stata but when I tried parallel lines test using the command oparallel it responded me in some explanatory variables as hessian is not negative semi definite and in some it says full model can’t be estimated due to perfect prediction. how can I solve this problem, please? 2. I have two categorical variables which are ordinal, what is the best way to analyze my data either using ordinal logistic regression for each of the dependent variable or any one model to use in combination? sincerely!

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Ordinal Scale and Nonparametric Methods

As ordinal scales are frequently encountered in research studies, the usual parametric tests don't hold true because of two reasons. First, they assume a level of measurement of interval/ratio scales. Second, they assume that the samples are drawn from a population with a known distribution such as the normal distribution. Measurement of attitudes, consumer tastes and preferences, and ranking of attributes are very prevalent in research. You need exclusive hypothesis testing procedures that deal with ordinal scales. These fall under a set of elegant nonparametric methods.

This article attempts to give an illustrative account of nonparametric methods that are used in ordinal scales of measurement. The coverage is by no means exhaustive. However typical situations are discussed to throw light on how useful these tests are.

Next-Kolmogorov-Smirnov Test

Kolmogorov-Smirnov Test:

Kolmogorov-Smirnov test is a test of goodness of fit for the univariate case when the scale of measurement is ordinal. It is similar to the chi-square test of goodness of fit in the sense it also examines whether the observed frequencies are in accordance with the expected frequencies under a well defined null hypothesis. Of course the chi-square test involves nominal measurement. Kolmogorov-Smirnov test is more powerful than the chi-square test when ordinal data are encountered in any decision problem. In the concluding remarks, you will see the advantages of using Kolmogorov-Smirnov test over the chi-square test. To understand how this test works in practice, let us take an example.

A manufacturing company producing decorative paints is interested in knowing whether the consumers have distinct preferences for different shades in the context of a new decorative paint that it proposes to market. If the consumers have special preference for any particular shade, then the company would market only that shade. Else, it would plan to market all the shades. A sample of 150 consumers was interviewed and the data collected on shade preferences are given in the table below:

Table showing shade preferences

What are your conclusions?

  Next-Analysis and Interpretations                         previous

Analysis and Interpretations:

The test involves comparing the expected cumulative distribution function under the null hypothesis being true with that of observed cumulative distribution function. If we designate Fo(X) as the expected cumulative distribution function and Sn(X) as the observed cumulative distribution function, Kolmogorov-Smirnov D is calculated as D = Max |Fo(X)-Sn(X)| (D is the absolute difference between the expected cumulative proportion and the observed cumulative proportion). Please note that n is the sample size. The following table shows the necessary calculations.

Table1: Basic Calculations for the Example

The null hypothesis is that all shades are equally preferred

The alternative hypothesis is that they are not equally preferred

Computed D = Max |Fo(X)-Sn(X)| = 0.1667. The critical D value for a level of significance of 5% is given by

Next- Concluding Remarks on Kolmogorov-Smirnov Test                        previous

Concluding Remarks on Kolmogorov-Smirnov Test

You could very well have used the chi-square test of goodness of fit for testing the hypothesis of equal preference for all shades in this example instead of the Kolmogorov-Smirnov test. When the data measurement are ordinal, Kolmogorov-Smirnov test is more powerful than the chi-square test for the following reasons. 

Median Test

Median Test:

Median test is used for testing whether two groups differ in their median value. In simple terms, median test will focus on whether the two groups come from populations with the same median. This test stipulates the measurement scale is at least ordinal and the samples are independent (not necessary of the same sample size). The null hypothesis structured is that the two populations have the same median. Let us take an example to appreciate how this test is useful in a typical practical situation.

Example:   A private bank is interested in finding out whether the customers belonging to two groups differ in their satisfaction level. The two groups are customers belonging to current account holders and savings account holders. A random sample of 20 customers of each category was interviewed regarding their perceptions of the bank's service quality using a Likert-type (ordinal scale) statements. A score of "1" represents very dissatisfied and a score of "5" represents very satisfied. The compiled aggregate scores for each respondent in each groupare tabulated be given below:

What are your conclusions regarding the satisfaction level of these two groups?

Next-Analysis and Interpretations                             previous

The first task in the median test is to obtain the grand median. Arrange the combined data of both the groups in the descending  order of magnitude. That is rank them from the highest to the lowest. Select the middle most observation in the ranked data. In this case, median is the average of 20th and 21st observation in the array that has been arranged in the descending order of magnitude. 

Table showing descending order of aggregate score and rank in the combined sample

Grand median is the average of 20th and 21st observation = (62+61)/2 =61.5. Please note that in the above table, average rank is taken whenever the scores are tied. The next step is to prepare a contingency table of two rows and two columns. The cells represent the number of observations that are above and below the grand median in each group. Whenever some observations in each group coincide with the median value, the accepted practice is to first count the observations that are strictly above grand median and put the rest under below grand median. In other words, below grand median in such cases would include less than or equal to grand median.

Scores of Current Account Holders and Savings Account Holders as compared with Grand Median

Null Hypothesis: There is no difference between the current account holders and savings account holders in the perceived satisfaction level.

alternative Hypothesis: There is difference between the current account holders and savings account holders in the perceived satisfaction level.

The test statistic to be used is given by

on substituting the values of  a, b, c, d, and n we have

Critical chi-square for 1 d.f at 5% level of significance = 3.84. Since the computed chi-square(0.90) is less than critical chi-square(3.84), we have no convincing evidence to reject the null hypothesis. Thus the the data are consistent with the null hypothesis that there is no difference between the current account holders and savings account holders in the perceived satisfaction level.

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Ordinal Data

A type of data in which the data values follow a natural order

What is Ordinal Data?

In statistics, ordinal data are the type of data in which the values follow a natural order. One of the most notable features of ordinal data is that the differences between the data values cannot be determined or are meaningless. Generally, the data categories lack the width representing the equal increments of the underlying attribute.

In some cases, the values of interval or ratio data can be grouped together to obtain the data’s characteristics. For example, the ranges of income are considered ordinal data while the income itself is the ratio data.

Unlike interval or ratio data, ordinal data cannot be manipulated using mathematical operators. Due to this reason, the only available measure of central tendency for datasets that contain ordinal data is the median.

Uses of Ordinal Data

Ordinal data are commonly employed in various surveys and questionnaires. The Likert scale that you may find in many surveys is one example. The Likert scale lists the categories of the psychometric scale such as “Strongly Agree,” “Agree,” etc.

Various examples of this data type can be frequently encountered in finance and economics. Consider an economic report that investigates the GDP levels of different countries. If the report ranks the countries according to their GDP figures, the ranks are examples of ordinal data.

How to Analyze the Data?

The simplest way to analyze ordinal data is to use visualization tools . For instance, the data may be presented in a table in which each row indicates a distinct category. In addition, they can also be visualized using various charts. The most commonly used chart for representing such types of data is the bar chart.

Ordinal data can also be analyzed using advanced statistical analysis tools such as hypothesis testing . Note that the standard parametric methods such as t-test or ANOVA cannot be applied to such types of data. The hypothesis testing of the data can be carried out only using nonparametric tests such as the Mann-Whitney U test or Wilcoxon Matched-Pairs test.

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Hypothesis testing and estimation in ordinal data under a simple crossover design

Affiliation.

• 1 Department of Mathematics and Statistics, College of Sciences, San Diego State University, San Diego, California 92182-7720, USA. [email protected]
• PMID: 23075013
• DOI: 10.1080/10543406.2011.574326

Since each patient serves as his/her own control, the crossover design can be of use to improve power as compared with the parallel-groups design in studying noncurative treatments to certain chronic diseases. Although the research studies on the crossover design have been quite intensive, the discussions on analyzing ordinal data under such a design are truly limited. We propose using the generalized odds ratio (GOR) for paired sample data to measure the relative effect on patient responses for both treatment and period in ordinal data under a simple crossover trial. Assuming the treatment and period effects are multiplicative, we note that one can easily derive the maximum likelihood estimator (LE) in closed forms for the GOR of treatment and period effects. We develop asymptotic and exact procedures for testing treatment and period effects. We further derive asymptotic and exact interval estimators for the GOR of treatment and period effects. We use the data taken from a crossover trial to assess the clarity of leaflet instructions between two devices among asthma patients to illustrate the use of these test procedures and estimators developed here.

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Hypothesis test on ordinal data

• Thread starter ultimatejester
• Start date Nov 2, 2007
• Tags Data Ordinal Test
• Nov 2, 2007
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A PF Mountain

You can; with caveats. Try googling chi-square test ( http://en.wikipedia.org/wiki/Chi-square_test )  

I could do proportions and use test statistic for proportions but the neutral category causes problem because that splits the customer base into categories. -Satisfied -Neutral -Dissatisfied.  

You could use binary variables: Is the customer highly satisfied? Is the customer at least satisfied? Is the customer at least neutral? Is the customer at least dissatisfied?  

ultimatejester said: Thank you for replying(much sooner than i expected :) I am familiar with chi distribution but i don't think it would work here

• Nov 5, 2007
• Nov 8, 2007

hey this may sound weird but i have the same question well it seems the same.. and I am stuck I am just wondering how you did it? did u use the population proportion?  

Yes, i used proportions. You can change the code for success from 1,2,3,4,5 but the one you are really interested in is the "highly satisfied" or 1. I did a one tail test since the company wants to know where its is meeting the required 95% level. To get you started H0: u=95% H1: U<95% Good Luck  

satisfaction question so would you recommend using the t-test, since the std. dev. is unknown? also, wat would the null and alternative hypotheses be? thanks a lot!  

• Nov 12, 2007

The null and alternative hypothesis are mentioned in my above post. You can't use the t-test. All you are allowed to do on data in count the occurences. Use proportions. Post here if you need more help. Thanks. Adil.A  

• May 15, 2010

Hi, I was wondering if u could post how u did that question. I'm having the same problem and i still can't figure it out :( Any help would be really appreciated! Thanks! Dona.  

Related to Hypothesis test on ordinal data

A hypothesis test on ordinal data is a statistical analysis that is used to determine whether there is a significant difference between two or more groups on an ordinal scale. It is commonly used in social sciences and other fields where data is ranked in a specific order.

A hypothesis test on ordinal data is different from other types of hypothesis tests because it uses data that is ranked in a specific order instead of numerical values. This type of test also uses non-parametric statistical methods, which do not assume a normal distribution of the data.

Ordinal data can include things like rankings, ratings, and categories such as "strongly agree", "agree", "neutral", "disagree", and "strongly disagree". It can also include data that is in a specific order, such as education level (e.g. high school, college, graduate school).

The steps involved in conducting a hypothesis test on ordinal data include: 1) defining the research question and hypothesis, 2) selecting an appropriate statistical test, 3) collecting and organizing the data, 4) performing the statistical test, 5) interpreting the results, and 6) drawing conclusions and making recommendations based on the results.

One limitation of hypothesis tests on ordinal data is that they do not provide information about the size or magnitude of the difference between groups. They also cannot determine causation, only association. Additionally, the results of these tests may be affected by the way the data is ranked or categorized.

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Test on Ordinal data with interval

I have a query regarding statistical analysis. I have collected data on percentage of savings done by individuals. These data are ordinal, grouped into equal intervals with the exception of the final category representing all savings of 50% or higher. Because of this, I gather I can't apply a parametric test.

I want to study whether there is significant difference in saving habits of two groups. Please suggest an appropriate test. I have a screen shot of the data.

Would a t-test or Mann Whitney U-test be appropriate?

• hypothesis-testing
• wilcoxon-mann-whitney-test

2 Answers 2

When the intervals are equally spaced, the data are perhaps ordinal by name only. The T-test summarizes the actual mean difference in % savings. Alas, the largest threshold group is not equal in spacing to the prior groups. So you must reframe the question slightly.

If a non-parametric test is a consideration, then perhaps you are less interested in quantifying the actual % difference between men and women. In that case, you could use the T-test anyway and comment on the statistical significance as suggesting whether one group saves more or less than the other. This "non-parametric T test" is asymptotically equivalent to the Mann-Whitney test.

Another approach is to use ordinal logistic regression. This is also called a proportional odds model. Expand the data using a full data matrix of 252+98 observations or use a weighted dataset for the 10 tabular values:

Then fit the model with Savings as a response, sex as the main predictor, and supply the Weight to obtain a semi-parametric test of statistical significance.

Lastly, you are not forbidden from using a parametric model. Simply declare an underlying parametric model for the distribution of savings, and use expectation-maximization to estimate the latent beta (or other appropriate) distribution of savings in males and in females, then estimate the T-test of association.

Following up on the response by @AdamO, in my experience with this kind of data, a Wilcoxon-Mann-Whitney test usually corresponds to ordinal regression pretty well.

I have some plots on this question near the bottom of this webpage .

As always, it's best not to put too much emphasis on the p -value, but to also assess the size of the effect. Looking at the proportions in each category for each sex, it's clear that Male has a higher proportion in "below 10%" than does Female, and that Female has higher proportions in some of the higher categories. But the overall size of the differences is relatively small: Vargha and Delaney's A suggests that the probability of Female having a higher observation than Male is only 0.57 (95% confidence interval, 0.502, 0.629, not shown below). Whether this effect size has practical importance is up to you.

For this data in R:

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