writing a null hypothesis for chi square

How to Write a Null Hypothesis (5 Examples)

A hypothesis test uses sample data to determine whether or not some claim about a population parameter is true.

Whenever we perform a hypothesis test, we always write a null hypothesis and an alternative hypothesis, which take the following forms:

H 0 (Null Hypothesis): Population parameter =,  ≤, ≥ some value

H A  (Alternative Hypothesis): Population parameter <, >, ≠ some value

Note that the null hypothesis always contains the equal sign .

We interpret the hypotheses as follows:

Null hypothesis: The sample data provides no evidence to support some claim being made by an individual.

Alternative hypothesis: The sample data  does provide sufficient evidence to support the claim being made by an individual.

For example, suppose it’s assumed that the average height of a certain species of plant is 20 inches tall. However, one botanist claims the true average height is greater than 20 inches.

To test this claim, she may go out and collect a random sample of plants. She can then use this sample data to perform a hypothesis test using the following two hypotheses:

H 0 : μ ≤ 20 (the true mean height of plants is equal to or even less than 20 inches)

H A : μ > 20 (the true mean height of plants is greater than 20 inches)

If the sample data gathered by the botanist shows that the mean height of this species of plants is significantly greater than 20 inches, she can reject the null hypothesis and conclude that the mean height is greater than 20 inches.

Read through the following examples to gain a better understanding of how to write a null hypothesis in different situations.

Example 1: Weight of Turtles

A biologist wants to test whether or not the true mean weight of a certain species of turtles is 300 pounds. To test this, he goes out and measures the weight of a random sample of 40 turtles.

Here is how to write the null and alternative hypotheses for this scenario:

H 0 : μ = 300 (the true mean weight is equal to 300 pounds)

H A : μ ≠ 300 (the true mean weight is not equal to 300 pounds)

Example 2: Height of Males

It’s assumed that the mean height of males in a certain city is 68 inches. However, an independent researcher believes the true mean height is greater than 68 inches. To test this, he goes out and collects the height of 50 males in the city.

H 0 : μ ≤ 68 (the true mean height is equal to or even less than 68 inches)

H A : μ > 68 (the true mean height is greater than 68 inches)

Example 3: Graduation Rates

A university states that 80% of all students graduate on time. However, an independent researcher believes that less than 80% of all students graduate on time. To test this, she collects data on the proportion of students who graduated on time last year at the university.

H 0 : p ≥ 0.80 (the true proportion of students who graduate on time is 80% or higher)

H A : μ < 0.80 (the true proportion of students who graduate on time is less than 80%)

Example 4: Burger Weights

A food researcher wants to test whether or not the true mean weight of a burger at a certain restaurant is 7 ounces. To test this, he goes out and measures the weight of a random sample of 20 burgers from this restaurant.

H 0 : μ = 7 (the true mean weight is equal to 7 ounces)

H A : μ ≠ 7 (the true mean weight is not equal to 7 ounces)

Example 5: Citizen Support

A politician claims that less than 30% of citizens in a certain town support a certain law. To test this, he goes out and surveys 200 citizens on whether or not they support the law.

H 0 : p ≥ .30 (the true proportion of citizens who support the law is greater than or equal to 30%)

H A : μ < 0.30 (the true proportion of citizens who support the law is less than 30%)

Additional Resources

Introduction to Hypothesis Testing Introduction to Confidence Intervals An Explanation of P-Values and Statistical Significance

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Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike.  My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.

2 Replies to “How to Write a Null Hypothesis (5 Examples)”

you are amazing, thank you so much

Say I am a botanist hypothesizing the average height of daisies is 20 inches, or not? Does T = (ave – 20 inches) / √ variance / (80 / 4)? … This assumes 40 real measures + 40 fake = 80 n, but that seems questionable. Please advise.

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Null Hypothesis In Chi Square

Unraveling the null hypothesis in chi-square analysis.

In the vast landscape of statistical analysis, where numbers dance and patterns emerge, the chi-square test stands as a stalwart, helping researchers discern the significance of observed data. At its heart lies a critical concept: the null hypothesis. Let us embark on a journey to demystify this cornerstone of chi-square analysis, exploring its essence, implications, and applications.

Null Hypothesis in Chi-Square:

Unveiling the Essence

At its core, the null hypothesis in chi-square analysis posits that there is no significant difference between the observed and expected frequencies of a categorical variable. In simpler terms, it suggests that any deviation between what we expect to observe and what we actually observe is due to chance alone, rather than any true effect or relationship.

This hypothesis serves as the null point against which researchers gauge the validity of their findings. It embodies skepticism, challenging researchers to substantiate any claims of association or difference in frequencies within their data. By subjecting their hypotheses to rigorous scrutiny, researchers ensure that their conclusions are grounded in empirical evidence rather than mere conjecture.

Understanding Chi-Square Analysis

Before delving deeper into the null hypothesis, let’s acquaint ourselves with the chi-square test itself. Named for the Greek letter “χ²” (chi-square), this statistical method evaluates the distribution of categorical data and assesses whether any observed differences are statistically significant.

In essence, chi-square analysis compares observed frequencies in different categories to the frequencies we would expect to see if there were no association between the variables being studied. It quantifies the extent of deviation from expected frequencies, providing researchers with a measure of the likelihood that such deviation occurred purely by chance.

Embracing the Null Hypothesis:

A Guiding Principle

Within the realm of chi-square analysis, the null hypothesis serves as both a guiding principle and a formidable adversary. Its assertion of no significant difference challenges researchers to scrutinize their data rigorously, employing statistical tools to discern genuine patterns from random fluctuations.

When conducting a chi-square test, researchers formulate two hypotheses: the null hypothesis (H₀) and the alternative hypothesis (H₁). The null hypothesis posits no relationship or difference between variables, while the alternative hypothesis suggests the presence of such a relationship or difference.

In the context of chi-square analysis, the null hypothesis typically takes the form: “There is no significant difference between the observed and expected frequencies of the categorical variable.” Conversely, the alternative hypothesis might propose that a relationship exists, such as: “There is a significant difference between the observed and expected frequencies of the categorical variable.”

Interpreting Chi-Square Results:

The Dance of P-Values

Once the chi-square test is conducted, researchers turn their attention to the p-value—a numerical measure that quantifies the strength of evidence against the null hypothesis. A low p-value suggests that the observed deviation from expected frequencies is unlikely to occur purely by chance, leading researchers to reject the null hypothesis in favor of the alternative.

Conversely, a high p-value indicates that the observed deviation could plausibly occur by chance alone, failing to provide sufficient evidence to reject the null hypothesis. In such cases, researchers refrain from making definitive claims of association or difference, acknowledging the possibility that their findings may be due to random variation.

Applications and Extensions:

Beyond the Basics

While the null hypothesis in chi-square analysis finds widespread application in fields ranging from biology to social sciences, its significance extends beyond the confines of traditional statistical testing. Researchers often employ extensions of the chi-square test to analyze complex data sets, adapting its principles to suit the unique demands of their research questions.

From contingency table analysis to goodness-of-fit tests, the null hypothesis remains a steadfast companion, guiding researchers through the intricacies of categorical data analysis. Its resilience in the face of uncertainty underscores the importance of skepticism and empirical rigor in the pursuit of scientific knowledge.

In the realm of chi-square analysis, the null hypothesis serves as a beacon of skepticism, challenging researchers to scrutinize their findings with precision and diligence. By subjecting their hypotheses to rigorous testing, researchers ensure that their conclusions are anchored in empirical evidence rather than mere speculation.

As we navigate the intricate landscape of statistical analysis, let us heed the call of the null hypothesis, embracing its skepticism as a cornerstone of scientific inquiry. In doing so, we honor the pursuit of truth and uphold the integrity of empirical research for generations to come.

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14.4: Example of How to Test a Hypothesis Using Chi-squared Goodness of Fit

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Suppose a café is preparing to offer four different flavors of latte for the fall season but they think the preferences of customers for the four flavors will be uneven. They do a test run wherein they have 120 participants try each latte and select their favorite. Let’s test the hypothesis that the counts of customers who prefer each of the four different flavors will be uneven using Data Set 14.1. We will use this information to follow the steps in hypothesis testing.

Steps in Hypothesis Testing

In order to test a hypothesis, we must follow these steps:

1. State the hypothesis.

A summary of the research hypothesis and corresponding null hypothesis in sentence and symbol format are shown below. However, researchers often only state the research hypothesis using a format like this: It is hypothesized that the counts of preferences will be uneven across different drink flavors.

2. Choose the inferential test (formula) that best fits the hypothesis.

The counts of categories for a qualitative variable are being tested so the appropriate test is chi-squared goodness of fit.

3. Determine the critical value.

In order to determine the critical value for chi-square, we need to know the alpha level and the degrees of freedom. The alpha level is often set at .05 unless there is reason to adjust it such as when multiple hypotheses are being tested in one study or when a Type I Error could be particularly problematic. The default alpha level can be used for this example because only one hypothesis is being tested and there is no clear indication that a Type I Error would be especially problematic. Thus, alpha can be set to 5%, which can be summarized as \(\alpha\) = .05.

The degrees of freedom for chi-squared goodness of fit are computed using the following formula:

\[d f=k-1 \nonumber \]

Where \(k\) stands for the number of categories. In the current hypothesis there are 4 drink flavors so \(k\) = 4. Thus, the calculation for \(df\) for this example is as follows:

\[\begin{gathered} d f=4-1 \\ d f=3 \end{gathered} \nonumber \]

The alpha level and \(df\) are used to determine the critical value for the test. Below is the \(\chi^2\) critical values tables that fits the current hypothesis and data. Under the conditions of an alpha level of .05 and \(df\) = 3, the critical value is 7.815.

The critical value represents the value which must be exceeded in order to declare a result significant. It represents the threshold of evidence needed to be confident a hypothesis is true. The obtained \(\chi^2\) - value must be greater than 7.815 to be declared significant when using Data Set 14.1.

4. Calculate the test statistic.

In order to use a goodness of fit test, we first must find the observed and expected counts. Observed counts are based on a data set, however, expected counts must be computed. To find the expected counts, we need to know what the hypothesized proportions are. In the hypothesis, it states that the counts of the four groups will be uneven . This means that the null will state the counts are even. There are four categories being compared; if they had even counts (as stated by the null), it would mean that 25% of the sample would be in each of the four categories (because 25% is one-fourth). Now, we must find the total sample size and multiply it by 25% (which is 0.25 when written in decimal form) to find the expected counts.

Notice that all the expected counts are the same. This will occur anytime we are testing whether counts are even or not. Now that we have the observed and expected counts for all categories, we can plug these values into the formula and solve. The computations for this example, shown in formula format, are as follows:

5. Apply a decision rule and determine whether the result is significant.

Assess whether the obtained value for \(\chi^2\) exceeds the critical value as follows:

The critical value is 7.815

The obtained \(\chi^2\) - value is 11.40

The obtained \(\chi^2\)-value exceeds (i.e. is greater than) the critical value, thus, the result is significant.

Keep in mind that obtained values are often rounded to the hundredths place when reported.

6. Calculate the effect sizes and any secondary analyses.

The chi-squared goodness of fit test is an omnibus test which can tell us whether, overall, the observed counts are different from the expected counts; it does not, however, always allow us to determine which category counts are different from their expected counts and which are not when not all counts are different than expected. Thus, post-hoc tests are sometimes desired when a chi-squared goodness of fit result is significant. When two categories are being compared, a post-hoc test is not generally used. However, when three or more category counts are being compared, various post-hoc tests (such as using a chi-squared test for independence with a Bonferroni correction to compare each pair of categories) may be desired and used. However, as the focus of this chapter is the omnibus test, we will not go into a detailed review of these secondary analyses. For our purposes, therefore, secondary analyses will not be used to test the hypothesis and the use of both the chi-squared goodness of fit test and the chi-squared test for independence will only be reviewed for use as omnibus tests.

7. Report the results in American Psychological Associate (APA) format.

Results for inferential tests are often best summarized using a paragraph that states the following:

• the hypothesis and specific inferential test used,
• the main results of the test and whether they were significant,
• any additional results that clarify or add details about the results,
• whether the results support or refute the hypothesis.

There are no means or standard deviations to report for chi-squared because it is non parametric. It is, however, necessary to include the observed counts for each category in the results. Finally, it is customary to reported the total sample size using “\(N\) =” to the right of the \(df\) in the parenthesis of the evidence string. Following this, the results for our hypothesis with Data Set 14.1 can be written as shown in the summary example below.

APA Formatted Summary Example

A chi-squared goodness of fit was used to test the hypothesis that the counts of preference for different drink flavors would be significantly uneven. Consistent with the hypothesis, the counts of preference for non-flavored (\(n\) = 30), vanilla (\(n\) = 39), chocolate (\(n\) = 36), and pumpkin (\(n\) = 15) were significantly uneven, \(\chi^2(3, N = 120) = 11.40\), \(p\) < .05.

As always, the APA-formatted summary provides a lot of detail in a particular order. For a brief review of the structure for the APA-formatted summary of the omnibus test results, see the summary below.

Anatomy of the Evidence String

The following breaks down what each part represents in the evidence string for the chi-squared results in the APA-formatted paragraph above:

Reading Review 14.3

• How is \(df\) calculated for a chi-squared goodness of fit test?
• What is reported within the parenthesis next to \(df\) in the evidence string for chi-squared?
• How are expected counts for each category calculated when testing whether counts are even or uneven using a chi-squared goodness of fit test?
• What detail about each category should be included in the APA-formatted summary for a chi-squared goodness of fit test?

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Mastering the Chi-Square Test: A Comprehensive Guide

The Chi-Square Test is a statistical method used to determine if there’s a significant association between two categorical variables in a sample data set. It checks the independence of these variables, making it a robust and flexible tool for data analysis.

Introduction to Chi-Square Test

The  Chi-Square Test  of Independence is an important tool in the statistician’s arsenal. Its primary function is determining whether a significant association exists between two categorical variables in a sample data set. Essentially, it’s a test of independence, gauging if variations in one variable can impact another.

This comprehensive guide gives you a deeper understanding of the Chi-Square Test, its mechanics, importance, and correct implementation.

• Chi-Square Test assess the association between two categorical variables.
• Chi-Square Test requires the data to be a random sample.
• Chi-Square Test is designed for categorical or nominal variables.
• Each observation in the Chi-Square Test must be mutually exclusive and exhaustive.
• Chi-Square Test can’t establish causality, only an association between variables.

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Case Study: Chi-Square Test in Real-World Scenario

Let’s delve into a real-world scenario to illustrate the application of the  Chi-Square Test . Picture this: you’re the lead data analyst for a burgeoning shoe company. The company has an array of products but wants to enhance its marketing strategy by understanding if there’s an association between gender (Male, Female) and product preference (Sneakers, Loafers).

To start, you collect data from a random sample of customers, using a survey to identify their gender and their preferred shoe type. This data then gets organized into a contingency table , with gender across the top and shoe type down the side.

Next, you apply the Chi-Square Test to this data. The null hypothesis (H0) is that gender and shoe preference are independent. In contrast, the alternative hypothesis (H1) proposes that these variables are associated. After calculating the expected frequencies and the Chi-Square statistic, you compare this statistic with the critical value from the Chi-Square distribution.

Suppose the Chi-Square statistic is higher than the critical value in our scenario, leading to the rejection of the null hypothesis. This result indicates a significant association between gender and shoe preference. With this insight, the shoe company has valuable information for targeted marketing campaigns.

For instance, if the data shows that females prefer sneakers over loafers, the company might emphasize its sneaker line in marketing materials directed toward women. Conversely, if men show a higher preference for loafers, the company can highlight these products in campaigns targeting men.

This case study exemplifies the power of the Chi-Square Test. It’s a simple and effective tool that can drive strategic decisions in various real-world contexts, from marketing to medical research.

The Mathematics Behind Chi-Square Test

At the heart of the  Chi-Square Test  lies the calculation of the discrepancy between observed data and the expected data under the assumption of variable independence. This discrepancy termed the Chi-Square statistic, is calculated as the sum of squared differences between observed (O) and expected (E) frequencies, normalized by the expected frequencies in each category.

In mathematical terms, the Chi-Square statistic (χ²) can be represented as follows: χ² = Σ [ (Oᵢ – Eᵢ)² / Eᵢ ] , where the summation (Σ) is carried over all categories.

This formula quantifies the discrepancy between our observations and what we would expect if the null hypothesis of independence were true. We can decide on the variables’ independence by comparing the calculated Chi-Square statistic to a critical value from the Chi-Square distribution. Suppose the computed χ² is greater than the critical value. In that case, we reject the null hypothesis, indicating a significant association between the variables.

Step-by-Step Guide to Perform Chi-Square Test

To effectively execute a  Chi-Square Test , follow these methodical steps:

State the Hypotheses:  The null hypothesis (H0) posits no association between the variables — i.e., independent — while the alternative hypothesis (H1) posits an association between the variables.

Construct a Contingency Table:  Create a matrix to present your observations, with one variable defining the rows and the other defining the columns. Each table cell shows the frequency of observations corresponding to a particular combination of variable categories.

Calculate the Expected Values:  For each cell in the contingency table, calculate the expected frequency assuming that H0 is true. This can be calculated by multiplying the sum of the row and column for that cell and dividing by the total number of observations.

Compute the Chi-Square Statistic:  Apply the formula χ² = Σ [ (Oᵢ – Eᵢ)² / Eᵢ ] to compute the Chi-Square statistic.

Compare Your Test Statistic:  Evaluate your test statistic against a Chi-Square distribution to find the p-value, which will indicate the statistical significance of your test. If the p-value is less than your chosen significance level (usually 0.05), you reject H0.

Interpretation of the results should always be in the context of your research question and hypothesis. This includes considering practical significance — not just statistical significance — and ensuring your findings align with the broader theoretical understanding of the topic.

Assumptions, Limitations, and Misconceptions

The  Chi-Square Test , a vital tool in statistical analysis, comes with certain assumptions and distinct limitations. Firstly, it presumes that the data used are a  random sample  from a larger population and that the variables under investigation are nominal or categorical. Each observation must fall into one unique category or cell in the analysis, meaning observations are mutually  exclusive  and  exhaustive .

The Chi-Square Test has limitations when deployed with small sample sizes. The  expected frequency  of any cell in the contingency table should ideally be 5 or more. If it falls short, this can cause distortions in the test findings, potentially triggering a Type I or Type II error.

Misuse and misconceptions about this test often center on its application and interpretability. A standard error is using it for continuous or ordinal data without appropriate  categorization , leading to misleading results. Also, a significant result from a Chi-Square Test indicates an association between variables, but it doesn’t infer  causality . This is a frequent misconception — interpreting the association as proof of causality — while the test doesn’t offer information about whether changes in one variable cause changes in another.

Moreover, more than a significant Chi-Square test is required to comprehensively understand the relationship between variables. To get a more nuanced interpretation, it’s crucial to accompany the test with a measure of  effect size , such as Cramer’s V or Phi coefficient for a 2×2 contingency table. These measures provide information about the strength of the association, adding another dimension to the interpretation of results. This is essential as statistically significant results do not necessarily imply a practically significant effect. An effect size measure is critical in large sample sizes where even minor deviations from independence might result in a significant Chi-Square test.

Conclusion and Further Reading

Mastering the  Chi-Square Test  is vital in any data analyst’s or statistician’s journey. Its wide range of applications and robustness make it a tool you’ll turn to repeatedly.

For further learning, statistical textbooks and online courses can provide more in-depth knowledge and practice. Don’t hesitate to delve deeper and keep exploring the fascinating world of data analysis .

• Effect Size for Chi-Square Tests
• Assumptions for the Chi-Square Test
• Assumptions for Chi-Square Test (Story)
• Chi Square Test – an overview (External Link)

Understanding the Null Hypothesis in Chi-Square

• What is the Difference Between the T-Test vs. Chi-Square Test?
• How to Report Chi-Square Test Results in APA Style: A Step-By-Step Guide

Frequently Asked Questions (FAQ)

It’s a statistical test used to determine if there’s a significant association between two categorical variables.

The test is suitable for categorical or nominal variables.

No, the test can only indicate an association, not a causal relationship.

The test assumes that the data is a random sample and that observations are mutually exclusive and exhaustive.

It measures the discrepancy between observed and expected data, calculated by χ² = Σ [ (Oᵢ – Eᵢ)² / Eᵢ ].

The result is generally considered statistically significant if the p-value is less than 0.05.

Misuse can lead to misleading results, making it crucial to use it with categorical data only.

Small sample sizes can lead to wrong results, especially when expected cell frequencies are less than 5.

Low expected cell frequencies can lead to Type I or Type II errors.

Results should be interpreted in context, considering the statistical significance and the broader understanding of the topic.

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“Questo viene calcolato moltiplicando il totale di riga e colonna per quella cella e dividendo per il totale complessivo.” Siccome la frase é ambigua non ho capito cosa bisogna fare esattamente. Aspettavo un esempio semplice numerico che non é arrivato.

Grazie per il tuo commento! Per chiarire, il calcolo si basa sulla formula: (Expected Frequency) = (Totale della Riga × Totale della Colonna) / Totale Complessivo.

Un esempio semplice: Supponiamo di avere una tabella 2×2 con i seguenti totali:

Totale della Riga 1 = 50 Totale della Colonna 1 = 30 Totale Complessivo = 100 La frequenza attesa per la cella nella Riga 1, Colonna 1 sarebbe: (Expected Frequency) = (50 × 30) / 100 = 15.

Se hai ulteriori domande, fammi sapere!

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S.4 chi-square tests, chi-square test of independence section  .

Do you remember how to test the independence of two categorical variables? This test is performed by using a Chi-square test of independence.

Recall that we can summarize two categorical variables within a two-way table, also called an r × c contingency table, where r = number of rows, c = number of columns. Our question of interest is “Are the two variables independent?” This question is set up using the following hypothesis statements:

 \[E=\frac{\text{row total}\times\text{column total}}{\text{sample size}}\]

We will compare the value of the test statistic to the critical value of \(\chi_{\alpha}^2\) with the degree of freedom = ( r - 1) ( c - 1), and reject the null hypothesis if \(\chi^2 \gt \chi_{\alpha}^2\).

Example S.4.1 Section  

Is gender independent of education level? A random sample of 395 people was surveyed and each person was asked to report the highest education level they obtained. The data that resulted from the survey are summarized in the following table:

Question : Are gender and education level dependent at a 5% level of significance? In other words, given the data collected above, is there a relationship between the gender of an individual and the level of education that they have obtained?

Here's the table of expected counts:

So, working this out, \(\chi^2= \dfrac{(60−50.886)^2}{50.886} + \cdots + \dfrac{(57 − 48.132)^2}{48.132} = 8.006\)

The critical value of \(\chi^2\) with 3 degrees of freedom is 7.815. Since 8.006 > 7.815, we reject the null hypothesis and conclude that the education level depends on gender at a 5% level of significance.