## Chi-Square Test In Excel

The

Chi-Square Test in Excelis a distribution-free test that follows a statistical procedure for comparing and finding the relationship between categorical variables. We can derive a set of expected results using the given data, which is called the observed data, and then evaluate if there exists a relation between the observed and the expected results.

For example, the below table shows the observed orders delivered frequencies of two branch offices. And row 9 shows each column’s sum, and column D shows each row’s sum.

Consider we take a random sample of 1000 orders delivered by Branch Offices 1 and 2 together.

And the null and alternative hypotheses (H_{0} and H_{1}) are that the number of orders delivered does not depend and depends on the day of the week, respectively. Assume the significance threshold is **0.05**.

We must conclude whether to accept or reject the null hypothesis (H_{0}), and then **calculate the Chi-Square Test in Excel** as depicted below to make the decision.

As the first step in the above example, we must calculate the expected orders delivered frequencies for the two branch offices. Next, calculate the **Chi-Square Test in Excel** data points (refer to the third table), and adding the values will give us the **Chi-Square (χ ^{2})**,

**21.34549532**. And then using the

**CHISQ.INV.RT()**, calculate the

**Chi-Square**critical value,

**9.487729037**. Also, as the source data contains five rows and two columns of data (cell range B4:C8), the degree of freedom is

**4**.

Finally, apply the **CHISQ.TEST()** to determine the p-value, **0.000270432**.

Thus, the above calculations show the χ^{2} statistic for the given data is **21.34549532**, with **4** degrees of freedom. And the **Chi-Square** statistic, **21.34549532**, exceeds the determined critical value, **9.487729037**. Thus, we can reject the null hypothesis and consider the alternative hypothesis (H_{1}) that the number of orders delivered depends on the day of the week.

Further, the **CHISQ.TEST()** returns the **p-value for Chi-Square Test in Excel**, **0.000270432**. It is the probability of the χ^{2} statistic being as high as the calculated value of **21.34549532**, by chance, with the assumed independence.

And the p-value of **0.000270432** is less than the significance threshold of **0.05**. So, we can reject the null hypothesis.

###### Key Takeaways

- The
**Chi-Square Test in Excel**is a nonparametric statistical test that assesses the relationship between categorical variables. It also helps check if the variation of observed frequencies from the expected results is purely by chance or due to the relation between them. - When performing the Goodness of Fit Test, determine the data points to calculate the
**Chi-Square**(χ^{2}) statistic. Next, find the degree of freedom and the**Chi-Square Test**p-value using the function**CHISQ.DIST.RT()**. - When conducting a
**Chi-Square Test**for Independence, calculate the expected frequencies and the data points to calculate the**Chi-Square**(χ^{2}) statistic. Next, compute the degree of freedom and the**Chi-Square**critical value using**CHISQ.INV.RT()**. And finally, calculate the**Chi-Square Test**p-value using the function**CHISQ.TEST()**. - If the calculated
**Chi-Square**(χ^{2}) exceeds the**Chi-Square**critical value, or the**Chi-Square**p-value is lesser than or equal to the significance threshold. Then, we can reject the null hypothesis (H_{0}) and accept the alternative hypothesis (H_{1}).

**Chi-Square Goodness Of Fit Test**

The statistical hypothesis test, the **Chi-Square** **Goodness of Fit Test **helps us make decisions about population distribution based on a sample, and we can assess whether a categorical variable adheres to a hypothesis distribution.

The following expression mathematically represents the **Chi-Square** **Goodness of Fit Test**.

Where,

We can use this **Chi-Square Test in Excel** to test the hypothesis concerning the distribution of only one categorical variable.

**Chi-Square Test For Independence**

The **Chi-Square Test for Independence **is a Pearson’s **Chi-Square Test in Excel**.It helps evaluate whether the given data differs significantly from the expected value.

This type of **Chi-Square Test in Excel** works on observed frequencies, defining observations in every combined group. And it compares the observed frequencies to the expected frequencies when the test involves unrelated variables.

The below expression is the mathematical representation of the **Chi-Square Test** for Independence.

Where,

We can use this **Chi-Square Test in Excel** when performing a hypothesis test about the relation between two categorical variables.

**How To Perform The Chi-Square Test in Excel?**

We will understand with examples to perform the **Chi-Square Test in Excel **using the**,**

**Chi-Square Goodness of Fit Test.****Chi-Square Test for Independence.**

**Chi-Square Goodness of Fit Test**

A professor claims that the same number of students attend the History session every week. We will conduct the **Chi-Square Test in Excel **and check if the collected data aligns with the professor’s claim.

The table below shows the weekly attendance recorded across five weeks required to test the above mentioned hypothesis.

Assume the total number of students attending the History is 1000. Thus, according to the professor’s claim, 200 students attend the session every week. And the significance threshold is 0.05.

The steps to conduct the **Chi-Square Test in Excel **are**,**

**1:** We will introduce new columns for displaying the expected weekly student count and the **Chi-Square Test in Excel** calculations.

**2:** We must determine the **Chi-Square** (χ^{2}) statistic data points in column D. So, select cell **D3**, enter the formula ** =((B3-C3)^2)/C3**, and press

**Enter**. The result is ‘

**2**’, as shown below.

Drag the formula from cell **D3 **to** D7** using the excel fill handle.

**3:** We need to apply the mathematical expression of the **Chi-Square** **Goodness of Fit Test** in cell B10, which is the sum of the calculated **Chi-Square** (χ^{2}) statistic data points. So, select cell **B10**, enter the below formula ** =SUM(D3:D7)**, and press

**Enter**.

**4:** As Column B contains the weekly observations in 5 rows, enter the below formula to determine the degree of freedom, a value one less than the number of observations (n). So, select cell **B11**, enter the formula ** =(5-1), **and press “

**Enter**”.

**5: **Select the target cell **B12**, enter the **CHISQ.DIST.RT() formula =CHISQ.DIST.RT(B10,B11) **to determine the right-tailed probability (p-value) of the

**Chi-Square**distribution for the given test statistic of

**4.625**and

**4**degrees of freedom, and press

**Enter**.

[**Output Observation:** The **CHISQ.DIST.RT()** syntax is:

The function takes two mandatory arguments, **x **and **deg_freedom**. They are the calculated χ^{2} Statistic value and the degree of freedom, respectively.

Thus, the **p-value for Chi-Square Test **cell B12 is **0.327981915**. It exceeds the significance threshold of 0.05, so we cannot reject the null hypothesis. It suggests a lack of evidence of a disparity between the students’ true distribution and the professor’s claimed statistic.]

**Chi-Square Test for Independence**

Consider a random sample of 100 employees in an organization. And these employees achieved the sales ratings mentioned under the heading **Sales** **Rating**.

Suppose the null hypothesis is that the sales rating does not depend on the third-party product quality. And the alternative hypothesis is that the sales rating depends on the third-party product quality.

Further, when conducting the **Chi-Square Test for Independence**, use a contingency table (cross-tabulation). It shows the observed frequencies in each group’s combinations and includes rows and columns sums.

The observations are available in the first table as 15 data points (cell range B4:D8).

The steps to conduct the **Chi-Square Test in Excel **are**,**

**1: **First, determine the sum of the columns and the rows values in the first table. So, select cell **B9**, enter the formula** =SUM(B4:B8), **and press

**Enter**.

Next, select cell **E4**, enter the formula ** =SUM(B4:D4), **and press

**Enter**.

**2: **As shown in the image below, using the fill handle, drag the formula from,

- Cell
**B9**to**D9,**horizontally or row-wise. - Cell
**E4**to**E9,**vertically or column-wise.

**3: **Next, calculate the expected frequencies in the second table. We will need to apply the E_ij** **formula provided in the **Chi-Square Test for Independence** section. So, select cell **B14**, enter the formula ** =$B$9*E4/$E$9**, and press

**Enter**.

Next, select cell **C14**, enter the formula ** =$C$9*E4/$E$9**, and press

**Enter**.

And select cell **D14**, enter the formula ** =$D$9*E4/$E$9**, and press

**Enter**.

Next, select cells **B14:D14** and drag the formula from these 3 cells simultaneously using the fill handle to **B18:D18.**

**4: **Next, we need to calculate the **Chi-Square** data points in the third table to determine the result of the **Chi-Square Test in Excel** cell L11. So, select cell **G14**, enter the formula ** =((B4-B14)^2)/B14**, and press

**Enter**.

To fill in the formula in the remaining cells of the third table, drag the formula from cell **G14 **to** G18** using the fill handle.

**5: **Select cell **L11**, enter the formula **=SUM(G14:I18) **to apply the mathematical expression of the** Chi-Square Test for Independence**, and press **Enter**.

**6: **Select cell **L12**, enter the formula ** =(5-1)*(3-1)** to determine the degree of freedom based on the observed data in the first table, and press

**Enter**.

As the first table contains the observations in 5 rows (rows 4 to 8) and three columns (columns B:D), the above formula considers the total rows and columns as 5 and 3, respectively.

**7: **Select cell **L13**, enter the **CHISQ.INV.RT()** formula ** =CHISQ.INV.RT(0.05,L12) **to get the

**Chi-Square**critical value, and press

**Enter**.

The **CHISQ.INV.RT() **syntax is:

It accepts two mandatory arguments, **probability **and **deg_freedom**, as input. Typically, the first argument is the significance value, **0.05**. And in this case, the second argument value is **8 **(cell L12)

**8:** Select the target cell **L14**, enter the **CHISQ.TEST() **formula ** =CHISQ.TEST(B4:D8,B14:D18)** to determine the p-value of the

**Chi-Square Test**, and press

**Enter**.

[**Output Observation:** The **CHISQ.TEST() **syntax is:

The function takes two mandatory arguments, **actual_range** and **expected_range**, as input. They take the cell ranges of observed and expected frequencies, respectively.

As the determined **Chi-Square** (χ^{2}) value, **24.36195813**, is more than the critical value, **15.50731306**, we can reject the null hypothesis and accept the alternative hypothesis. Thus, the sales rating depends on the third-party product quality.

Also, the calculated p-value, **0.001992359**, is less than the significance threshold of **0.05**, implying that we can dismiss the null hypothesis and accept the alternative hypothesis.]

**The Uses Of The Goodness Of Fit Test**

We can use this test in the following scenarios:

- To determine the association between sales person’s performance and their salary.
- To check borrowers’ credit worthiness based on their salary package and current debts.
- To assess the impact of marketing tactics on a selected target audience.

**The Uses Of The Chi-Square Test For Independence**

We can use this test in the following scenarios:

- To assess the relationship between categorical variables in a cross-tabulation.
- To perform a
**Chi-Square Test**involving unmeasurable variables such as the cause of fluctuations in rainfall in a specified region.

**Important Things To Note**

- To evaluate the
**Chi-Square Test in Excel**for only one categorical variable, or to access the sample data’s fitness in a hypothesis distribution, we must use the**Goodness of Fit Test**. - For evaluating the relationship between two categorical variables, we use the
**Chi-Square Test for Independence.** - When performing a
**Chi-Square Test**for Independence, use the contingency table to display the observed frequencies in each group’s combinations, with the rows and columns totals.

**Frequently Asked Questions**

**1. Which Excel function returns the test for independence?**

The **CHISQ.TEST** function returns the test for independence. We can select the target cell and go to **Formulas** – **More Functions** – **Statistical** – **CHISQ.TEST** to apply the function in the selected cell.

**2. A dealer claims to sell 100 units of a product each month. How to test the hypothesis based on the given five months’ observations?**

For instance, if a dealer claims to sell 100 units of a product each month, we will see how we can test the hypothesis based on the given five months’ observations.

Assume the significance threshold (α) is 0.05.

In the table below, Column A has the 5 months, Column B shows the five months’ observations, and column C represents the dealer’s claim.

The steps to perform the Goodness of Fit Test are as follows:**Step 1: **We need to calculate **Chi-Square** statistic data points to determine the **Chi-Square** statistic value in cell **B10**. So, select cell **D3**, enter the formula ** =((B3-C3)^2)/C3**, and press

**Enter**.

Drag the formula from cell

**D3**to

**D7**using the fill handle.

**Step 2:**Select cell

**B10**, enter the formula

**=SUM(D3:D7)**to add the data points and get the

**Chi-Square**statistic, and press

**Enter**.

**Step 3:**Next, as the observations are in 5 rows and 1 column, we can calculate the degree of freedom in the following way. Select cell

**B11**, enter the formula

**, and press**

*=(5-1)***Enter**.

**Step 4:**Select cell

**B12**, enter the

**CHISQ.DIST.RT()**formula

**to get the**

*=CHISQ.DIST.RT(B10,B11)***Chi-Square Test**p-value, and press

**Enter**.

[

**Output Observation:**The

**CHISQ.DIST.RT()**takes the calculated

**Chi-Square**statistic and degree of freedom as the input. It returns the right-tailed probability (p-value) of the

**Chi-Square**distribution for the test statistic

**16.5**as

**0.002416642**. And as the determined p-value is less than the significance threshold of 0.05, we can reject the null hypothesis. It means that the actual number of product units the dealer sells per month differs from the count he claims to sell.]

**3. How to interpret the Chi-Square Test p-value?**

We can interpret the **Chi-Square Test** p-value in the following way.

• If the p-value is lesser than or equal to the significance threshold, we can dismiss the null hypothesis and accept the alternative hypothesis.

• If the p-value exceeds the significance threshold. Then, we cannot reject the null hypothesis.

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