# Calculate the P-Value from Chi-Square Statistic in R

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Calculate the P-Value from Chi-Square Statistic in R, You’ll get a Chi-Square test statistic every time you run a Chi-Square test.

The p-value associated with this test statistic can then be used to assess if the test findings are statistically significant.

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The pchisq() function in R can be used to find the p-value that corresponds to a Chi-Square test statistic using the following syntax:

pchisq(q, df, lower.tail = TRUE)

where:

q: The Chi-Square test statistic

df: The degrees of freedom

lower.tail: If TRUE, the probability in the Chi-Square distribution to the left of q is returned. If FALSE, the probability in the Chi-Square distribution to the right of q is returned. TRUE is the default value.

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The examples below demonstrate how to utilize this function in practice.

## Example 1: Chi-Square Goodness of Fit Test

Every day, an equal amount of consumers enter a gold shop, according to the owner.

An independent researcher records the number of customers who come into the shop during a specific week to test this hypothesis and discovers the following.

60 customers on Monday, 60 customers on Monday, 50 customers on Wednesday, 40 customers on Thursday, and 50 customers on Friday

The researcher discovers the following after completing a Chi-Square Goodness of Fit Test.

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Suppose the Chi-Square Test Statistic (X2): 4 and the degrees of freedom: (df): 4

We can use the following R function to find the p-value associated with this Chi-Square test statistic and degrees of freedom.

To find the p-value for the Chi-Square test statistic

pchisq(q=4, df=4, lower.tail=FALSE) 0.4060058

The p-value is determined to be 0.406. The null hypothesis is not rejected since the p-value is not less than 0.05.

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This indicates we don’t have enough evidence to conclude that the genuine client distribution differs from the one claimed by the gold shop owner.

## Example 2: Chi-Square Test of Independence

The researchers want to determine if gender has anything to do with political party preference.

The poll 700 voters in a simple random sample to determine their political party preferences. They discover the following after doing a Chi-Square Test of Independence.

Chi-Square Test Statistic (X2): 0.7 and the degrees of freedom: (df): 2

We can use the following R function to find the p-value associated with this Chi-Square test statistic and degrees of freedom.

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To find the p-value for the Chi-Square test statistic

pchisq(q=0.7, df=2, lower.tail=FALSE) 0.7046881

The p-value for this case is 0.7046881. The null hypothesis is not rejected since the p-value is not less than 0.05.

This suggests that we don’t have enough evidence to claim that gender and political party choice are linked.

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