Chi-Square Test in Statistics Using R

 

Experiment 

Chi-Square Test in Statistics Using R

1. Experiment Title

Hypothesis Testing Using the Chi-Square Goodness-of-Fit Test in R

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2. Aim

To study and perform a Chi-Square goodness-of-fit test using R to determine whether the observed frequencies differ significantly from the expected frequencies.

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3. Objectives

After completing this experiment, students should be able to:

1. Understand the purpose of the Chi-Square test.
2. Formulate null and alternative hypotheses.
3. Calculate expected frequencies.
4. Calculate the Chi-Square statistic manually.
5. Perform a Chi-Square test using R.
6. Interpret the Chi-Square statistic and p-value.
7. Make a statistical decision based on the test result.

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4. Theory

4.1 What is a Chi-Square Test?

The Chi-Square ($\chi^2$) test is a non-parametric statistical test commonly used to analyze categorical data.

It compares:

• Observed frequencies ($O$) – the actual frequencies obtained from data.
• Expected frequencies ($E$) – the frequencies expected according to a hypothesis.

The test helps us determine whether the difference between the observed and expected frequencies is due to random variation or represents a statistically significant difference.

In this experiment, we demonstrate the Chi-Square Goodness-of-Fit Test.

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4.2 Chi-Square Goodness-of-Fit Test

The goodness-of-fit test determines whether an observed frequency distribution fits an expected distribution.

For example, suppose a die is assumed to be fair. If it is rolled many times, each face should occur approximately the same number of times. A Chi-Square test can determine whether the observed frequencies are significantly different from the expected frequencies.

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4.3 Hypotheses

Null Hypothesis ($H_0$)

$$H_0:$$

There is no significant difference between the observed and expected frequencies.

The observed data follow the expected distribution.

Alternative Hypothesis ($H_1$)

$$H_1:$$

There is a significant difference between the observed and expected frequencies.

The observed data do not follow the expected distribution.

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5. Problem Statement

A researcher wants to determine whether a six-sided die is fair.

The die is rolled 60 times, and the following frequencies are observed:

Face of Die	Observed Frequency
1	8
2	12
3	9
4	11
5	10
6	10

For a fair die, each face has an equal probability of occurring.

Therefore, the expected frequency for each face is:

$$E=\frac{60}{6}=10$$

Using a significance level of:

$$\alpha=0.05$$

determine whether the die can be considered fair.

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6. Manual Calculation

The Chi-Square statistic is calculated by comparing the observed and expected frequencies:

The sample does not provide enough evidence that the four proportions differ

For this problem:$$E=10$$

for every category.

Step 1: Create the Calculation Table

Die Face	Observed $O$	Expected $E$	$O-E$	$(O-E)^2$	$\frac{(O-E)^2}{E}$
1	8	10	-2	4	0.40
2	12	10	2	4	0.40
3	9	10	-1	1	0.10
4	11	10	1	1	0.10
5	10	10	0	0	0.00
6	10	10	0	0	0.00

Therefore:

$$\chi^2=0.40+0.40+0.10+0.10+0+0$$ $$\boxed{\chi^2=1.00}$$

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Step 2: Calculate Degrees of Freedom

For a goodness-of-fit test:

$$df=k-1$$

where $k$ is the number of categories.

Here:

$$k=6$$

Therefore:

$$df=6-1=\boxed{5}$$

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Step 3: Determine the Decision

The calculated Chi-Square statistic is:

$$\chi^2=1.00$$

Degrees of freedom:

$$df=5$$

At the 5% significance level, the result can be evaluated using the p-value obtained from R.

If:

• p-value < 0.05 → Reject $H_0$
• p-value ≥ 0.05 → Fail to reject $H_0$

Since the observed frequencies are very close to the expected frequencies, we expect no statistically significant difference.

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7. R Program

# Chi-Square Goodness-of-Fit Test

# Observed frequencies of the six faces of a die
observed <- c(8, 12, 9, 11, 10, 10)

# Expected frequencies for a fair die
expected <- rep(sum(observed) / 6, 6)

# Display observed and expected frequencies
data <- data.frame(
  Face = 1:6,
  Observed = observed,
  Expected = expected
)

print(data)


# Perform Chi-Square Goodness-of-Fit Test
result <- chisq.test(observed)

print(result)


# Significance level
alpha <- 0.05


# Decision based on p-value
if (result$p.value < alpha) {
  
  cat("\nDecision: Reject the Null Hypothesis\n")
  cat("Conclusion: The die does not appear to be fair.\n")
  
} else {
  
  cat("\nDecision: Fail to Reject the Null Hypothesis\n")
  cat("Conclusion: There is not enough evidence to conclude that the die is unfair.\n")
}

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8. Explanation of the R Function

The main function used is:

chisq.test()

For this experiment:

chisq.test(observed)

Since no expected probabilities are specified, R assumes that all categories have equal probabilities.

Therefore, for a fair six-sided die:

$$P(1)=P(2)=P(3)=P(4)=P(5)=P(6)=\frac{1}{6}$$

and the expected frequency for each category is:

$$\frac{60}{6}=10$$

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9. Expected Output

Face Observed Expected
1    1        8       10
2    2       12       10
3    3        9       10
4    4       11       10
5    5       10       10
6    6       10       10

	Chi-squared test for given probabilities

data:  observed
X-squared = 1, df = 5, p-value = 0.9626


Decision: Fail to Reject the Null Hypothesis
Conclusion: There is not enough evidence to conclude that the die is unfair.

10. Interpretation

The exact interpretation is:

Measure	Value
Chi-Square statistic	1.0
Degrees of freedom	5
Significance level	0.05
p-value	Approximately 0.96

Since:

$$p\text{-value}>0.05$$

we fail to reject the null hypothesis.

The observed frequencies of the six faces are not significantly different from the frequencies expected for a fair die.

Therefore, based on this sample:

There is insufficient evidence to conclude that the die is unfair.

It is important to say "fail to reject the null hypothesis" rather than "accept the null hypothesis."

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11. Result

Thus, the Chi-Square goodness-of-fit test was performed using R to compare observed and expected frequencies. The Chi-Square statistic, degrees of freedom, and p-value were calculated. Since the p-value was greater than the significance level of 0.05, the null hypothesis was not rejected, indicating that the observed frequencies were consistent with those expected from a fair die.