5.10 Chi-Square Goodness of Fit Test

Learning Objectives

By the end of this chapter, you will be able to:

• Understand the difference between goodness of fit and independence tests
• Determine expected frequencies from hypothesized distributions
• Perform goodness of fit tests step by step
• Test for uniform and other distributions
• Interpret results correctly

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Goodness of Fit vs. Independence

Goodness of Fit	Independence
One categorical variable	Two categorical variables
Tests if distribution matches expected	Tests if variables are related
Single row of data	Contingency table
df = k - 1	df = (r-1)(c-1)

flowchart TD
    A[Chi-square Test]
    B{How many<br/>variables?}
    C[Goodness of Fit]
    D[Test of Independence]

    A --> B
    B -->|One variable| C
    B -->|Two variables| D

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The Goodness of Fit Test

Purpose

Tests if observed frequencies follow a hypothesized distribution:

• Uniform distribution (all categories equal)
• Known proportions
• Theoretical distribution (binomial, normal, etc.)

Hypotheses

• H₀: Observed distribution matches expected distribution
• H₁: Observed distribution does NOT match expected distribution

Test Statistic

\[\chi^2 = \sum \frac{(O - E)^2}{E}\]

Degrees of Freedom

\[df = k - 1\]

Where k = number of categories

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Step-by-Step Example 1: Uniform Distribution

Problem: A die is rolled 120 times with these results:

Face	1	2	3	4	5	6	Total
Observed	25	17	22	15	19	22	120

Test at α = 0.05 if the die is fair (uniform distribution).

Solution:

Step 1: State hypotheses

• H₀: Die is fair (all faces equally likely)
• H₁: Die is not fair

Step 2: Calculate expected frequencies

For a fair die, each face should appear: \(E = \frac{120}{6} = 20\)

Face	O	E
1	25	20
2	17	20
3	22	20
4	15	20
5	19	20
6	22	20

Step 3: Calculate chi-square

Face	O	E	(O-E)²	(O-E)²/E
1	25	20	25	1.25
2	17	20	9	0.45
3	22	20	4	0.20
4	15	20	25	1.25
5	19	20	1	0.05
6	22	20	4	0.20
Total				3.40

\[\chi^2 = 3.40\]

Step 4: Find critical value

• df = 6 - 1 = 5
• α = 0.05
• From chi-square table: χ²* = 11.070

Step 5: Decision

• χ² = 3.40 < 11.070
• Fail to Reject H₀

Step 6: Conclusion At the 0.05 level of significance, there is insufficient evidence to conclude that the die is unfair. The observed frequencies are consistent with a fair die.

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Step-by-Step Example 2: Specified Proportions

Problem: A marketing manager claims preferences for 4 products are in ratio 4:3:2:1. Survey of 200 consumers shows:

Product	A	B	C	D	Total
Observed	90	55	35	20	200

Test at α = 0.05 if the claim is correct.

Solution:

Step 1: State hypotheses

• H₀: Preferences are in ratio 4:3:2:1
• H₁: Preferences are not in this ratio

Step 2: Calculate expected frequencies

Total ratio = 4 + 3 + 2 + 1 = 10

Product	Proportion	Expected
A	4/10 = 0.4	200 × 0.4 = 80
B	3/10 = 0.3	200 × 0.3 = 60
C	2/10 = 0.2	200 × 0.2 = 40
D	1/10 = 0.1	200 × 0.1 = 20

Step 3: Calculate chi-square

Product	O	E	(O-E)²	(O-E)²/E
A	90	80	100	1.25
B	55	60	25	0.417
C	35	40	25	0.625
D	20	20	0	0.000
Total				2.292

\[\chi^2 = 2.292\]

Step 4: Find critical value

• df = 4 - 1 = 3
• α = 0.05
• From chi-square table: χ²* = 7.815

Step 5: Decision

• χ² = 2.29 < 7.815
• Fail to Reject H₀

Step 6: Conclusion At the 0.05 level of significance, the data is consistent with the claimed ratio 4:3:2:1.

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Step-by-Step Example 3: Day of Week Distribution

Problem: Emergency calls received on different days:

Day	Mon	Tue	Wed	Thu	Fri	Sat	Sun	Total
Observed	65	58	62	60	70	85	100	500

Test at α = 0.01 if calls are uniformly distributed across days.

Solution:

Step 1: State hypotheses

• H₀: Calls are uniformly distributed (same for each day)
• H₁: Calls are not uniformly distributed

Step 2: Calculate expected frequencies

For uniform distribution: \(E = \frac{500}{7} = 71.43\)

Step 3: Calculate chi-square

Day	O	E	(O-E)²	(O-E)²/E
Mon	65	71.43	41.3	0.578
Tue	58	71.43	180.6	2.528
Wed	62	71.43	89.0	1.246
Thu	60	71.43	130.6	1.829
Fri	70	71.43	2.0	0.028
Sat	85	71.43	184.0	2.576
Sun	100	71.43	816.0	11.423
Total				20.21

\[\chi^2 = 20.21\]

Step 4: Find critical value

• df = 7 - 1 = 6
• α = 0.01
• From chi-square table: χ²* = 16.812

Step 5: Decision

• χ² = 20.21 > 16.812
• Reject H₀

Step 6: Conclusion At the 0.01 level of significance, there is strong evidence that emergency calls are NOT uniformly distributed across days. The pattern shows more calls on weekends.

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Step-by-Step Example 4: Testing Percentages

Problem: A city claims the distribution of employees by department is:

• Administration: 30%
• Technical: 45%
• Support: 25%

A sample of 200 employees shows:

Department	Admin	Technical	Support	Total
Observed	70	80	50	200

Test at α = 0.05 if the actual distribution matches the claim.

Solution:

Step 1: State hypotheses

• H₀: Distribution is 30%, 45%, 25%
• H₁: Distribution is not as claimed

Step 2: Calculate expected frequencies

Department	Claimed %	Expected
Admin	30%	200 × 0.30 = 60
Technical	45%	200 × 0.45 = 90
Support	25%	200 × 0.25 = 50

Step 3: Calculate chi-square

Dept	O	E	(O-E)²	(O-E)²/E
Admin	70	60	100	1.667
Technical	80	90	100	1.111
Support	50	50	0	0.000
Total				2.778

\[\chi^2 = 2.778\]

Step 4: Find critical value

• df = 3 - 1 = 2
• α = 0.05
• From chi-square table: χ²* = 5.991

Step 5: Decision

• χ² = 2.78 < 5.991
• Fail to Reject H₀

Step 6: Conclusion At the 0.05 level of significance, the observed distribution is consistent with the claimed percentages.

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Common Expected Distributions

Distribution Type	How to Calculate E
Uniform	E = Total ÷ k for each category
Given proportions	E = Total × proportion
Given percentages	E = Total × (percentage/100)
Given ratios	E = Total × (ratio/sum of ratios)

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Assumption Check

flowchart TD
    A[Calculate Expected<br/>Frequencies]
    B{All E ≥ 5?}
    C[Proceed with<br/>chi-square test]
    D{Can combine<br/>adjacent categories?}
    E[Combine categories]
    F[Use alternative<br/>methods]

    A --> B
    B -->|Yes| C
    B -->|No| D
    D -->|Yes| E
    E --> A
    D -->|No| F

Rule: All expected frequencies should be at least 5 for the chi-square approximation to be valid.

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Chi-Square Critical Values (Reference)

df	α = 0.10	α = 0.05	α = 0.01
1	2.706	3.841	6.635
2	4.605	5.991	9.210
3	6.251	7.815	11.345
4	7.779	9.488	13.277
5	9.236	11.070	15.086
6	10.645	12.592	16.812
7	12.017	14.067	18.475
8	13.362	15.507	20.090
9	14.684	16.919	21.666
10	15.987	18.307	23.209

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Practice Problems

Problem 1

A coin is flipped 100 times: Heads = 58, Tails = 42. Test at α = 0.05 if the coin is fair.

Problem 2

Customer arrivals at a service center:

Time	9-10am	10-11am	11-12pm	12-1pm	Total
Observed	45	60	75	70	250

Test at α = 0.05 if arrivals are uniformly distributed.

Problem 3

A company claims market share is:

• Brand A: 40%
• Brand B: 35%
• Brand C: 25%

A survey of 300 consumers shows: A = 110, B = 120, C = 70. Test the claim at α = 0.01.

Problem 4

Blood type distribution in a sample of 500:

Type	O	A	B	AB	Total
Observed	220	180	70	30	500

Population proportions are: O = 45%, A = 40%, B = 11%, AB = 4%. Test at α = 0.05 if the sample matches the population.

Problem 5

Student preferences for exam format:

Format	Multiple Choice	Short Answer	Essay	Total
Observed	90	45	15	150

Test at α = 0.05 if preferences are in ratio 5:3:2.

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Summary

Component	Details
Purpose	Test if observed distribution matches expected
Test statistic	$\chi^2 = \sum \frac{(O-E)^2}{E}$
Degrees of freedom	df = k - 1
Expected (uniform)	E = Total/k
Expected (proportions)	E = Total × p
Assumption	All E ≥ 5
Decision rule	Reject H₀ if χ² > critical value

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Summary: Goodness of Fit Steps

1. State hypotheses about the distribution
2. Calculate expected frequencies based on the hypothesized distribution
3. Calculate χ² using the formula
4. Find critical value with df = k - 1
5. Compare and decide
6. State conclusion in context

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Next Topic

In the next chapter, we will study the Kruskal-Wallis Test - a non-parametric alternative for comparing more than two groups.