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Table of Contents

 


 

Chi-Square Goodness-of-Fit Test

Used to compare frequencies (counts) among multiple categories of nominal or ordinal level data for one-sample (univariate analysis).

 

Problem: You wish to evaluate variations in the proportion of defects produced from five assembly lines. A random sample of 100 defective parts from the five assembly lines produced the following contingency table.

Line A

Line B

Line C

Line D

Line E

24

15

22

20

19

 

Assumptions

Independent random sampling

Nominal or Ordinal level data

State the Hypothesis

Ho: There is no significant difference among the assembly lines in the observed frequencies of defective parts.

Ha: There is a significant difference among the assembly lines in the observed frequencies of defective parts.

Set the Rejection Criteria

Determine degrees of freedom (df) = k – 1   where k equals the number of categories        df=5-1 or df=4

Establish the confidence level (.05, .01, etc.)

Use the chi-square distribution table to establish the critical value

At alpha .05 and 4 degrees of freedom, the critical value from the chi-square distribution is 9.488

 

Compute the Test Statistic

             where             

and . . .

n = sample size

k = number of categories or cells

Fo = observed frequency

 

 

Line A

Line B

Line C

Line D

Line E

 

Fo

24

15

22

20

19

 

Fe (100/5=20)

20

20

20

20

20

.8

1.25

.2

0

.05

2.30

 

 

Decide Results of Null Hypothesis

Since the chi-square test statistic 2.30 does not meet or exceed the critical value of 9.488, you cannot conclude there is a statistically significant difference among the assembly lines in the observed frequencies of defective parts.


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