Hypothesis Tests for Variance Case II

Χ2 Test Hypothesis Tests for Variance Case II

The chi-squared (Χ2) test provides the basis for the second case of hypothesis tests for variance. In this case we want to compare observed and expected frequencies, or counts, of outcomes when there is no defined variance, in other words we are working with attribute data.

Contingency table or row and column analysis are other common names for this analysis. It is useful when comparing results from different treatments or processes.


There is a six step procedure for this analysis:

  1. Determine the observed frequencies (O) for the various conditions under comparison. For example, the number of defects from each of three processes. Layout value in a table.

  2. Assuming no differences exist among the processes (conditions under scrutiny) calculate the expected frequencies (E). The expected value therefore for each condition (cell in table)  becomes the(  row total x column total ) divided by the grand total.

  3. Compare the observed and expected frequencies using the following calculation for each condition:

\displaystyle \frac{{{\left( O-E \right)}^{2}}}{E}

  1. Sum all the conditions to determine the Χ2 Statistic.

  2. Determine the critical value using the Χ2 table or probability calculation. The degrees of freedom is the number of rows minus 1 times the number of columns minus one.

  3. Compare the test statistic to the critical value and determine if a convincing difference exists at a specific confidence level.


A bicycle manufacturer wants to determine if three inspection stations provide the same results or not. Each station should detect a specific dimension fault on a finished bicycle frame. In order to evaluate the test stations we collected 90 frame of which 30 have the fault. Each test station evaluated each of the 90 frames in random order. The observed results are below.

Test Stations Observed Results
#1 #2 #3
Defects Detected 27 25 22 74
Defects Not Detected 3 5 8 16
Totals 30 30 30 90

Is there any convincing difference in the test stations (use 95% confidence).


Null Hypothesis: H0: p1 = p2 = p3, or there is no difference among the three test stations.

Alternative hypothesis: Ha: p1 ≠ p2 ≠ p3

Step 1. The observations are in the table above.

Step 2. The expected values are in the table below.

#1 #2 #3
Defects Detected 24.67 24.67 24.67 74
Defects Not Detected 5.33 5.33 5.33 16
Totals 30 30 30 90

Step 3. Compare the observed and expected frequencies.

The following table has the results of the ( O – E )2 / E

#1 #2 #3
Defects Detected 0.220 0.004 0.289
Defects Not Detected 1.019 0.020 1.338

Step 4. Total the expected vs observed values.

The tally is the Χ2 statistic

Χ2 = 0.220 + 0.004 + 0.289 + 1.019 + 0.020 + 1.338 = 2.89

Step 5. Determine the critical value.

The degrees of freedom = (rows -1)(columns – 1) = (  2 -1 )( 3 – 1 ) = 2.

Using a Χ2 table or calculation we find there is only a 5% chance that the calculated value will exceed 5.99.

Step 6. Compare the test statistic and critical value.

Since the calculated value, 2.89, is less then the critical value, 5.99, there is not convincing evidence the test stations are different.

This entry was posted in II. Probability and Statistics for Reliability and tagged by Fred Schenkelberg. Bookmark the permalink.

About Fred Schenkelberg

I am an experienced reliability engineering and management consultant with FMS Reliability, a consulting firm I founded in 2004. I left Hewlett Packard (HP)’s Reliability Team, where I helped create a culture of reliability across the organization, to assist other organizations. Given the scope of my work, I am considered an international authority on reliability engineering. My passion is working with teams to improve product reliability, customer satisfaction, and efficiencies in product development; and to reduce product risk and warranty costs. I have a Bachelor of Science in Physics from the United States Military Academy and a Master of Science in Statistics from Stanford University.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s