Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have items made from two different materials. For each material I took a set of 250 samples and subjected them to a test which either broke them or did not.

Which test could answer the question - is there a significant difference between the breakage rates of the two materials. I constructed a contingency table but am not sure how to proceed - some sort of chi-squared test?

$$ \begin{matrix} & \text{Broken} & \text{Unbroken}\\ \text{Material A} & 2 & 248 \\ \text{Material B} & 1 & 249\\ \end{matrix} $$

The numbers of broken samples was small, as shown in the table.

share|improve this question
Based on these data, if some statistical test showed that material A and material B were different, would you believe it? – John Dec 17 '12 at 17:40
up vote 4 down vote accepted

Pearson's chi-squared test for association can be used for this sort of problem. For tables with low expected values, like your one, Fisher's Exact Test is better. But you don't need to do any statistical test to see that you need more data to tell whether A or B has the highest breakage rate.

share|improve this answer
+1 One way to provide intuition concerning your last remark is to point out that under the null hypothesis (of equal breakage rates), we may view the distribution of the three broken samples among the two rows (materials) as being a matter of pure chance, as if a fair coin had been flipped for each breakage to determine which row it would be assigned to. We would see some imbalance in the assignments (because $3$ is odd) and even the most extreme imbalance where all breakages occur for a single material would have a chance of $1/8+1/8=1/4$, too big to be considered "significant." – whuber Dec 17 '12 at 17:09
Agreed. Now I know the appropriate test I can estimate how much data I would need to start to see differences. – Peter Hull Dec 19 '12 at 9:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.