One last thing: degrees of freedom
When doing a chi-square goodness of fit test, there is one last wrinkle to iron out, called degrees of freedom.
When I told you that 42 out of 100 sickdays were on Mondays or Fridays, you automatically knew that 58
had to be in the middle of the week, right? I was "free" to specify how many were on Monday/Friday,
but then I was NOT "free" to decide how many were on non-Monday/Friday. So we say that, in this problem,
there is only 1 degree of freedom.
Say you flip a coin 100 times. If we want to do a chi-square test to determine whether a coin is fair (lands equally on heads and tails), how many degrees of freedom would the test have?
(To make this problem interactive, turn on javascript!)
- I need a hint ... :If I tell you the number of heads, do you also know the number of tails?
- ...another hint ... : How many variables are "free" to vary?
I think I have the answer: There are two variables here -- number of
heads and number of tails. But only 1 is free to vary -- once I tell you
how many heads there were, you know how many tails there were,
or vice versa.
It is possible to do chi-square tests using more than 2 variables. For example, let's say I got data on how many sickdays fell on EACH of the five weekdays:
| day | observed | expected |
| mon | 22 | 20 |
| tues | 19 | 20 |
| wed | 19 | 20 |
| thurs | 20 | 20 |
| fri | 20 | 20 |
We could do a chi-square test to check whether the distribution of sickdays matched our expectations for ALL FIVE weekdays
How many degrees of freedom would this test have?
(To make this problem interactive, turn on javascript!)
- I need a hint ... : There are 5 weekdays -- how many of those am I "free"
to specify data for?
- ...another hint ... : If I knew that there were 20 sickdays each on Monday
through Thursday, is Friday still "free" to vary?
I think I have the answer: Once I know how many sickdays occurred
on 4 of the 5 days, the fifth day is no longer "free" to vary.
Therefore there are only 4 degrees of freedom.
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