# Comparing percentages

#### amperes

##### New Member
I'm having trouble wrapping my head around some of this. I need to determine whether or not there is a statistical difference between two percentages. The data are as follows:

In population A there are 525 teeth and out of these, 114 have cavities, thus 21.71%.
In population B there are 413 teeth and 50 have cavities, thus 12.10%.

Which statistical function would I use to determine if the difference between the two populations is statistically significant? From what I can tell, a chi square test would work, but I am not sure. After fudging around with the numbers in a horrible medley of chi-square formulas, I've determined that there is statistical difference between the two groups, but none of my results match one another so I'm pretty lost...

Any help would be greatly appreciated, please let me know if any other data would be helpful or necessary.

#### Mean Joe

##### TS Contributor
You can do a chi-square test. I suspect you are not calculating the expected frequencies correctly. You should make sure you have that down.

As a check, for your answer, note that the z-test for two proportions ( see here) is equivalent: both give p=.00012

#### amperes

##### New Member
For expected frequencies I followed an example that you gave in http://talkstats.com/showthread.php?t=8203 and ended up with 91.77 expected for population A and 72.19 expected for population B.

Ask me what any of this means and I have no idea. Also, how would I go about plugging this into excel? I've mostly been doing this by hand.

#### amperes

##### New Member
Using some resources/references found around the web, I've come up with the attached excel spreadsheet for my data chi square, but I'm not sure if I am even on the right track.

#### Mean Joe

##### TS Contributor

As for what you're doing:
If you combine populations A and B, then you have 164 cavities in 938 teeth (p=.1748).
So if populations A and B have the same incidence of cavities, our best guess would be that 17.48% have cavities.

In population A there are 525 teeth, so 17.48% of 525 is 91.79
In population B there are 413 teeth, so 17.48% of 413 is 72.21

#### amperes

##### New Member
I definitely messed up the Group B cavity number, just another reason why doing stats while tired is a bad idea. Thanks a lot for catching that.

Knowing that a p-value below 0.05 is generally considered statistically significant, how would I report this information? Would it be something like this:

"The difference between cavities found in Group A and Group B is statistically significant, X²(1, n=938) = ?, p<.05"?

Or is there an easier way? Perhaps something like this:
"Using a chi-square test, the difference between cavities found in Group A and Group B is statistically significant (p=0.000120166)"

#### amperes

##### New Member
If I know how many cavities each individual in both groups has, would an independent student t-test make more sense for determining statistical significance?

#### Mean Joe

##### TS Contributor

Reporting: "Using a chi-square test..." is a sound way to report it. It's good sense to state how you tested the data.

Alternative test: In this instance (two groups x two outcomes being cavity/no cavity) the chi-square test is equivalent to the test of two proportions.

#### numbersgame

##### New Member
This may sound stupid because I am a high school student in AP stats still learning, but what is the chi-square test? I just got sent off to boarding school and they are drilling me with new information. Have you ever heard of Chrysalis School Montana, is that kind of thing normal at boarding school? Thanks in advance guys.

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#### Cyndi Armstrong Persily

##### New Member
Does anyone have advice on how to compare percentages from four unequal groups that have been weighted to reflect complex sampling techniques to determine if any are significantly different from each other? Z scores? Here is what I have:
n % 95% CI

Urban 168,093 21.9 (21.5 , 22.3)
Micropolitan 44,186 21.3 (20.3 , 22.3)
Small rural 22,172 20.9 (19.1 , 22.7)
Remote rural 19,190 22.0 (20.0 , 24.0)

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