It sounds like you have a series of 2 x 2 tables: indigenous/non-indigenous and custodial/non-custodial sentence, with one such table for each crime. If that is correct, then it should be straightforward to do the chi-square analyses. But please clarify if there is more to it than that.

Just to add from what David L Morgan wrote, and if I'm correct (which is never guaranteed), you do effectively have (well it can be calculated) expected vs actual. For example, here with the total number of each group of people and total number of penalties for each group, you can calculate (well R will do this for you) what is the expected number of penalties if each group are equally likely to get penalties. So with equal numbers of people in each group (to keep it simple):

Ind: 50 total people in the population, 4 given custodial, (46 not given custodial)

Non: 50 total people in the population, 8 given custodial, (42 not given custodial)

Overall you have 100 people and 12 custodial so if they are equally likely (the null) then each group of 50 (Ind an Non) you would expect 6 custodial each and so the test will see if the observed significantly differs from this. It is slightly more complicated when you have 95 people in one group and 5 in the other but the logic is the same (and R should handle this automatically - though I dont use R too often).

Splitting out by crime is great as you account for some groups may be more/less likely to commit certain crimes that may be more/less likely to be punished with a custodial sentence but remember the above analyses are only showing 1 group are more/less likely to be given custodial sentences relative to their total population. It doesn't therefore necessarily show that they are more likely to commit certain crimes (e.g. chi-square of total population and number of crimes) or that they are more likely to get custodial sentences than non-custodial (e.g. chi-square of the number of crimes and number of custodial sentences). There may be an aggregation issue, for example, if one group are more likely to live in one area where they are clamping down on crime, more of that group would likely be given those sentences etc.

Hope this helps

Hi David and Michael.

Thank you both for getting back to me. Neither of my groups are of equal numbers, so for example I might have 100 Indigenous people but 1005 non-Indigenous people. Some of my tables are simple 2x2, but others are more complex. For example:

I NI

RNCR 136 1491

RCR 19 96

FNCR 339 3306

FCR 876 2485

PNCR 159 847

PCR 574 1108

CNCR 23 70

CCR 139 142

WSS 417 809

(sorry, doesn't paste that well). My thoughts are whether the chi sq will still work? T-test doesn't work for it according to our statistical expert here at the university.

For the above in R my chi sq result was 'X-squared = 978.27, df = 8, p-value < 2.2e-16', which suggests an extremely high statistical difference. So does this mean a stat diff between I and NI or between the Types of Penalties (the left letters :-))?

I am so very new to stats and am getting a bit confused, but unfortunately I need to use quantitative to help substantiate and provide context for my qualitative data.

Thanks

Kirstie

Chi-sq can be used with any type of table not just 2x2 And it doesn’t require equal numbers of observations. To interpret the results, use percentages that add to 100 within each category of your independent variable of indigenous or not.

Hi @DavidLMorgan

Thanks, that is what I have been doing and using Mosaic Plots to show the results. I just wasn't sure I was on the right track.

So, the chi sq result indicates a significance or not for all the variables as a whole rather than just a difference between Indigenous or Non Indigenous?

It indicates that indigeneous and non differ with regard to the “dependent” variable.

Awesome. Thanks so much for your help. Stats and numbers are my biggest weakness.

Hi David L Morgan

I have another query in relation to these states (I feel my head is going round and round in circles right now). So I have the following table:

I NI

Prison 10 8

ICO 2 0

PSS 1 1

Dism 2 7

My chi squ result is:

Pearson's Chi-squared test

data: QMC01

X-squared = 4.9729, df = 3, p-value = 0.1738

BUT as there are

I am not sufficiently familiar with mosaic plots to interpret the one you included. If you believe that the difference is located within a subset of your categories, you can re-examine the variable in question and run the analysis on just those categories.