Chi Squared

Last Stats test - although probably my least favourite!!!
 Chi Squared = used to show whether there is a siginficant difference between the expected frequencies and observed frequencies in one or more categories
---> Could be used in Poole, for number of people walking in the Hamworthy regeneration site, before and after the brdige is built
Worked Example (the one we did in class)
As with the others, the first two things to do are to set a null and positive hypothesis (there will/won't be a difference between expected and observed frequencies) and then set the data in a table. O represents the observed data, so this is the relevant figures from the table in the AIB, or if you were to conduct fieldwork i.e a traffic survey, the results of your survey. Then you have to add up the totals of the columns and rows....
For row SOA = 76 + 16 + 8
                      = 100
For column 'Owner Occupied' = 76 + 94
                                                 = 170
.....The most important total to get correct is the GRAND TOTAL. This is the sum of all the row totals which, if you have done it correctly, will also equal the sum of all the column totals and vice versa
Sum of Row totals = 100 + 100
                               = 200
Sum of Column totals = 170 + 18 + 12
                                    = 200
Next you have to calculate the expected values, E.
E = row total x column total
For the first 'box':
E = 100 x 170 = 85
           200
Note, that it is just coincidence (or clever planning by the teachers!) that the E values are the same down the columns - this is not always the case.
Now you have both O and E and so can use the equation. For each 'box' you need to minus E from O, square the answer and then divide by E (O – E)²
                                                                    E
For the first box:
O = 76 and E = 85
So, O - E = 76 - 85 = -9
-9 x -9 = 81
81 = 81
 E     85
= 0.95 (to 2.d.p)
The final step is to add all those numbers up that you have just calculated (the red numbers) as, refering back to the formula, this gives you the value of Chi Squared
0.95 + 5.44 + 0.67 + 0.95 + 5.44 + 0.67 = 14.12
This is your value of Chi Squared but before you can 'use' it you have to compare the result with the critical values according to degrees of freedom. The degree of freedom for Chi Squared can be calculated by:
Degree of Freedom = (number of columns- 1) x (number of rows -1)
With this example, we can say with 99% statistical significance that there is a difference between the observed and expected values.

Well, that is the last of the statistical tests we need to know - if anyone would like any more examples from the AIB then just ask; I realise that the above example is one we did in class, so at some stage I may add another example from the AIB. Next up I am going to focus on fieldwork, like sampling techniques etc and also summarising some general information about Poole and the regeneration project - I hope the revision is going well!