**ASSUMPTIONS**

A**. **Independent variables

**Example of non-independence: Height after experiment
by height before experiment*

**Q: Can Chi square be used on this data?**

**A: No, because the variables are not independent (height before
and after are**

dependent).

***Note: This will generally not be a problem
in this course.**

- Every fe
__>__5 recall

***The smallest sample size you can possibly
have and meet this assumption is 20.**

CHURCH AFFILIATION

ALIENATION SCORES | V | D | Q | R | T | |

10 | 0 | 0 | 0 | 1 | 1 | 2 |

8 | 0 | 0 | 1 | 2 | 2 | 5 |

7 | 0 | 4 | 2 | 0 | 0 | 6 |

6 | 1 | 3 | 6 | 1 | 0 | 11 |

3 | 0 | 2 | 3 | 1 | 0 | 6 |

2 | 1 | 1 | 1 | 0 | 0 | 3 |

2 | 10 | 13 | 5 | 3 | 33 |

**Q:** Can Chi square be used on this data?

**A:** = __(fo - fe)__^{2}

------------fe

Where: fo = number in a cell

fe = __(row total)(col total)
__-------------N

But computing every fe and then summing them up is quite time
consuming. Instead compute the smallest fe and, if it is greater
than five, then all others will satisfy the assumption.

Using table above: Smallest fe = __2(2)__ = __4 __= .12

---------------------------------33----33

**Steps for meeting chi square assumptions:**

1. Compute smallest **fe **=(__smallest row total)(smallest
column total)
__------------------------------------------ N

If smallest **fe** is less than 5, it is necessary to collapse
the table, assuming it is larger than a

2 x 2 table.

2. Locate the smallest subtotal in the entire table (i.e. the smallest row total or the smallest column total). If this smallest subtotal is for a subcategory of a nominal variable, then combine it with the next smallest subtotal for a subcategory of a nominal variable.

3. If the smallest subtotal is for a subcategory of an ordinal or interval variable, then combine it with the adjacent subcategory of the variable with the smallest subtotal.

4. After making a combination of a row total with another row
total or a column with another column, check again to see if the
smallest **fe **is greater than 5. If not, proceed to collapse
the table again by the procedure described above, assuming it
is still larger than a 2 x 2 table.

5. Continue collapsing the table only until the smallest **fe**
is greater than 5. Stop collapsing at this point and compute Chi
square. Be sure to compute the degrees of freedom on the collapsed
table rather than the original table.

6. If the table is collapsed until it is a 2 x 2 table, and the
smallest **fe** is still smaller than 5, then it can be collapsed
no further. It is not legitimate to use Chi square on the data.

Notice: This **fe** is < 5 so look for the smallest subtotal
in the table. If the data is nominal, combine the smallest subtotal
with the next smallest in that category. If the data is ordinal,
combine the subtotal with the row or column on either side. This
is called "collapsing the table for Chi Square". It
is a very frequent task in SPSS with large range variables .

CHURCH AFFILIATION

ALIENATION SCORES | V&T
| D | Q | R | |

10 | 1 | 0 | 0 | 1 | 2 |

8 | 2 | 0 | 1 | 2 | 5 |

7 | 0 | 4 | 2 | 0 | 6 |

6 | 1 | 3 | 6 | 1 | 11 |

3 | 0 | 2 | 3 | 1 | 6 |

2 | 1 | 1 | 1 | 0 | 3 |

5 | 10 | 13 | 5 | 33 |

Here column V and column T were combined. Now figure fe again:

Smallest fe= [2(5)]/33 = 10/33 = .33, but this is still <5
so recombine the smallest sub-total.

CHURCH AFFILIATION

ALIENATION SCORES | V&T | D | Q | R | |

10 & 8 | 3 | 0 | 1 | 3 | 7 |

7 | 0 | 4 | 2 | 0 | 6 |

6 | 1 | 3 | 6 | 1 | 11 |

3 | 0 | 2 | 3 | 1 | 6 |

2 | 1 | 1 | 1 | 0 | 3 |

5 | 10 | 13 | 5 | 33 | |

Here row 10 and row 8 were combined. Note that this is ordinal data so it could have only been combined with either the row above or below this. (Since 10 was the top of the category, combine the row with number 8, which was directly below 10.) The smallest fe = [(3)(5)]/33 = 15/33 = too small; combine again!

CHURCH AFFILIATION

ALIENATION SCORES | V&T | D | Q | R | |

10 & 8 | 3 | 0 | 1 | 3 | 7 |

7 | 0 | 4 | 2 | 0 | 6 |

6 | 1 | 3 | 6 | 1 | 11 |

3 & 2 | 1 | 3 | 4 | 1 | 9 |

5 | 10 | 13 | 5 | 33 |

Here rows 3 & 2 were combined.

fe = [5(5)]/33 = 25/33 = too small! Do again . . .

CHURCH AFFILIATION

ALIENATION SCORES | VT&R
| D | Q | |

10 & 8 | 6 | 0 | 1 | 7 |

7 | 0 | 4 | 2 | 6 |

6 | 2 | 3 | 6 | 11 |

3 & 2 | 2 | 3 | 4 | 9 |

10 | 10 | 13 | 33 |

Combine columns V, T and R. fe = [6(10)]/33 < 2

This fe is still less than 5, so combine again.

CHURCH AFFILIATION

ALIENATION SCORES | VT&R | D | Q | |

10,8, & 7 |
6 | 4 | 3 | 13 |

6 | 2 | 3 | 6 | 11 |

3 & 2 | 2 | 3 | 4 | 9 |

10 | 10 | 13 | 33 |

Combine rows 10, 8 and 7.

fe < 3

CHURCH AFFILIATION

ALIENATION SCORES | VT&R | D | Q | |

10,8, & 7 | 6 | 4 | 3 | 13 |

6,3, & 2 |
4 | 6 | 10 | 20 |

10 | 10 | 13 | 33 |

Combine rows 6, 3, and 2.

fe < 4

CHURCH AFFILIATION

ALIENATION SCORES | VTR&D | Q | |

10,8, & 7 | 10 | 3 | 13 |

6,3, & 2 | 10 | 10 | 20 |

20 | 13 | 33 |

Combine columns V, R, T and D.

fe = [13(13)]/33 > 5 so done. Notice that this is a 2 x 2 table.

Some things to remember when collapsing tables:

Much information is lost when this is done. Make the fewest number
of combinations so that fe __>__ 5.

Do not immediately assume that the table will be 2 x 2.

Short cut: Don't rewrite table after each combination, just do it often enough to keep track of the data.