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BuiltWithNOF
Chi-Squared 
Test of Independence

Mathematics & Statistic Tutor Perth - SPSS Help   

    Chi-Square Test of Independence 

    Words to look  for          Related, Independence, Relationship, Depends, Expected

    Layout to look for          Table with one variable across the top and one down the side.                                                                 

    These will be the variables that replace the dots in the hypotheses

     Hypotheses

     Null Hypothesis:                 ............... and ............... are independent.
     Alternative Hypothesis     ............... and ............... are not independent.

Notice the first line has one word “ are ” and the second line has two words “ are not ”

The dots are replaced by whatever the variables are in the table i. e. the two variables which in the question.

         Test Data


Finding Expected values by using formula as well as subtraction. This formula calculates the Expected value if the two variables are indeed independent.

The formula is:

        Expected Value = (Row Total)(Column Total)
                                                   Grand Total

Subtraction will save some time but you may use the formula for all expected values. 
What do we mean by this?  The row and column totals of Observed and Expected values  are equal, so we can subtract to obtain other Expected values.

In the following tables, I have starred the values that you may calculate using the formula and the blank ones you would get by subtraction.

*

*

 

 

 

 

*

*

 

*

*

 

 

 

 

Ok, now put some numbers in so we can see how it works.

    The Blue numbers are the row and column totals.
    The Red numbers are the expected values obtained by Formula.
    The Green numbers are the expected values obtained by subtraction.

 

26

15

9

50

21

10

 

 

18

 

 

 

65

 

 

125

 So 26 is obtained by     65 x 50
                                                            125


 
So 18 is obtained by   65(26 + 21)
 So  is obtained by   50(26 + 15)

 

      Test Statistic


Evaluating the chi-square statistic, chi square back

There is no short cut this time apart from a calculator but I still find it handy to write some values down. I tend to find it handy to write the values in the table format.
This saves time when you are checking as you can tell whether you got the same answer last time.  I have shown the the actual values below, but in the examination perhaps just the answers would do as long as you put the formula somewhere.

The Observed values are in Red.
The Expected values are in Blue

26    20

15   17

  13

50

 21    23

10   13

 

 

18

 

 

 

65

 

 

125

      chi square back  = (O - E) 2      =     (26 -   20) (15 -  17 )(9 - 13) 2
                              E                            20                     17                   13

                                (21 - 23) 2   +  (10 - 13) 2   +  etc ...
                                        23                       13

      P-value

The value is found in
chi square back tables
with degrees of freedom being (number of rows - 1) times (number of columns - 1)


When the level of significance is 0.05 then the following applies:

      If the p-value bigger  than 0.05   then  Do not reject the Null Hypothesis

      If the p-value smaller than 0.05 then  Reject the Null Hypothesis

 

     Decision and Conclusion.

If we Reject the Null Hypothesis,

then the results are statistically significant suggesting that there is sufficient evidence
to indicate that the two variables are not independent at the 5% significance level.

 

If we Do Not  Reject the Null Hypothesis,

then the results are not statistically significant suggesting that there is insufficient evidence
to indicate that the two variables are not independent at the 5% significance level.

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