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:
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
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15
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9
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50
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21
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10
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18
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65
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125
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So 26 is obtained by 65 x 50 125
So 18 is obtained by 65 - (26 + 21) So 9 is obtained by 50 - (26 + 15)
Test Statistic
Evaluating the chi-square statistic,
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
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15 17
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9 13
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50
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21 23
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10 13
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18
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65
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125
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= (O - E) 2 = (26 - 20) 2 + (15 - 17 )2 + (9 - 13) 2 E 20 17 13
P-value
The value is found in 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
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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