CK-12 Probability and Statistics - Advanced

10.2. Test of Independence http://www.ck12.org

TABLE10.9: Number Rolled on the Potentially Loaded Dice

1 2 3 4 5 6 Totals

Dice 1 6 1 2 2 3 6 20

Dice 2 4 1 3 3 1 8 20

Totals 10 2 5 5 4 14 40

Like the other Chi-Square tests, we first need to establish a hypothesis based on a research question. In this case,
our research question would look something like: “Is the probability of rolling a specific number the same for Dice
1 and Dice 2?” This would give us the following hypotheses:

Null Hypothesis H 0 :O=E (The probabilities are the same for both die)

Alternative Hypothesis Ha:O 6 =E (The probabilities differ for both die)

Similar to the other test, we need to calculate the expected values for each cell and the total number of Degrees
of Freedom. To get the expected frequency for each cell, we use the same formula as we used for the Test of
Independence:

Expected Frequency=

(Row Total)(Column Total)

Total Number of Observations

The following table has includes the Expected Frequency (in parenthesis) for each cell along with the Chi-Square
statistic((O−E)^2 /E)in a separate column.

Number Rolled on the Potentially Loaded Dice

TABLE10.10:

1 X^22 X^23 X^24 X^25 X^26 X^2 X^2

Total

Dice

1

6 ( 7. 5 ) 0. 3 1 ( 1 ) 0 2 ( 2. 5 ). 1 2 ( 2. 5 ). 1 3 ( 2 ). 5 6 ( 7 ). 2 1. 2

Dice

2

4 ( 7. 5 ) 1. 6 1 ( 1 ) 0 3 ( 2. 5 ). 1 3 ( 2. 5 ). 1 1 ( 2 ). 5 8 ( 7 ). 2 2. 5

Totals 10 2 5 5 4 14

and the Degrees of Freedom= (C− 1 )(R− 1 )

df= ( 6 − 1 )( 2 − 1 ) = 5

Using the same Chi-Square formula and the information from the table above, we find that:

X^2 =. 1. 2 + 2. 5 = 3. 7

Using an alpha level of .05,we look under the column for.05 and the row for Degrees of Freedom(d f= 5 ). Using
the standard Chi-Square distribution table, we see that the critical value for Chi-Square is 11.07. Therefore we would
reject the null hypothesis if the Chi-Square statistic is greater than 11.07.

Reject(H 0 :O)ifX^2 > 11. 07

Since our calculated Chi-Square value of 3.7 is not greater than 11.07, we fail to reject the null hypothesis. Therefore,
we can conclude that each number is just as likely to be rolled on one die as the other. This means that if the dice
are loaded, they are probably loaded in the same way or were made by the same manufacturer.