1. A gambler suspects that a die is loaded. She rolls the die 60 times and gets the following results.

Face

1

2

3

4

5

6

Count

24

7

8

9

8

4

 

H0: The die is not loaded. It is fair.

HA: The die is not fair.

If the null hypothesis is true and the die is fair, then each of the faces should be as likely to come up as any other face. The probability of any face should then be 1/6. 1/6 of 60 is 10, so the expected number of occurrences for each face is 10. We compute chi squared.

Face

Obs

Exp

O - E

(O-E)2

(O-E)2/E

1

24

10

14

196

19.6

2

7

10

-3

9

0.9

3

8

10

-2

4

0.4

4

9

10

-1

1

0.1

5

8

10

-2

4

0.4

6

4

10

-6

36

3.6

Totals

25.0

With 6 cells, there will be 5 degrees of freedom. A value of 25 for chi square is off the chart to the right for that many degrees of freedom. We conclude that we should reject the null hypothesis and conclude that the die is not a fair die. It looks like a 1 comes up way more than we would expect. A TI 84+ calcualtes the p-value to be approximately 0.0001393. This probability that you could be this far from what you would expect if the null hypothesis were true is so small that we should not assume that the die is fair.