Question

You suspect that an unscrupulous employee at a casino has tampered with a die; that is, he is using a loaded die. In order to test this claim, you roll the die 282 times and obtain the following frequencies: (You may find it useful to reference the appropriate table: chi-square table or F table)

      
Category   1   2   3   4   5   6
Frequency   64   57   46   38   38   39

Click here for the Excel Data File

a. Choose the appropriate alternative hypothesis to test if the population proportions differ.

- All population proportions differ from 1/6.
- Not all population proportions are equal to 1/6.

b. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.)

c. Find the p-value.

0.01 p-value < 0.025
p-value < 0.01
p-value 0.10
0.05 p-value < 0.10
0.025 p-value < 0.05

d. At the 10% significance level, can you conclude that the die is loaded?

No since the p-value is more than significance level.
Yes since the p-value is more than significance level.
No since the p-value is less than significance level.
Yes since the p-value is less than significance level.

Answer 1

a

- Not all population proportions are equal to 1/6.

b)

applying chi square test:

	relative	observed	Expected	residual	Chi square
category	frequency	Oi	Ei=total*p	R2i=(Oi-Ei)/√Ei	R2i=(Oi-Ei)2/Ei
1	1/6	64.000	47.00	2.48	6.149
2	1/6	57.000	47.00	1.46	2.128
3	1/6	46.000	47.00	-0.15	0.021
4	1/6	38.000	47.00	-1.31	1.723
5	1/6	38.000	47.00	-1.31	1.723
6	1/6	39	47.00	-1.17	1.362
total	1.000	282	282		13.106

value of the test statistic X² = 13.106

c)

0.01 p-value < 0.025

d)

Yes since the p-value is less than significance level.