Question

Jorge was at the park playing with friends. He found a typical die with 6 sides on the ground. He took it home and rolled it 100 times and recorded the results (found in the table below). He wanted to see if the die was a 'fair die' or if it was weighted on one side so somone could cheat when playing games!

Is this a 'fair die' or has it been tampered with? Test at the α=0.05 level of significance.

Which would be correct hypotheses for this test?

H0:μ1=μ2

; H1:μ1≠μ2
H0:
The die is a fair die; H1:
The die has been tampered with
H0:p1=p2
; H1:p1≠p2
H0:
The die has been tampered with; H1:

The die is a fair die

Roll count:

Rolled   Count
1   1
2   5
3   4
4   6
5   9
6   75

Test Statistic:

Give the P-value:

Which is the correct result:

Reject the Null Hypothesis
Do not Reject the Null Hypothesis

Which would be the appropriate conclusion?

There is enough evidence to suggest that the die has been tampered with.
There is not enough evidence to suggest that the die has been tampered with.

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: p₁ = p₂ = p₃ = p₄ = p₅ = p₆

Alternative hypothesis: At least one of the proportions in the null hypothesis is false.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.

Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = k - 1 = 6 - 1
D.F = 5
(E_i) = n * p_i

Rolled	Observed	Expected	[(Or,c - Er,c)2/ Er,c]
1	1	16.667	14.7269988
2	5	16.667	8.166970001
3	4	16.667	9.6269808
4	6	16.667	6.826956801
5	9	16.667	3.526902802
6	75	16.667	204.1602501
Total	100	100	247.0350593

X² = 247.04

where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, E_i is the expected frequency count for level i, O_i is the observed frequency count for level i, and X² is the chi-square test statistic.

The P-value is the probability that a chi-square statistic having 5 degrees of freedom is more extreme than 247.04.

We use the Chi-Square Distribution Calculator to find P(X² > 247.04) = 0.00

Interpret results. Since the P-value (0.00) is less than the significance level (0.05), we have to reject the null hypothesis.

Reject the Null Hypothesis.
There is enough evidence to suggest that the die has been tampered with.