A person drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 200 times. Here are the observed frequencies for the outcomes of​ 1, 2,​ 3, 4,​ 5, and​ 6, respectively: 28​, 29​, 40​, 41​, 28​, 34. Use a 0.10 significance level to test the claim that the outcomes are not equally likely. Does it appear that the loaded die behaves differently than a fair​ die

Solution :

The objective of the study is to test the claim that the loaded die behaves at a different way than a fair die.

Null hypothesis, [tex]$H_0:p_1=p_1=p_3=p_4=p_5=p_6=\frac{1}{6}$[/tex]

That is the loaded die behaves as a fair die.

Alternative hypothesis, [tex]$H_a$[/tex] : loaded die behave differently than the fair die.

Number of attempts , n = 200

Expected frequency, [tex]$E_i=np_i$[/tex]

                                        [tex]$=200 \times \frac{1}{6} = 33.333$[/tex]

Test statistics, [tex]$x^2= \sum^6_{i=1} \frac{(O_i-E_i)^2}{E_i} $[/tex]

                            [tex]$=\frac{(28-33.333)^2}{33.333}+\frac{(29-33.333)^2}{33.333}+\frac{(40-33.333)^2}{33.333}+\frac{(41-33.333)^2}{33.333}+\frac{(28-33.333)^2}{33.333}+$[/tex][tex]$\frac{(34-33.333)^2}{33.333}$[/tex]

≈ 5.8

Degrees of freedom, df = n - 1

                                       = 6 - 1

                                       = 5

Level of significance, α = 0.10

At α = 0.10 with df = 5, the critical value from the chi square table

[tex]$x^2_{\alpha}= \text{chi inv}(0.10,5)$[/tex]

     = 9.236

Thus the critical value is [tex]$x_{\alpha}^2=9.236$[/tex]

[tex]$P \text{ value} = P[x^2_{df} \geq x^2]$[/tex]

             [tex]$=P[x^2_5\geq 5.80]$[/tex]

             = chi dist (5.80, 5)

             = 0.3262

Decision : The value of test statistics 5.80 is not greater than the critical value 9.236, thus fail to reject [tex]$H_o$[/tex] at 10% LOS.

Conclusion : There is no enough evidence to support the claim that the loaded die behave in a different than a fair die.

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