Respuesta :
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.