Question

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

Answer 1

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,
`$H_0:p_1=p_1=p_3=p_4=p_5=p_6=(1)/(6)$`

That is the loaded die behaves as a fair die.

Alternative hypothesis,
`$H_a$` : loaded die behave differently than the fair die.

Number of attempts , n = 200

Expected frequency,
`$E_i=np_i$`

`$=200 * (1)/(6) = 33.333$`

Test statistics,
`$x^2= \sum^6_(i=1) ((O_i-E_i)^2)/(E_i) $`

`$=((28-33.333)^2)/(33.333)+((29-33.333)^2)/(33.333)+((40-33.333)^2)/(33.333)+((41-33.333)^2)/(33.333)+((28-33.333)^2)/(33.333)+$`
`$((34-33.333)^2)/(33.333)$`

≈ 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

`$x^2_(\alpha)= \text{chi inv}(0.10,5)$`

= 9.236

Thus the critical value is
`$x_(\alpha)^2=9.236$`

`$P \text{ value} = P[x^2_(df) \geq x^2]$`

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

= 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
`$H_o$` 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.