Question

The author 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: 27, 31, 42, 40, 28, and 32. Use a 0.05 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

`\chi^2 = ((27-33.33)^2)/(33.33)+((31-33.33)^2)/(33.33)+((42-33.33)^2)/(33.33)+((40-33.33)^2)/(33.33)+((28-33.33)^2)/(33.33)+((32-33.33)^2)/(33.33) =5.860`

`p_v = P(\chi^2_(5) >5.860)=0.32`

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

Explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Number: 1, 2 , 3 , 4 , 5 ,6

Frequency: 27, 31, 42, 40, 28, 32

We need to conduct a chi square test in order to check the following hypothesis:

H0: The outcomes are equally likely.

H1: The outcomes are not equally likely.

The level of significance assumed for this case is
`\alpha=0.05`

The statistic to check the hypothesis is given by:

`\chi^2 =\sum_(i=1)^n ((O_i -E_i)^2)/(E_i)`

The observed values are given:

`O_(1)=27`
`O_(2)=31`

`O_(3)=42`
`O_(4)=40`

`O_(5)=28`
`O_(6)=32`

The expected values are given by:

`E_(1) =(1)/(6)*200=33.33`
`E_(2) =(1)/(6)*200=33.33`

`E_(3) =(1)/(6)*200=33.33`
`E_(4) =(1)/(6)*200=33.33`

`E_(5) =(1)/(6)*200=33.33`
`E_(6) =(1)/(6)*200=33.33`

And now we can calculate the statistic:

`\chi^2 = ((27-33.33)^2)/(33.33)+((31-33.33)^2)/(33.33)+((42-33.33)^2)/(33.33)+((40-33.33)^2)/(33.33)+((28-33.33)^2)/(33.33)+((32-33.33)^2)/(33.33) =5.860`

Now we can calculate the degrees of freedom for the statistic given by:

`df=Categories-1=6-1=5`

And we can calculate the p value given by:

`p_v = P(\chi^2_(5) >5.860)=0.32`

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(5.860,5,TRUE)"

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.