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valentinak56 [21]
3 years ago
14

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
Mathematics
1 answer:
algol [13]3 years ago
4 0

Answer:

\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(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.

Step-by-step 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 \frac{(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} =\frac{1}{6}*200=33.33   E_{2} =\frac{1}{6}*200=33.33

E_{3} =\frac{1}{6}*200=33.33   E_{4} =\frac{1}{6}*200=33.33

E_{5} =\frac{1}{6}*200=33.33   E_{6} =\frac{1}{6}*200=33.33

And now we can calculate the statistic:

\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(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.

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