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In: Math

Use the following information to answer the next questions: A five-sided die is rolled 100 times....

Use the following information to answer the next questions: A five-sided die is rolled 100 times. Conduct a hypothesis test to determine if the die is fair. Use a 5% level of significance.
Observed Rolls: One=10; Two=29; Three=16, Four=15, Five=30
Expected Rolls: All the categories of rolls are the same

What test are you running?

What is the observed values of one for the rolled die?

What is the observed values of two for the rolled die?

What is the observed values of three for the rolled die?

What is the observed values of four for the rolled die?

What is the observed values of five for the rolled die?

What is the expected values of one,two,three,four,five for the rolled die?

What are the degrees of freedom?

What is the null hypothesis?

What is the alternative hypothesis?

What is the test statistic? Use one decimal place.

What is the p-value? Use three decimal places.

What is your conclusion based on the p-value and the level of significance?

At the 5% significance level, what can you conclude?

Solutions

Expert Solution

What test are you running?: chi square goodness of fit:

observed values of one for the rolled die=10

observed values of two for the rolled die=29

observed values of three for the rolled die=16

observed values of four for the rolled die=15

observed values of five for the rolled die=30

expected values of one,two,three,four,five for the rolled die =np =100*0.2 =20

degree of freedom =categories-1= 4

  null hypothesis: Ho:die is fair or all numbers appear with equal proportion

Alternate hypothesis: Ho:die is biased or all numbers does not appear with equal proportion

applying chi square goodness of fit test:  
           relative observed Expected residual Chi square likelihood ratio
category frequency(p) Oi Ei=total*p R2i=(Oi-Ei)/√Ei R2i=(Oi-Ei)2/Ei G2 =2*Oi*ln(Oi/Ei)
1 0.200 10.000 20.00 -2.24 5.000 -13.8629
2 0.200 29.000 20.00 2.01 4.050 21.5507
3 0.200 16.000 20.00 -0.89 0.800 -7.1406
4 0.200 15.000 20.00 -1.12 1.250 -8.6305
5 0.200 30.000 20.00 2.24 5.000 24.3279
total 1.000 100 100 16.1000 16.2446
test statistic X2 = 16.1000
p value = 0.003
since test statistic falls in rejection region we reject null hypothesis
we have sufficient evidence to conclude that,die is biased

milcah answered 7 months ago
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