Question

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=

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)	O_i	E_i=total*p	R²_i=(O_i-E_i)/√E_i	R²_i=(O_i-E_i)²/E_i	G² =2O_iln(O_i/E_i)
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 4 weeks ago

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