Question

Below gives a data set for movie budgets, and domestic and worldwide box office gross (all in millions) for several major movies over time. Test the claim, at the � = 0.05 level, that the domestic box office gross has a wider variation than the worldwide box office gross. Movie Data Sheet, h̶t̶t̶p̶s̶:̶/̶/̶w̶w̶w̶.̶s̶t̶a̶t̶c̶r̶u̶n̶c̶h̶.̶c̶o̶m̶/̶a̶p̶p̶/̶i̶n̶d̶e̶x̶.̶p̶h̶p̶?̶d̶a̶t̶a̶i̶d̶=̶2̶1̶8̶8̶6̶8̶4̶ (See below table instead)

1. Note that some of the values for gross are zero. What does this mean and does that effect how you should perform the test?
2. Perform the hypothesis test.
   1. Check Conditions
   2. Hypotheses
   3. Statistic
   4. Critical Value(s)
   5. Test Statistic
   6. P-Value
   7. Decision
   8. Conclusion

Column	n	Mean	Variance	Std. dev.	Std. err.	Median	Range	Min	Max	Q1	Q3
Domestic Gross($M)	5222	43.496456	4616.5913	67.945502	0.9402477	19.188847	936.66223	0	936.66223	2.694071	54.927174
Worldwide Gross($M)	5222	94.434584	30132.598	173.58744	2.4021485	31.16133	2783.919	0	2783.919	6.173485	102.22663

Answer 1

(a) Since the sample sizes are very large, normality can be assumed

(b) Hypothesis test:

Data:      

n1 = 5222     

n2 = 5222     

s1^2 = 4616.591     

s2^2 = 30132.6     

Hypotheses:     

Ho: σ1^2 = σ2^2     

Ha: σ1^2 > σ2^2     

Decision Rule:     

α = 0.05     

Numerator DOF = 5222 - 1 = 5221   

Denominator DOF = 5222 - 1 = 5221   

Critical F- score = 1.046585    

Reject Ho if F > 1.046585    

Test Statistic:     

F = s1^2 / s2^2 = 4616.5913/30132.598 = 0.1532   

p- value = 1     

Decision (in terms of the hypotheses):   

Since 0.153209 < 1.0466 we fail to reject Ho

Conclusion (in terms of the problem):   

There is no sufficient evidence that the domestic box office gross has a wider variation than the worldwide box office gross.

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