Problem Set 2

Problem Set 2 - Tatiana Nemirovsky Liza Levin BMGT452...

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Tatiana Nemirovsky Liza Levin BMGT452 Problem Set 2 1. ANOVA Test a. Q7c vs. Q13 a.i. Step 1 : Null: The mean importance ratings of phoning in for information as an information source for movies playing at movie theaters are the same across all classifications. Alternative: The mean importance ratings of phoning in for information as an information source for movies playing at movie theaters are different across all classifications. a.ii. Step 2 : ANOVA Test a.iii. Step 3 : significance level= 0.05 a.iv. Step 4 : use SPSS to calculate: F test statistic= 0.688 Degree of freedom for the denominator= 4 Degree of freedom for the numerator= 442 P-value= 0.600 a.v. Step 5 : Conclusion: Since the p-value is larger than 0.05, we fail to reject the hypothesis that the mean importance ratings of phoning in for information as an information source for movies playing at movie theaters is the same across all classifications. b. Q7d vs. Q13 b.i. Step 1 : Null: The mean importance ratings of television as an information source for movies playing at movie theaters are the same across all classifications. Alternative: The mean importance ratings of television as an information source for movies playing at movie theaters are different across all classifications. b.ii. Step 2 : ANOVA Test b.iii. Step 3 : significance level= 0.05
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b.iv. Step 4 : use SPSS to calculate: F test statistic= 3.662 Degree of freedom for the denominator= 4 Degree of freedom for the numerator= 442 P-value= 0.006 b.v. Step 5 : Conclusion: Since the p-value is less than 0.05, reject the hypothesis that the mean importance ratings of television as an information source for movies playing at movie theaters is the same across all classifications. 1.
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This note was uploaded on 12/15/2011 for the course BMGT 452 taught by Professor Staff during the Fall '08 term at Maryland.

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Problem Set 2 - Tatiana Nemirovsky Liza Levin BMGT452...

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