Answers to 2002 Final Exam

1. 12,870
    
2. a) mean=76.83, variance=181.37, standard error=5.498
   b) 1 sample t-test: t=3.06, DF=5, t-crit = 2.571 (two-tailed)
   Yes, it can be said that the 23-Skidoo users do swing dance over an hour on average. t-crit for one-tailed test. A one-tailed could also be defended, giving t-crit=2.015.
    
3. Regression: Independent = Age, Dependent = Amount of dancing time
    
4. Wilcoxon Paired-sample test: T-plus = 33, T-minus = 3, T-crit = 3 >
   There are significant differences among the dancing shoes worn.
    
5. Model III 2-factor ANOVA, Independent = Treatment (fixed) & Dance type (random), Dependent = Dancing time Ooh... Only partial credit. This would be a Model III 2-factor, with treatment as a fixed effect and dance type as a random effect. > 6.
    
6. Z=1.73, t-crit=1.960 >
   No significant difference between the two
    
7. (1/256) = 0.0039 or 0.39% Right - or 0.0039. I should have made this more challenging by asking for the number of groups with 3 & 1 or 2 & 2. > 8.
    
8. A nested ANOVA was done. Group F-ratio = 1.06, F-crit = 7.71, P > 0.25
   No significant differences among the two groups, only significant differences among the different nursing homes. Right on most of it. F-crit for DF of 1 & 4 is 7.71 (one-tailed!), and p > 0.25 (would have been p > 0.50 for your 2-tailed test). > 9.
    
9. 2 × 2 contingency table
   Chi-square = 0.287 (uncorrected) or 0.101 (Haber's correction)
   Log-likelihood = 0.287 (uncorrected) or 0.047 (Yate's correction)
   Chi-squared-crit = 2.841 - Does not significantly improve chances of being asked to dance.
    
10. a) A survival table analsysi would be one approach. Another would be a 2 × 2 contingency table with Drug/No drug and Died/Didn't die as the categories.
    b) A logistic regression would be most appropriate, with Use/No use as the bionomial dependent variable.