2. (33 pts total) Use the data in the SPSS file ‘ sleepdeprived-spring2017 ’ and
ID: 3248560 • Letter: 2
Question
2. (33 pts total) Use the data in the SPSS file ‘sleepdeprived-spring2017’ and refer to Pagano, Chapter 15, Problem 21. (Note: Older editions of the book have a different problem number, but the problems are the same: “A sleep researcher… Note also that the numbers in the SPSS file have been slightly changed from those in the book.) Normal sleep = group 1; sleep-deprived for 24 hours = group 2; sleep-deprived for 48 hours = group 3. Analyze the data using SPSS. Answer the following questions:
A. (2 pts) What is the omnibus null hypothesis? What is the alternative hypothesis?
B. (3 pts) Report the mean and standard deviation for each condition.
C. (2 pts) Report the full ANOVA table for this analysis (including the p value from SPSS).
D. (3 pts) What is the critical value for this F-test for a = 0.05? (Be sure to state the degrees of freedom.)
E. (2 pts) Based on this critical value, should you reject or fail to reject the omnibus null hypothesis? Why or why not? In plain English, say substantively what the result means.
F. (18 pts) In SPSS, do pairwise comparisons of each pair of means using Fisher’s LSD, Bonferroni, Tukey’s HSD, Scheffé, and Newman-Keuls, in each case with a = 0.05. For each comparison, report which pairs of means are reliably different (you may want to organize this information in a small table). Based on the pattern of number of pairwise comparisons that are statistically reliable for each kind of test, what can you say about how conservative each kind of test is? That is, which tests are more conservative, or less conservative?
G. (3 pts) By hand, calculate and interpret the effect size, omega-squared, for the omnibus F test.
See HW #1 for general instructions. For by-hand calculations show your work. For SPSS, append the relevant output.
1. (25 pts t
87.00 1.00
88.00 1.00
76.00 1.00
64.00 1.00
77.00 1.00
61.00 2.00
60.00 2.00
75.00 2.00
50.00 2.00
61.00 2.00
60.00 3.00
48.00 3.00
38.00 3.00
44.00 3.00
50.00 3.00
Explanation / Answer
A.Null hypothesis All means are equal
Alternative hypothesis At least one mean is different
b. The mean and variance are presented in the next table
C. The full ANOVA analysis
Source DF Adj SS Adj MS F-Value P-Value
group 2 2321.2 1160.60 14.47 0.001
Error 12 962.4 80.20
Total 14 3283.6
D. The critial vale is 0.001
E. p value is lower than the levele of significance (0.05), so we reject the null hypothesis, meaning that statistically we can't assume that all means are equal.
group Mean Variance 1 78.4 9.76 2 61.4 8.91 3 48 8.12Related Questions
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