Mathematics, 02.08.2021 22:00 debk
One of the advantages of a repeated-measures design is that it removes the individual differences from the error variance and increases the likelihood of rejecting the null hypothesis. The following data were obtained from a research study comparing three treatment conditions.
Treatment 1 Treatment 2 Treatment 3
6 8 10 G=48
5 5 5
1 2 3
0 1 2
T=12S S=26 T=16S S=30 T=20S S=38
a. Assume that the data are from an independent-measures study (like in Chapter 12 normal ANOVA) using three separate samples, each with n=4 participants, and use an independent-measures ANOVA with a=.05 to test for significant differences among the three treatments.
b. Assume that the data are from a repeated-measures study using one sample of n=4 participants and use a repeated-measures ANOVA with a=.05 to test for significant differences among the three treatments.
c. Compare your results from part A and part B.
Answers: 2
Mathematics, 21.06.2019 13:30, Angeldelissa
Which expression is factored form? x^2+14x+48 a) (x-8)(x-6) b) (x+6)(x-8) c) (x+8)(x-6) d) (x+6)(x+8)
Answers: 1
Mathematics, 21.06.2019 15:30, gui00g7888888888888
Match each equation with the operation you can use to solve for the variable. subtract 10. divide by 10. divide by 5. subtract 18. multiply by 10. add 18. add 10. multiply by 5. 5 = 10p arrowright p + 10 = 18 arrowright p + 18 = 5 arrowright 5p = 10 arrowright
Answers: 3
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