topic 6.docx

B false topic 6 exercises when the researcher notes

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(b) False?

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TOPIC 6 EXERCISES When the researcher notes the null hypothesis is false, both estimates would reflect random error. The estimate for between groups would also reflect the treatment effect. Chapter 18 question 18.8 For the two-factor experiment described in the previous question, assume that, as shown, mean bar press rates of either 4 or 8 are identified with three of the four cells in the 2 3 2 table of outcomes. Furthermore, just for the sake of this question, ignore sampling variability and assume that effects occur whenever any numerical differences correspond to either food deprivation, reward amount, or the interaction. Indicate whether or not effects occur for each of these three components if the empty cell in the 2 3 2 tables are occupied by a mean of: (a) 12 Effects for food deprivation and reward amount (b) 8 Effects for food deprivation, reward amount, and interaction (c) 4 Effects for interaction only Chapter 18 question 18.11 In what sense does a two-factor ANOVA use observations more efficiently than a one-factor ANOVA does? It is noted the two-factor ANOVA is able to measure the effects of two factors simultaneously. With this, the ability to contain more sources of variability compared to a one-factor ANOVA, thereby making it more efficient. Chapeter 18 question 18.12 A psychologist employs a two-factor experiment to study the combined effect of sleep deprivation and alcohol consumption on the performance of automobile drivers. Before the driving test, the subjects go without sleep for various time periods and then drink a glass of orange juice laced with controlled amounts of vodka. Their performance is measured by the number of errors made on a driving simulator. Two subjects are randomly assigned to each cell,
TOPIC 6 EXERCISES that is, each possible combination of sleep deprivation (either 0, 24, 48, or 72 hours) and alcohol consumption (either 0, 1, 2, or 3 ounces), yielding the following results: (a) Summarize the results with an ANOVA table. SS column = ∑ T 2 column rn G 2 n [ ( 25 ) 2 8 + ( 30 ) 2 8 + ( 64 ) 2 8 + ( 67 ) 2 8 ¿ ( 186 ) 2 32 [ 625 8 + 900 8 + 4096 8 + 4489 8 ¿ 34596 32 [78.13 + 112.5 + 512 + 561.13] – 1081.13 df column = 3 SS row = T 2 row cn G 2 n [ 841 8 + 1296 8 + 2809 8 + 4624 8 ¿ 34596 32 [105.13 + 162 + 351.13 + 578] – 1081.13 1196.26 – 1081.13 = 115.14 SS between ∑x 2 - ∑ n T cell 2 1466 – 1399 = 67 Df within = n – ( c ) (r ) 32 – (4) (4) = 32 – 16 =16 SS interaction = SS between – (SS column + SS row ) 317.87 – [182.63 + 115.14] 317.87 – 297.77 = 20.10 Df interaction = 9

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TOPIC 6 EXERCISES Source SS Df Ms F ratio Column 182.63 3 60.88 14.53 Row 115.14 3 38.38 9.16 Interaction 20.10 9 2.23 0.53 Within 67 16 4.19 Total 384.87 31 We will reject the two null hypotheses because they exceed the critical value of 3.24. The third one will be retained.
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• Fall '14
• dargon
• Null hypothesis, Cohen, researcher, Statistical significance, Statistical power

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