PSY520 – Module 6

Chapter 16:

16.9 Given the aggression scores below for Outcome A of the sleep deprivation experiment, verify that, as suggested earlier, these mean differences shouldn’t be taken seriously by testing the null hypothesis at the .05 level of signiﬁcance. Use the computation formulas for the various sums of squares and summarize results with an ANOVA table.

16.10 Another psychologist conducts a sleep deprivation experiment. For reasons beyond his control, unequal numbers of subjects occupy the different groups. (Therefore, when calculating in SS between and SS within , you must adjust the denominator term, n , to reﬂect the unequal numbers of subjects in the group totals.)

16.12 For some experiment, imagine four possible outcomes, as described in the following ANOVA table.

16.14 The F test describes the ratio of two sources of variability: that for subjects treated differently and that for subjects treated similarly. Is there any sense in which the t test for two independent groups can be viewed likewise?

Chapter 17:

17.6 Return to the study ﬁrst described in Question 16.5 on page 336, where a psychologist tests whether shy college students initiate more eye contacts with strangers because of training sessions in assertive behavior. Use the same data, but now assume that eight subjects, coded as A, B, . . . G, H, are tested repeatedly after zero, one, two, and three training sessions. (Incidentally, since the psychologist is interested in any learning or sequential effect, it would not make sense—indeed, it’s impossible, given the sequential nature of the independent variable—to counterbalance the four sessions.) The results are expressed as the observed number of eye contacts:

17.7 Recall the experiment described in Review Question 16.11 on page 314, where errors on a driving simulator were obtained for subjects whose orange juice had been laced with controlled amounts of vodka. Now assume that repeated measures are taken across all ﬁve conditions for each of ﬁve subjects. (Assume that no lingering effects occur because sufﬁcient time elapses between successive tests, and no order bias appears because the orders of the ﬁ ve conditions are equalized across the ﬁve subjects.)

17.8 While analyzing data, an investigator treats each score as if it were contributed by a different subject even though, in fact, scores were repeated measures.

Chapter 18:

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 X 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.

18.11 In what sense does a two-factor ANOVA use observations more efﬁciently than a one-factor ANOVA does?

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, 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: