exercise 5.37 introduced an experiment 5.51 Prison isolation experiment, Part II

Question

exercise 5.37 introduced an experiment 5.51 Prison isolation experiment, Part II. Exercise 5.3 conducted with the goal of identifying a treatment that reduces subjects' psychopat scores, where this score measures a person's need for control or his rebellion against Exercise 5.37 you evaluated the success of each treatment individually. An alternative analysis involves comparing the success of treatments. The relevant ANOVA output is given below. 7 introduced an experiment that was deviant T control. I Df Sum Sq Mean SqF value Pr) treatment 2 639.48 319.74 3.33 Residuals 39 3740.43 0.0461 95.91 8pooled = 9.793 on df = 39 (a) What are the hypotheses? (b) what is the conclusion of the test? Use a 5% significance level. (c) If in part (b) you determined that the test is significant, conduct pairwise tests to determine which groups are different from each other. If you did not reject the null hypothesis in part (b), recheck your answer.

Explanation / Answer

we are given above the table of ANOVA,

The hypotheses of interest in an ANOVA are as follows:

where k = the number of independent comparison groups.

the formulas for calculation for ANOVA,

table is given as,

conclusion:

p- value is p<o.o5 , so we reject the null hypothesis and conclude that the effect all treatment are differ from each other.

for pairwise difference please provide the data of all mean values so we can use the Tukey test for pairwise analysis. we reject the null hypothesis i.e there is difference between the mean. normal conclusion we can give, if pair difference is large then there is significant difference between the two variable.

thanks

df ss mss f p treatment 2 639.48 319.74 3.33 0.0461 residual 39 3740.43 95.91

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