Question

"Bullying," according to noted expert Dan Olweus, "poisons the educational environment and affects the leaming of every child." Bullying and victimization are evident as early as preschool, with the problem peaking in middle school Suppose you are interested in the emotional well-being of not only the victims but also bystanders, bullies, and those who bully but who are also victims (bully-victims). You decide to measure depression in a group of bullies and a group of bully-victims using a 26-item, 3-point depression scale. Assume scores on the depression scae are normally distributed and that the variances of the depression scores are the same among bullies and bully-victims. The group of 30 bullies soored an average of 21.5 with a sample standard deviation of 10 on the depression scale. The group of 27 bully-victims scored an average of 25.8 with a sample standard deviation of 9 on the same scale. You do not have any presupposed assumptions about whether bullies or bully-victims will be more depressed, so you formulate the null and altemative hypotheses as: Ho: ?bulles-ubully-victims 0 H1 : ?bulles-ubully-victims * 0 You conduct an independent-measures t test. Given your null and altemative hypotheses, this is a test. To use the Distributions tool to find the critical region, you first need to set the degrees of freedom. The degrees of freedom is ' Degrees of Freedom 68 -4.0 3.0 2.0 1.0 0.0 1.0 2.0 3.0 4.0 The critical t-scores that form the boundaries of the critical region for ? 0.01 are ± In order to calculate the t statistic, you first need to calculate the standard error under the assumption that the null hypothesis is true. In order to calculate the standard error, you first need to calculate the pooled variance. The pooled variance is s- The standard error is s(M M2) - 91.0182 0.3556 The t statisic is 89.3901

Answer 1

group 1 group 2 25.800 9.0000 27.000 21.500 X2 S1 10.0000 S 30.000 pooled SD S2,-( (n1-1)s21t(m-1)*s*2)/htn2-2)- 91.02 Point estimate -4.300 degree of freedom v (n,+n2-2) 55.000 std error of difference-Se Sp(1/n^+1/n2) 2.5308 test stat t = (x1-X2-Ap)/se 1.70 p value:- 0.0950

this is a 2 independent sample t test.

degree of freedom =n1+n2-2= 30+27-2=55

Sp² =91.0182 ; std error is s(m1 -m2)=2.5308

t statsitic is -1.70 ;

the t statistic des not lie in the critical region............null hypothesis is not rejected....

you can not conclude ...........not significantly different.......