Question

Help please!

3. Notation for comparing two population means Aa Aa "Bullying," according to noted expert Dan Olweus, "poisons the educational environment and affects the learning 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 not only of the victims but also of 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 bystanders using an 18-item, 5-point depression scale. Assume scores on the depression scale are normally distributed The group of 23 bullies scored an average of 40.1 with a sample standard deviation of 10.1 on the depression scale. The group of 28 bystanders scored an average of 46.8 with a sample standard deviation of 11.6 on the same scale. Suppose you intend to conduct a hypothesis test on the difference in population means. In preparation, you call the sample of bullies sample 1 and the sample of bystanders sample 2. Organize the provided data by completing the following table: Sample 1 Sample2

Answer 1

	Sample 1	Samplen 2
n=	23	28
mean=	40.10	46.80
s=	10.1000	11.6000
s^2/n	4.4352	4.8057

Ho:there is no significant difference in the means of two groups

H1: there is significant difference in the means of two groups

T =
= (Xbar1-Xbar2)/sqrt(s1^2/n1 + s2^2/n2)
= (40.1-46.8)/sqrt(4.4352+4.8057)
-2.2040

df=
(s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1))
(4.4352+4.8057)^2/(4.4352^2/(23-1) + 4.8057^2/(28-1))
80

alpha= 0.05

t(a/2, df) =t(0.05/2,80) =t.INV.2T(0.05,80) = 1.9901

P-value
2*(1-P(T<|t|)
2*(1-P(T<abs(-2.204))
T.DIST.2T(abs(-2.204),80)
0.030398524

With (t=-2.2040, p<5%), I reject the null hypothesis at 5% level of significance and conclude that there is significant difference in the means of two groups