5. Thet test for two independent samples Two-tailed example Bullying,\" accordin

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5. Thet test for two independent samples Two-tailed example 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 of not only the victims but also bystanders, bullies, and those who bully but who are also victims (bully-victims). You decide to measure anxiety in a group of bullies and a group of bystanders using an 18-item, 5-point anxiety scale. Assume scores on the anxiety scale are normally distributed and that the variances of the anxiety scores are the same among bullies and bystanders. The group of 39 bullies scored an average of 51.6 with a sample standard deviation of 9 on the anxiety scale. The group of 31 bystanders scored an average of 45.2 with a sample standard deviation of 12 on the same scale. You do not have any presupposed assumptions about whether bullies or bystanders will be more anxious, so you formulate the null and alternative hypotheses as: Ho: bullies-ubystanders-0 You conduct an independent-measures t test. Given your null and alternative 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 t Distribution Degrees of Freedom = 78 -4.0 -3.0 2.0 0.0 1.0 3.0 4.0 The critical t-scores that form the boundaries of the critical region for = 0.05 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) - The t statistic is The t statistic in the critical region. Therefore, the null hypothesis is You conclude that bullies have a different mean anxiety score than bystanders. Thus, it can be said that these two means are different from one another

Explanation / Answer

Given that,
mean(x)=51.6
standard deviation , s.d1=9
number(n1)=39
y(mean)=45.2
standard deviation, s.d2 =12
number(n2)=31
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =1.995
since our test is two-tailed
reject Ho, if to < -1.995 OR if to > 1.995
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (38*81 + 30*144) / (70- 2 )
s^2 = 108.794118
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=51.6-45.2/sqrt((108.794118( 1 /39+ 1/31 ))
to=6.4/2.509797
to=2.550007
| to | =2.550007
critical value
the value of |t | with (n1+n2-2) i.e 68 d.f is 1.995
we got |to| = 2.550007 & | t | = 1.995
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != 2.55 ) = 0.013
hence value of p0.05 > 0.013,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 2.550007
critical value: -1.995 , 1.995
decision: reject Ho
p-value: 0.013
we have evidence that bullies have a different mean anxiety score than bystanders.

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