The group of 23 bullies scored an average of 40.1 with a sample standard deviation of 10 on the anxiety scale. The group of 28 victims scored an average of 46.8 with a sample standard deviation of 11 on the same scale. You do not have any presupposed assumptions about whether bullies or victims will be more anxious, so you formulate the null and alternative hypotheses as: H00: μbulliesbullies – μvictimsvictims = 0 H11: μbulliesbullies – μvictimsvictims ≠ 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 rejection region, you first need to set the degrees of freedom. The degrees of freedom is . t Distribution Degrees of Freedom = 58 -4.0-3.0-2.0-1.00.01.02.03.04.0t The critical t-scores that form the boundaries of the rejection region for α = 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 s2pp2 = . The standard error is s(M1 – M2)(M1 – M2) = . The t statistic is . The t statistic in the rejection region. Therefore, the null hypothesis is . You conclude that bullies have a different mean anxiety score than victims. Thus, it can be said that these two means are different from one another.

MATLAB: An Introduction with Applications

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ISBN:9781119256830

Author:Amos Gilat

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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 victims 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 victims.

The group of 23 bullies scored an average of 40.1 with a sample standard deviation of 10 on the anxiety scale. The group of 28 victims scored an average of 46.8 with a sample standard deviation of 11 on the same scale. You do not have any presupposed assumptions about whether bullies or victims will be more anxious, so you formulate the null and alternative hypotheses as:

H00: μbulliesbullies – μvictimsvictims	=	0
H11: μbulliesbullies – μvictimsvictims	≠	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 rejection region, you first need to set the degrees of freedom. The degrees of freedom is .

t Distribution

Degrees of Freedom = 58

-4.0-3.0-2.0-1.00.01.02.03.04.0t

The critical t-scores that form the boundaries of the rejection region for α = 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 s2pp2 = . The standard error is s(M1 – M2)(M1 – M2) = .

The t statistic is .

The t statistic in the rejection region. Therefore, the null hypothesis is . You conclude that bullies have a different mean anxiety score than victims. Thus, it can be said that these two means are different from one another.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution

Step 1

The two given populations are the anxiety scores for bullies and victims.

It is also said that they follow normal distribution and have equal variances.

n1=23, sample mean for bullies = 40.1 and SD for bullies = 10

n2= 28 , sample mean for victims =46.8 and SD for victims = 11

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