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 victims and a group of bystanders using a 26-item, 3-point anxiety scale. Assume scores on the anxiety scale are normally distributed and that the variances of the anxiety scores are the same among victims and bystanders. The group of 30 victims scored an average of 21.5 with a sample standard deviation of 10 on the anxiety scale. The group of 27 bystanders 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 victims or bystanders will be more anxious, so you formulate the null and alternative hypotheses as: H00: μvictimsvictims – μbystandersbystanders = 0 H11: μvictimsvictims – μbystandersbystanders ≠ 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 = 68 -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 = 0.01 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 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 victims have a different mean anxiety score than bystanders. Thus, it can be said that these two means are different from one another.

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 victims and a group of bystanders using a 26-item, 3-point anxiety scale. Assume scores on the anxiety scale are normally distributed and that the variances of the anxiety scores are the same among victims and bystanders. The group of 30 victims scored an average of 21.5 with a sample standard deviation of 10 on the anxiety scale. The group of 27 bystanders 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 victims or bystanders will be more anxious, so you formulate the null and alternative hypotheses as: H00: μvictimsvictims – μbystandersbystanders = 0 H11: μvictimsvictims – μbystandersbystanders ≠ 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 = 68 -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 = 0.01 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 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 victims have a different mean anxiety score than bystanders. Thus, it can be said that these two means are different from one another.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Concept explainers

Question

The group of 30 victims scored an average of 21.5 with a sample standard deviation of 10 on the anxiety scale. The group of 27 bystanders 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 victims or bystanders will be more anxious, so you formulate the null and alternative hypotheses as:

H00: μvictimsvictims – μbystandersbystanders	=	0
H11: μvictimsvictims – μbystandersbystanders	≠	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 = 68

-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 = 0.01 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 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 victims have a different mean anxiety score than bystanders. 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

Given Information:

Sample size of victims (n₁) = 30

Average score of victims (x₁) = 21.5

Standard deviation of victims (s₁) = 10

Sample size of bystanders (n₂) = 27

Average score of bystanders (x₂) = 25.8

Standard Deviation of bystanders (s₂) = 9

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Continuous Probability Distribution$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.