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6. The t 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 depression in a group of victims and a group of bully-victims using an 18-item, 5-point depression scale. Assume scores on the depression scale are normally distributed and that the variances of the depression scores are the same among victims and bully-victims. The group of 39 victims scored an average of 51.6 with a sample standard deviation of 9 on the depression scale. The group of 31 bully-victims 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 victims or bully-victims will be more depressed, so you formulate the null and alternative hypotheses as: H0: μvictims - μbully-victims = 0 H1: μvictims - μbully-victims ≠ 0 You conduct an independent-measures t test. Given your null and alternative hypotheses, this is a two-tailed 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 68. t Distribution Degrees of Freedom = 68 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 t The critical t-scores that form the boundaries of the rejection region for α = 0.05 are ± 1.994. 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 s2p = 102.25. The standard error is s(M1 - M2) = 2.428. The t statistic is (51.6 - 45.2) / 2.428 = 2.636. The t statistic is in the rejection region. Therefore, the null hypothesis is rejected. You conclude that victims have a different mean depression score than bully-victims. Thus, it can be said that these two means are significantly different from one another.

Name: 6 the t 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 31515
Uploaded: 2022-11-19T20:09:04-08:00
Duration: 6 min 5 s
Channel: Jon Southam
Description: 6 the t 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 31515

6. The t 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 depression in a group of victims and a group of bully-victims using an 18-item, 5-point depression scale. Assume scores on the depression scale are normally distributed and that the variances of the depression scores are the same among victims and bully-victims.

The group of 39 victims scored an average of 51.6 with a sample standard deviation of 9 on the depression scale. The group of 31 bully-victims 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 victims or bully-victims will be more depressed, so you formulate the null and alternative hypotheses as:

H0: μvictims - μbully-victims = 0
H1: μvictims - μbully-victims ≠ 0

You conduct an independent-measures t test. Given your null and alternative hypotheses, this is a two-tailed 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 68.

t Distribution
Degrees of Freedom = 68
-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 t

The critical t-scores that form the boundaries of the rejection region for α = 0.05 are ± 1.994.

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 s2p = 102.25. The standard error is s(M1 - M2) = 2.428.

The t statistic is (51.6 - 45.2) / 2.428 = 2.636.

The t statistic is in the rejection region.

Therefore, the null hypothesis is rejected. You conclude that victims have a different mean depression score than bully-victims. Thus, it can be said that these two means are significantly different from one another.

Added by Cynthia T.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Jon Southam

11/19/2022

Step 1

The first step is to formulate the null and alternative hypotheses. In this case, the null hypothesis (H0) is that the mean depression score of victims and bully-victims is the same, while the alternative hypothesis (H1) is that the mean depression scores of Show more…

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6. The t 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 depression in a group of victims and a group of bully-victims using an 18-item, 5-point depression scale. Assume scores on the depression scale are normally distributed and that the variances of the depression scores are the same among victims and bully-victims. The group of 39 victims scored an average of 51.6 with a sample standard deviation of 9 on the depression scale. The group of 31 bully-victims 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 victims or bully-victims will be more depressed, so you formulate the null and alternative hypotheses as: H0: μvictims - μbully-victims = 0 H1: μvictims - μbully-victims ≠ 0 You conduct an independent-measures t test. Given your null and alternative hypotheses, this is a two-tailed 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 68. t Distribution Degrees of Freedom = 68 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 t The critical t-scores that form the boundaries of the rejection region for α = 0.05 are ± 1.994. 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 s2p = 102.25. The standard error is s(M1 - M2) = 2.428. The t statistic is (51.6 - 45.2) / 2.428 = 2.636. The t statistic is in the rejection region. Therefore, the null hypothesis is rejected. You conclude that victims have a different mean depression score than bully-victims. Thus, it can be said that these two means are significantly different from one another.

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00:01 All right, so we are measuring the amount of depression in groups of victims and group of bully victims.

00:08 So our null hypothesis is that the mean of the victim group, the mean depression level based upon the test that they took, minus that of the bully victims group is equal to zero.

00:24 And the alternative is that the difference is not equal to zero.

00:30 This is just saying we don't know which is high.

00:34 Higher or lower.

00:35 We have no pre -existing knowledge about this.

00:38 We just want to see, is there a difference between these two groups of people? and so we're going to use a two independent samples t -test.

00:50 And so for that, we're going to get the, so here's our test statistic.

00:57 It's going to be x bar 1.

00:59 We'll say v to zone.

01:03 It's the mean score of the victim group.

01:06 And then we have x bar bv, the bully victims group, divided by this term here.

01:11 This is the standard error.

01:14 And we're going to do the pooled variance.

01:16 We're instructed to use the pooled variance times one over the sample size of the victim group, plus one over the sample size of the bully victim group.

01:24 So the pooled variance.

01:26 So we need to get the standard error.

01:28 This whole thing here.

01:28 This is called the standard error.

01:31 And you'll see it abbreviated as se.

01:33 So we need the pooled variance.

01:36 Let's go and get that.

01:37 Well, that's equal to the sample size of the victim.

01:41 Group minus 1 multiplied by the variance of that group plus the sample size of the bully victim group minus 1 times the variance of that group all divided by the sum of these sample sizes minus 1 or excuse me minus 2 all right and that's our pooled variance and we toss all the values in here and we're given we're given the sample sizes of 39 and 31 for the victims and bully victim's so we test that in here.

02:20 So we get 39 minus 1 times 9 squared plus 31 minus 1 times 12 squared all over 39 plus 31 minus 2.

02:39 And it gives us a pooled variance of this, 108 .79.

02:44 All right.

02:45 So that's going to go into our formula here...

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