Take your time answer the questions, this is all one question, please answer correctly, thank you very much, will write an wonderful review when solutions are done. 8. A one-tailed hypothesis test with the t statisticAntisocial personality disorder (ASPD) is characterized by deceitfulness, reckless disregard for the well-being of others, a diminished capacity for remorse,superficial charm, thrill seeking, and poor behavioral control. ASPD is not normally diagnosed in children or adolescents, but antisocial tendencies cansometimes be recognized in childhood or early adolescence. James Blair and his colleagues have studied the ability of children with antisocial tendencies torecognize facial expressions that depict sadness, happiness, anger, disgust, fear, and surprise. They have found that children with antisocial tendencieshave selective impairments, with significantly more difficulty recognizing fearful and sad expressions.Suppose you have a sample of 45 14-year-old children with antisocial tendencies and you are particularly interested in the emotion of disgust. Theaverage 14-year-old has a score on the emotion recognition scale of 13.85. (The higher the score on this scale, the more strongly an emotion has to bedisplayed to be correctly identified. Therefore, higher scores indicate greater difficulty recognizing the emotion). Assume that scores on the emotionrecognition scale are normally distributed.You believe that children with antisocial tendencies will have a harder time recognizing the emotion of disgust (in other words, they will have higher scoreson the emotion recognition test).What is your null hypothesis stated using symbols?What is your alternative hypothesis stated using symbols?This is atailed test. Given what you know, you will evaluate this hypothesis using aUsing the Distributions tool, locate the critical region for a = 0.05.statistic. t DistributionDegrees of Freedom = 330 0 0ДАМ.0050-3.0The t statistic-2.733-2.0-1.0.5000.9900.5000lie in the critical region. Therefore, you0.00.0001.02.0In order to use the t distribution, you will first need to determine the degrees of freedom (df) for a = 0.05. The degrees of freedom (df)isThe critical value of t is(.0050Your sample of 14-year-old children with antisocial tendencies has an average score of 14.60 with a standard deviation of 2.96.3.02.733 tThe tCalculate the t statistic. To do this, you will first have to calculate the estimated standard error. The estimated standard error isstatistic is(Hint: For the most precise results, retain four significant figures from your calculation of the standard error to calculatethe t statistic. Round your final answer to four decimal places, and then round it again to two decimal places for your answer selection.)reject the null hypothesis.Based on the results of this test, there▼ enough evidence to conclude that children with antisocial tendencies have greater difficultyrecognizing disgust than do children without antisocial tendencies.

Given information: The average score on the emotion recognition scale is 13.85.

In order to use the t distribution, you will first need to determine the degrees of freedom (df) for a = 0.05. The degrees of freedom (df) . The critical value of tis is Your sample of 14-year-old children with antisocial tendencies has an average score of 14.60 with a standard deviation of 2.96. ▼. The t Calculate the t statistic. To do this, you will first have to calculate the estimated standard error. The estimated standard error is (Hint: For the most precise results, retain four significant figures from your calculation of the standard error to calculate the t statistic. Round your final answer to four decimal places, and then round it again to two decimal places for your answer selection.) statistic is The t statistic lie in the critical region. Therefore, you reject the null hypothesis. Based on the results of this test, there enough evidence to conclude that children with antisocial tendencies have greater difficulty recognizing disgust than do children without antisocial tendencies.

In order to use the t distribution, you will first need to determine the degrees of freedom (df) for a = 0.05. The degrees of freedom (df) . The critical value of tis is Your sample of 14-year-old children with antisocial tendencies has an average score of 14.60 with a standard deviation of 2.96. ▼. The t Calculate the t statistic. To do this, you will first have to calculate the estimated standard error. The estimated standard error is (Hint: For the most precise results, retain four significant figures from your calculation of the standard error to calculate the t statistic. Round your final answer to four decimal places, and then round it again to two decimal places for your answer selection.) statistic is The t statistic lie in the critical region. Therefore, you reject the null hypothesis. Based on the results of this test, there enough evidence to conclude that children with antisocial tendencies have greater difficulty recognizing disgust than do children without antisocial tendencies.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

Take your time answer the questions, this is all one question, please answer correctly, thank you very much, will write an wonderful review when solutions are done.

8. A one-tailed hypothesis test with the t statistic
Antisocial personality disorder (ASPD) is characterized by deceitfulness, reckless disregard for the well-being of others, a diminished capacity for remorse,
superficial charm, thrill seeking, and poor behavioral control. ASPD is not normally diagnosed in children or adolescents, but antisocial tendencies can
sometimes be recognized in childhood or early adolescence. James Blair and his colleagues have studied the ability of children with antisocial tendencies to
recognize facial expressions that depict sadness, happiness, anger, disgust, fear, and surprise. They have found that children with antisocial tendencies
have selective impairments, with significantly more difficulty recognizing fearful and sad expressions.
Suppose you have a sample of 45 14-year-old children with antisocial tendencies and you are particularly interested in the emotion of disgust. The
average 14-year-old has a score on the emotion recognition scale of 13.85. (The higher the score on this scale, the more strongly an emotion has to be
displayed to be correctly identified. Therefore, higher scores indicate greater difficulty recognizing the emotion). Assume that scores on the emotion
recognition scale are normally distributed.
You believe that children with antisocial tendencies will have a harder time recognizing the emotion of disgust (in other words, they will have higher scores
on the emotion recognition test).
What is your null hypothesis stated using symbols?
What is your alternative hypothesis stated using symbols?
This is a
tailed test. Given what you know, you will evaluate this hypothesis using a
Using the Distributions tool, locate the critical region for a = 0.05.
statistic.

t Distribution
Degrees of Freedom = 33
0 0 0
ДАМ
.0050
-3.0
The t statistic
-2.733
-2.0
-1.0
.5000
.9900
.5000
lie in the critical region. Therefore, you
0.0
0.000
1.0
2.0
In order to use the t distribution, you will first need to determine the degrees of freedom (df) for a = 0.05. The degrees of freedom (df)
is
The critical value of t is
(.0050
Your sample of 14-year-old children with antisocial tendencies has an average score of 14.60 with a standard deviation of 2.96.
3.0
2.733 t
The t
Calculate the t statistic. To do this, you will first have to calculate the estimated standard error. The estimated standard error is
statistic is
(Hint: For the most precise results, retain four significant figures from your calculation of the standard error to calculate
the t statistic. Round your final answer to four decimal places, and then round it again to two decimal places for your answer selection.)
reject the null hypothesis.
Based on the results of this test, there
▼ enough evidence to conclude that children with antisocial tendencies have greater difficulty
recognizing disgust than do children without antisocial tendencies.

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