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 30 10-year-old children with antisocial tendencies and you are particularly interested in the emotion of sadness. The average 10-year-old has a score on the emotion recognition scale of 8.95. (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 sadness (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 statistic. Using the Distributions tool, locate the critical region for α = 0.05. t Distribution Degrees of Freedom = 33 -3.0-2.0-1.00.01.02.03.0x In order to use the t distribution, you will first need to determine the degrees of freedom (df) for α = 0.05. The degrees of freedom (df) is . The critical value of t is . Your sample of 10-year-old children with antisocial tendencies has an average score of 9.65 with a standard deviation of 3.47. Calculate the t statistic. To do this, you will first have to calculate the estimated standard error. The estimated standard error is . The t 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.) 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 sadness than do children without antisocial tendencies.

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 30 10-year-old children with antisocial tendencies and you are particularly interested in the emotion of sadness. The average 10-year-old has a score on the emotion recognition scale of 8.95. (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 sadness (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 statistic. Using the Distributions tool, locate the critical region for α = 0.05. t Distribution Degrees of Freedom = 33 -3.0-2.0-1.00.01.02.03.0x In order to use the t distribution, you will first need to determine the degrees of freedom (df) for α = 0.05. The degrees of freedom (df) is . The critical value of t is . Your sample of 10-year-old children with antisocial tendencies has an average score of 9.65 with a standard deviation of 3.47. Calculate the t statistic. To do this, you will first have to calculate the estimated standard error. The estimated standard error is . The t 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.) 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 sadness 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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Suppose you have a sample of 30 10-year-old children with antisocial tendencies and you are particularly interested in the emotion of sadness. The average 10-year-old has a score on the emotion recognition scale of 8.95. (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 sadness (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 statistic.

Using the Distributions tool, locate the critical region for α = 0.05.

t Distribution

Degrees of Freedom = 33

-3.0-2.0-1.00.01.02.03.0x

In order to use the t distribution, you will first need to determine the degrees of freedom (df) for α = 0.05. The degrees of freedom (df) is . The critical value of t is .

Your sample of 10-year-old children with antisocial tendencies has an average score of 9.65 with a standard deviation of 3.47.

Calculate the t statistic. To do this, you will first have to calculate the estimated standard error. The estimated standard error is . The t 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.)

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 sadness than do children without antisocial tendencies.

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.

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