I'm looking for help in approaching my question 11. 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 forremorse, superficial charm, thrill seeking, and poor behavioral control. ASPD is not normally diagnosed in children or adolescents, but antisocialtendencies can sometimes be recognized in childhood or early adolescence. James Blair and his colleagues have studied the ability of children withantisocial tendencies to recognize facial expressions that depict sadness, happiness, anger, disgust, fear, and surprise. They have found that childrenwith antisocial tendencies have selective impairments, with significantly more difficulty recognizing fearful and sad expressions.Suppose you have a sample of 40 16-year-old children with antisocial tendencies and you are particularly interested in the emotion of disgust. Theaverage 16-year-old has a score on the emotion recognition scale of 11.80. (The higher the score on this scale, the more strongly an emotion has tobe displayed 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 higherscores on the emotion recognition test).What is your null hypothesis stated using symbols? H,: Pantisocial S Hgeneral population ▼What is your alternative hypothesis stated using symbols? H,: Mantisocial Hgeneral populationThis is a one-tailed test. Given what you know, you will evaluate this hypothesis using a tUsing the Distributions tool, locate the critical region for a = 0.05.t Distributionstatistic. t DistributionDegrees of Freedom = 33OOO^^^-3.0-2.0-1.00.01.02.03.0xIn 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 39 ▼The critical value of t is 1.685Your sample of 16-year-old children with antisocial tendencies has an average score of 12.55 with a standard deviation of 3.28.The t statistic does not lie in the critical region. Therefore, you cannot reject the null hypothesis.Calculate the t statistic. To do this, you will first have to calculate the estimated standard error. The estimated standard error is 0.52. The tstatistic is 1.45 ▼ (Hint: For the most precise results, retain four significant figures from your calculation of the standard error to calculate the tstatistic.)Based on the results of this test, there is not enough evidence to conclude that children with antisocial tendencies have greater difficultyrecognizing disgust than do children without antisocial tendencies.

Question

I'm looking for help in approaching my question 11. 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 forremorse, superficial charm, thrill seeking, and poor behavioral control. ASPD is not normally diagnosed in children or adolescents, but antisocialtendencies can sometimes be recognized in childhood or early adolescence. James Blair and his colleagues have studied the ability of children withantisocial tendencies to recognize facial expressions that depict sadness, happiness, anger, disgust, fear, and surprise. They have found that childrenwith antisocial tendencies have selective impairments, with significantly more difficulty recognizing fearful and sad expressions.Suppose you have a sample of 40 16-year-old children with antisocial tendencies and you are particularly interested in the emotion of disgust. Theaverage 16-year-old has a score on the emotion recognition scale of 11.80. (The higher the score on this scale, the more strongly an emotion has tobe displayed 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 higherscores on the emotion recognition test).What is your null hypothesis stated using symbols? H,: Pantisocial S Hgeneral population ▼What is your alternative hypothesis stated using symbols? H,: Mantisocial > Hgeneral populationThis is a one-tailed test. Given what you know, you will evaluate this hypothesis using a tUsing the Distributions tool, locate the critical region for a = 0.05.t Distributionstatistic. t DistributionDegrees of Freedom = 33OOO^^^-3.0-2.0-1.00.01.02.03.0xIn 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 39 ▼The critical value of t is 1.685Your sample of 16-year-old children with antisocial tendencies has an average score of 12.55 with a standard deviation of 3.28.The t statistic does not lie in the critical region. Therefore, you cannot reject the null hypothesis.Calculate the t statistic. To do this, you will first have to calculate the estimated standard error. The estimated standard error is 0.52. The tstatistic is 1.45 ▼ (Hint: For the most precise results, retain four significant figures from your calculation of the standard error to calculate the tstatistic.)Based on the results of this test, there is not enough evidence to conclude that children with antisocial tendencies have greater difficultyrecognizing disgust than do children without antisocial tendencies.

Accepted Answer

To test:Null hypothesis H0: There is no difference in the average score of average children and…