7. A one-tailed hypothesis test with the t statistic Aa Aa Antisocial personalit

Question

7. A one-tailed hypothesis test with the t statistic Aa Aa 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 35 children with antisocial tendencies and you are particularly interested in the emotion of disgust. You find that while the average 12-year-old has a score of 13.10 on the emotion recognition scale, your sample of 12-year-old children with antisocial tendencies has an average score of 14.35 with a standard deviation of 4.63. A higher score represents that it took a higher degree of emotional expression for the child to correctly identify the emotion of disgust, indicating greater difficulty recognizing the emotion. Assume that scores on the emotion recognition scale are normally distributed. The null hypothesis is that your sample of children with antisocial tendencies would have no more difficulty recognizing emotion than the general population of 12-year-olds. Stated using symbols: Ho:Hantisocial2 generap pula on H1: Han social generalpopulation . This is a one- tailed test. Given what you know, you will evaluate this hypothesis using a t statistic

Explanation / Answer

All answers in detail. Please mind the correction in the direrction of null hypothesis and df calculation changes to your attempt:

Change the direction of your hypothesis, the question has the following as null hypothesis : 'no more diffculty ....general population of 12-year olds'

So, Ho: Muantisocial>=Mu general population
Ha:Muantisocial<Mu general population
  
df = n-1 = 35-1 = 34

The critical value one sided value of t is 1.691
the standard error formula is sigma/sqrt(n) = (4.63/sqrt(35)) = .7826
t = (14.35-13.10)/(4.63/sqrt(35)) = 1.5972

The t statistic does not lie in the critical region . Therefore you should not reject the null hypothesis

Hence, we conclude that the null hypothesis can' tbe rejected for lack of evidence.

Related Questions

Navigate

• Browse (All)
• Browse 7
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.