Question

11. A one-tailed hypothesis test with the 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 25 16 year old children with antisocial tendencies and you are particularly interested in the emotion of fear. The average 16 year old has a score on the emotion recognition scale of 12.50. (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 fear (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 a = 0.05. Distribution

Answer 1

Let of given that us bethe population mecin Ho = 12.50, no 25. Nun hypothesis но: ш - шо Alternative bpothesis - Hi: urlo a This is to eight tailed test we upe t-statistic zoo texting Hoxersex H. at level of significance a = o-os. I Test statistic t it given by t = ř - no atent 7 freedom t' i The deguees 7 - n-1 = 25-1 14 = 24 The croitical rollue azt is ttable trable 1.711 © 2ue uso given theet. Y = sample mean = 13.10 2-85.- stan dered deviation. Scanned with CamScanner

Estimated standard error - . 2.85 125 2.85 5 - 20- 57 . B tcau = T-Mo 6lvo = 13.10 - 12-5 0-57 0-6 0-57 tcail = 1.0526 e Themezore tradettatet e Here tcaul <ttable i.e. not lie in the critical Region. Therefore w€ can not ceject the nul hypothesis. Burped on the eesults 2 this test there are 95% evidence to conclude that children with antisocial trend: encies have spectes difticwty cecognizing 1 Zears that do Childrean without antisocal tendencies. CS Scannea CamScanner C .