Question

11. A one-tailed hypothesis test with the t statistic a 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 20 10-year-old children with antisocial tendencies and you are particularly interested in the emotion of fear. The average 10-year-old has a score on the emotion recognition scale of 12.95. (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 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 = .05. t Distribution Degrees of Freedom = 33 • OO AA -10 00 10 20 30

Answer 1

The null hypothesis using symbols is stated as follows: H: 1512.95 The alternate hypothesis using symbols is stated as follows: Ho:p>12.95 This is a one-tailed test. Given what you know, you will evaluate this hypothesis using a t statistic. Since it is given that 20 10-year children were considered, therefore, the degrees of freedom is computed as follows: df =n-1 =20-1 =19 Thus, the degrees of freedom are 19). The level of significance is given to be 0.05. The right-tailed critical value for t - distribution with 0.05 level of significance with 19 degrees of freedom can be computed in MS Excel as follows: fr =T.INV(1-0.05,19) 1.729133

Thus, the critical value is 1.7291). Now, for a sample of size 20, it is given that the antisocial tendencies have an average score of 14.5 with a standard deviation of 2.95. The estimate of the standard error is computed as follows: SESTI vn 2.95 V20 =0.6596 Thus, the test statistic can be computed as follows: t-x-M SE 14.5-12.95 0.6596 = 2.3498 Therefore, the test statistic is 2.3498. = The test statistic does not lie in the critical region. Therefore, you cannot reject the null hypothesis. Based on the result of this test, there is not enough evidence to conclude that children with antisocial tendencies have greater difficulty recognizing fear than do children without antisocial tendencies.