11. A one-tailed hypothesis test with the t statistic Aaa Aa Antisocial personal
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11. A one-tailed hypothesis test with the t statistic Aaa Aa Antisocial personality disorder (ASPD) is characterized by deceitfulness, reckless disregard for the well-being of others, a diminished capacity for remorse, superficial charm, th 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 40 10-year-old children with antisocial tendencies and you are particularly interested in the emotion of surprise. The average 10-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 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 surprise (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 05Explanation / Answer
let mu1 is mean value of the score for children with anti social tendencies
let mu2 is mean value of the score for children with non anti social tendencies
Null hypothesis : mu1 = mu2
Alernate hypothesis mu1 > mu2
This is 1 tail test as we interested in the directional test of HIGHER
we shall evaluate this using a t test
df is given as
n1 +n2 -2
= 20 +20 - 2 = 38
lets look at the t table to check the value of critcal t
for df = 38 and alpha = 0.05 , for 1 tail test is
1.685 , please keep the t tables handy
the standard error is given as
SD/sqrt(n)
3.28 / sqrt(20)= 0.7334
Please note that we can answer only 4 subparts of a question at a time
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