Question

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 40 14-year-old children with antisocial tendencies and you are particularly interested in the emotion of surprise. The average 14-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 = 0.05.

Answer 1

Answer:

Null Hypothesis, $H_0$ : $\mu$ = 11.80

Alternate Hypothesis, $H_A$ : $\mu$ > 11.80

Step-by-step explanation:

We are given that you have a sample of 40 14-year-old children with antisocial tendencies and you are particularly interested in the emotion of surprise. The average 14-year-old has a score on the emotion recognition scale of 11.80.

Assume that scores on the emotion recognition scale are normally distributed.

Let $\mu$ = true average score on the emotion recognition scale

So, Null Hypothesis, $H_0$ : $\mu$ = 11.80

Alternate Hypothesis, $H_A$ : $\mu$ > 11.80

This a right-tailed test because 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.

We can evaluate this hypothesis using One-sample t-test statistics or One-sample z-test statistics depends on the information given in the question.

If the z-test statistic is used, then the critical region at 0.05 level of significance will be an area more than the critical value of 1.645.

And if the t-test statistic is used, then the critical region at 0.05 level of significance with 39 degrees of freedom will be an area more than the critical value of 1.685.

Answer 2

The null hypothesis should be considered as the $H_0: \mu = 11.80$

The alternative hypothesis should be considered as the $H_A: \mu > 11.80$

It is the right-tailed test.

Calculation of the null hypothesis & alternative hypothesis:

Since The average 14-year-old has a score on the emotion recognition scale of 11.80.

So here the null hypothesis should be $H_0: \mu = 11.80$

And, here it is the right-tailed test since there should be higher the score on the given scale, due to this it should strongly an emotion that represent for rightly identified. Also high scores represent the high difficults for recording the emotion.

Learn more about hypothesis here: https://brainly.com/question/18831983