Questions: 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 10-year-old children with antisocial tendencies and you are particularly interested in the emotion of sadness. 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 sadness (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.
Transcript text: 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 10-year-old children with antisocial tendencies and you are particularly interested in the emotion of sadness. 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 sadness (in other words, they will have higher scores on the emotion recognition test).
What is your null hypothesis stated using symbols? $\qquad$
What is your alternative hypothesis stated using symbols? $\qquad$
This is a $\qquad$ tailed test. Given what you know, you will evaluate this hypothesis using a $\qquad$ statistic.
Solution
Solution Steps
Step 1: Define Hypotheses
We define the null and alternative hypotheses as follows:
Null Hypothesis (\(H_0\)): \( \mu = 11.80 \)
Alternative Hypothesis (\(H_1\)): \( \mu > 11.80 \)
This indicates that we are testing whether children with antisocial tendencies have a higher mean score on the emotion recognition scale compared to the average score of 11.80.
Step 2: Calculate Standard Error
The standard error (\(SE\)) is calculated using the formula:
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{2.0}{\sqrt{40}} \approx 0.3162
\]
Step 3: Calculate Test Statistic
The test statistic (\(Z_{test}\)) is calculated using the formula:
\[
Z_{test} = \frac{\bar{x} - \mu_0}{SE} = \frac{12.5 - 11.80}{0.3162} \approx 2.2136
\]
Step 4: Calculate P-value
For a right-tailed test, the p-value is calculated as:
\[
P = 1 - T(z) \approx 0.0134
\]
Step 5: Conclusion
Given that the p-value \(0.0134\) is less than the significance level \(\alpha = 0.05\), we reject the null hypothesis. This suggests that children with antisocial tendencies have a significantly higher mean score on the emotion recognition scale compared to the average score of 11.80.
Final Answer
\(\boxed{H_0 \text{ is rejected, indicating higher difficulty in recognizing sadness.}}\)