Question

A one-tailed hypothesis test with the t 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 40 16-year-old children with antisocial tendencies and you are particularly interested in the emotion of disgust. The average 16-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.
The null hypothesis is that your sample of children with antisocial tendencies would have no more difficulty recognizing emotion than the general population of 16-year-olds. Stated using symbols:_________. This is a __________ tailed test. Given what you know, you will evaluate this hypothesis using a________ statistic.

Answer 1

Step-by-step explanation

The answers in bold is what was required from the question.

the null hypothesis using symbol:-

H₀ : μ = 11.80

the alternative hypothesis using symbol:-

H₁:  μ ≠ 11.80

This is a one tailed test. Given what you know, you will evaluate this hypothesis using a t test statistic.

Alpha level = 0.05

Sample size = n

Degree of freedom = n-1 = 40-1 = 39

To use the t distribution we will have to find the critical value from the t table this value is 1.685

But the question does not have sample mean and sample standard deviation. So I was unable to solve for the standard error