Question

3. 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 12-year-old children with antisocial tendencies and you are particularly interested in the emotion of surprise. The average 12-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 α = 0.05.


In order to use the t distribution, you will first need to determine the degrees of freedom (df) for α = 0.05. The degrees of freedom (df) is . The critical value of t is .

Your sample of 12-year-old children with antisocial tendencies has an average score of 12.55 with a standard deviation of 3.28.

Calculate the t statistic. To do this, you will first have to calculate the estimated standard error. The estimated standard error is . The t statistic is . (Hint: For the most precise results, retain four significant figures from your calculation of the standard error to calculate the t statistic. Round your final answer to four decimal places, and then round it again to two decimal places for your answer selection.)

The t statistic lie in the critical region. Therefore, you reject the null hypothesis.

Based on the results of this test, there enough evidence to conclude that children with antisocial tendencies have greater difficulty recognizing surprise than do children without antisocial tendencies.

Answer 1

Null hypothesis (H0): Children with antisocial tendencies have the same difficulty recognizing the emotion of surprise as children without antisocial tendencies. Symbolically: μ = 11.80

Alternative hypothesis (H1): Children with antisocial tendencies have a harder time recognizing the emotion of surprise compared to children without antisocial tendencies. Symbolically: μ > 11.80

The critical region for α = 0.05 with the t distribution would be in the upper tail of the distribution, as this is a one-tailed test (since the alternative hypothesis is directional). The degrees of freedom (df) would need to be determined based on the sample size and specific statistical test being used, but the given information does not provide the value of df.

The estimated standard error would need to be calculated using the given sample size, sample mean, and sample standard deviation. The t statistic would then be calculated as the difference between the sample mean and the hypothesized mean (11.80), divided by the estimated standard error.

Once the t statistic is calculated, it would need to be compared to the critical value of t for the appropriate degrees of freedom and α level (0.05) to determine if it falls in the critical region, and therefore whether the null hypothesis is rejected or not. Based on the given information, the t statistic is not provided, so it is not possible to determine if the null hypothesis should be rejected or not.

Explanation: