Answer:
Null Hypothesis, :
= 11.80
Alternate Hypothesis, :
> 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 = <u><em>true average score on the emotion recognition scale</em></u>
So, Null Hypothesis, :
= 11.80
Alternate Hypothesis, :
> 11.80
<u>This a right-tailed test</u> 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.