Question 1054414: A family psychologist develops an elaborate training program to reduce the stress of childless men who marry women with adolescent children. It is known from previous research that such men, one month after moving in with their new wife and her children, have a stress level of 85 with a standard deviation of 15, and the stress levels are normally distributed. The training program is tried on one an randomly selected from all those in particular city who during the preceding month have married a woman with an adolescent child. After the training program, this man's stress level is 60. (a) Using the .05 level, what should the researcher conclude? Solve this problem explicitly using all five steps of hypothesis testing and illustrate your answer with a sketch showing the comparison distribution, the cutoff (or cutoffs), and the score of the sample on the distribution. (b) Then explain your answer to someone who has never had a course in statistics (but who is familiar with mean, standard deviation, and Z-scores).
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Ho: There is no change in stress
Ha: There is a change in stress
alpha=0.05 P{reject Ho|Ho true}
two-sided
critical value is |z|>1.96
test statistic z=(x=mean)/sd=(60-85)/15=-25/15=-1.6
This does not meet the critical value.
The z-score for significance has to be greater than a certain value or less than a certain value. This z-score is not, so the difference is not significance.
Draw a normal distribution with the sd s marked to 3.
Put a vertical line at +1.96 and -1.96
The man's score is at z=-1.64.
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