A family psychologist developed an elaborate training program to reduce the stre

Question

A family psychologist developed 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 man randomly selected from all those in a 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. 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 this distribution. 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). A researcher predicts that listening to music while solving math problems will make a particular brain area more active. To test this, a research participant has

Explanation / Answer

a)

This is a one tailed test since we are only estimating if the stress levels have gone down. The steps of hypothesis testing are as follows:-

1) State the null and alternate hypothesis.

H0: 85 (Stress level is greater than or equal to 85)

H1: < 85 (Stress level is less than 85)

2) Specify the desired significance level=0.05 and n=1

3) Determine the appropriate technique (Z test)

4) Determine the critical/cut off values (1.64)

5) Compute the test statistic

ZSTAT=X-µ//sqrt(n)

        =60-85/15/sqrt(1)

        =-25/15

       =-1.67

Z cut off/critical value is -1.64 and since our answer -1.67 is less than -1.64 (i.e. falls in the rejection region), we would reject the null hypothesis.

This proves that stress levels have reduced less than 85

H0: 85 (Stress level is greater than or equal to 85)

H1: < 85 (Stress level is less than 85)

ZSTAT=X-µ//sqrt(n)

        =60-85/15/sqrt(1)

        =-25/15

       =-1.67

Z cut off/critical value is -1.64 and since our answer -1.67 is less than -1.64 (i.e. falls in the rejection region), we would reject the null hypothesis.

This proves that stress levels have reduced less than 85

H0: 85 (Stress level is greater than or equal to 85)

H1: < 85 (Stress level is less than 85)

ZSTAT=X-µ//sqrt(n)

        =60-85/15/sqrt(1)

        =-25/15

        =-1.67

b)

Z cut off/critical value is -1.64 and since our answer -1.67 is less than -1.64 (i.e. falls in the rejection region), we would reject the null hypothesis.

This proves that stress levels have reduced less than 85

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