3. Practice
You recently finished giving 5 Villanova students the MMPI paranoia measure. Determine if Villanova students’ paranoia score is significantly ( = .10) different than the average paranoia of the population ( = 56.1)?

4. Scores

5. Step 1: Write out Hypotheses
Alternative hypothesis
H1: sample = 56.1
Null hypothesis
H0: sample = 56.1

6. Step 2: Calculate the Critical t
N = 5
df =4
 = .10
tcrit = 2.132

7. Step 3: Draw Critical Region
tcrit =
tcrit = 2.132

8. Step 4: Calculate t observed
tobs = (X - ) / Sx

9. Step 4: Calculate t observed
tobs = (X - ) / Sx
Sx = S / N

10. Step 4: Calculate t observed
tobs = (X - ) / Sx
S = -1
Sx = S / N

11. Step 4: Calculate t observed
tobs = (X - ) / Sx
276
4.21 =
15306 5
5 - 1
Sx = S / N

12. Step 4: Calculate t observed
tobs = (X - ) / Sx
1.88 = 4.21/ 5

13. Step 4: Calculate t observed
tobs = (X - ) / Sx
-.48 = ( ) / 1.88
1.88 = 4.21/ 5

14. Step 5: See if tobs falls in the critical region
tcrit =
tcrit = 2.132

15. Step 5: See if tobs falls in the critical region
tcrit =
tcrit = 2.132
tobs = -.48

16. Step 6: Decision If tobs falls in the critical region:
Reject H0, and accept H1
If tobs does not fall in the critical region:
Fail to reject H0

17. Step 7: Put answer into words
We fail to reject H0
The average paranoia of Villanova students not statistically different ( = .10) than the average paranoia of the population.

18. One-tailed test
In the examples given so far we have only examined if a sample mean is different than some value
What if we want to see if the sample mean is higher or lower than some value
This is called a one-tailed test

19. Remember
You recently finished giving 5 Villanova students the MMPI paranoia measure. Determine if Villanova students’ paranoia score is significantly ( = .10) different than the average paranoia of the population ( = 56.1)?

20. Hypotheses Alternative hypothesis Null hypothesis H1: sample = 56.1

21. What if. . .
You recently finished giving 5 Villanova students the MMPI paranoia measure. Determine if Villanova students’ paranoia score is significantly ( = .10) lower than the average paranoia of the population ( = 56.1)?

22. Hypotheses Alternative hypothesis Null hypothesis
H1: sample < 56.1
Null hypothesis
H0: sample = or > 56.1

23. Step 2: Calculate the Critical t
N = 5
df =4
 = .10
Since this is a “one-tail” test use the one-tailed column
Note: one-tail = directional test
tcrit =
If H1 is < then tcrit = negative
If H1 is > then tcrit = positive

24. Step 3: Draw Critical Region
tcrit =

25. Step 4: Calculate t observed
tobs = (X - ) / Sx

26. Step 4: Calculate t observed
tobs = (X - ) / Sx
-.48 = ( ) / 1.88
1.88 = 4.21/ 5

27. Step 5: See if tobs falls in the critical region
tcrit =

28. Step 5: See if tobs falls in the critical region
tcrit =
tobs = -.48

29. Step 6: Decision If tobs falls in the critical region:
Reject H0, and accept H1
If tobs does not fall in the critical region:
Fail to reject H0

30. Step 7: Put answer into words
We fail to reject H0
The average paranoia of Villanova students is not statistically less then ( = .10) the average paranoia of the population.