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)?

Scores

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

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

Step 3: Draw Critical Region tcrit = -2.132 tcrit = 2.132

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

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

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

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

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

Step 4: Calculate t observed tobs = (X - ) / Sx -.48 = (55.2 - 56.1) / 1.88 1.88 = 4.21/ 5

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

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

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

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.

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

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)?

Hypotheses Alternative hypothesis Null hypothesis H1: sample = 56.1

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)?

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

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 = -1.533 If H1 is < then tcrit = negative If H1 is > then tcrit = positive

Step 3: Draw Critical Region tcrit = -1.533

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

Step 4: Calculate t observed tobs = (X - ) / Sx -.48 = (55.2 - 56.1) / 1.88 1.88 = 4.21/ 5

Step 5: See if tobs falls in the critical region tcrit = -1.533

Step 5: See if tobs falls in the critical region tcrit = -1.533 tobs = -.48

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

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.