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 –H 1 :  sample = 56.1 Null hypothesis –H 0 :  sample = 56.1

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

Step 3: Draw Critical Region t crit = 2.132t crit =

Step 4: Calculate t observed t obs = (X -  ) / S x

Step 4: Calculate t observed t obs = (X -  ) / S x S x = S / N

Step 4: Calculate t observed t obs = (X -  ) / S x S x = S / N S =

Step 4: Calculate t observed t obs = (X -  ) / S x S x = S / N =

Step 4: Calculate t observed t obs = (X -  ) / S x 1.88 = 4.21/ 5

Step 4: Calculate t observed t obs = (X -  ) / S x -.48 = ( ) / = 4.21/ 5

Step 5: See if t obs falls in the critical region t crit = 2.132t crit =

Step 5: See if t obs falls in the critical region t crit = 2.132t crit = t obs = -.48

Step 6: Decision If t obs falls in the critical region: –Reject H 0, and accept H 1 If t obs does not fall in the critical region: –Fail to reject H 0

Step 7: Put answer into words We fail to reject H 0 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 –H 1 :  sample = 56.1 Null hypothesis –H 0 :  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 –H 1 :  sample < 56.1 Null hypothesis –H 0 :  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 t crit = –If H 1 is < then t crit = negative –If H 1 is > then t crit = positive

Step 3: Draw Critical Region t crit =

Step 4: Calculate t observed t obs = (X -  ) / S x

Step 4: Calculate t observed t obs = (X -  ) / S x -.48 = ( ) / = 4.21/ 5

Step 5: See if t obs falls in the critical region t crit =

Step 5: See if t obs falls in the critical region t crit = t obs = -.48

Step 6: Decision If t obs falls in the critical region: –Reject H 0, and accept H 1 If t obs does not fall in the critical region: –Fail to reject H 0

Step 7: Put answer into words We fail to reject H 0 The average paranoia of Villanova students is not statistically less then (  =.10) the average paranoia of the population.