Is this quarter fair?

How could you determine this? You assume that flipping the coin a large number of times would result in heads half the time (i.e., it has a.50 probability)

Is this quarter fair? Say you flip it 100 times 52 times it is a head Not exactly 50, but its close –probably due to random error

Is this quarter fair? What if you got 65 heads? 70? 95? At what point is the discrepancy from the expected becoming too great to attribute to chance?

Example You give 100 random students a questionnaire designed to measure attitudes toward living in dormitories Scores range from 1 to 7 –(1 = unfavorable; 4 = neutral; 7 = favorable) You wonder if the mean score of the population is different then the population mean at Haverford (which is 4)

Hypothesis Alternative hypothesis –H 1 :  sample = 4 –In other words, the population mean will be different than 4

Hypothesis Alternative hypothesis –H 1 :  sample = 4 Null hypothesis –H 0 :  sample = 4 –In other words, the population mean will not be different than 4

Results N = 100 X = 4.51 s = 1.94 Notice, your sample mean is consistent with H 1, but you must determine if this difference is simply due to chance

Results N = 100 X = 4.51 s = 1.94 To determine if this difference is due to chance you must calculate an observed t value

Observed t-value t obs = (X -  ) / S x

Observed t-value t obs = (X -  ) / S x This will test if the null hypothesis H 0 :  sample = 4 is true The bigger the t obs the more likely that H 1 :  sample = 4 is true

Observed t-value t obs = (X -  ) / S x S x = S / N

Observed t-value t obs = (X -  ) / = 1.94/ 100

Observed t-value t obs = (4.51 – 4.0) /.194

Observed t-value 2.63 = (4.51 – 4.0) /.194

t distribution

t obs = 2.63

t distribution t obs = 2.63 Next, must determine if this t value happened due to chance or if represent a real difference in means. Usually, we want to be 95% certain.

t critical To find out how big the t obs must be to be significantly different than 0 you find a t crit value. Calculate df = N - 1 Table D –First Column are df –Look at an alpha of.05 with two-tails

t distribution t obs = 2.63

t distribution t obs = 2.63 t crit = 1.98 t crit = -1.98

t distribution t obs = 2.63 t crit = 1.98 t crit = -1.98

t distribution t obs = 2.63 t crit = 1.98 t crit = If t obs fall in critical area reject the null hypothesis Reject H 0 :  sample = 4

t distribution t obs = 2.63 t crit = 1.98 t crit = If t obs does not fall in critical area do not reject the null hypothesis Do not reject H 0 :  sample = 4

Decision Since t obs falls in the critical region we reject H o and accept H 1 It is statistically significant, the average favorability of Villanova dorms is significantly different than the favorability of Haverford dorms. p <.05

We usually test for significance at the “.05 level” This means that the results we got in the previous example would only happen 5 times out of 100 if the true population mean was really 4

Hypothesis Testing Basic Logic 1) Want to test a hypothesis (called the research or alternative hypothesis). –“Second born children are smarter than everyone else (Mean IQ of everyone else = 100”) 2) Set up the null hypothesis that your sample was drawn from the general population –“Your sample of second born children come from a population with a mean of 100”

Hypothesis Testing Basic Logic 3) Collect a random sample –You collect a sample of second born children and find their mean IQ is 145 4) Calculate the probability of your sample mean occurring given the null hypothesis –What is the probability of getting a sample mean of 145 if they were from a population mean of 100

Hypothesis Testing Basic Logic 5) On the basis of that probability you make a decision to either reject of fail to reject the null hypothesis. –If it is very unlikely (p <.05) to get a mean of 145 if the population mean was 100 you would reject the null –Second born children are SIGNIFICANTLY smarter than the general population

Example You wonder if the average IQ score of Villanova students is significantly different (at alpha =.05)than the average IQ of the population (which is 100). To determine this you collect a sample of 54 students. N = 54 X = 130 s = 18.4

The Steps Try to always follow these steps!

Step 1: Write out Hypotheses Alternative hypothesis –H 1 :  sample = 100 Null hypothesis –H 0 :  sample = 100

Step 2: Calculate the Critical t N = 54 df = 53  =.05 t crit = 2.0

Step 3: Draw Critical Region t crit = 2.00t crit = -2.00

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 2.5 = 18.4 / 54

Step 4: Calculate t observed t obs = (X -  ) / S x 12 = ( ) / = 18.4 / 54

Step 5: See if t obs falls in the critical region t crit = 2.00t crit = -2.00

Step 5: See if t obs falls in the critical region t crit = 2.00t crit = t obs = 12

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 reject H 0 and accept H 1. The average IQ of Villanova students statistically different (  =.05) than the average IQ of the population.

Practice You wonder if the average agreeableness score of Villanova students is significantly different (at alpha =.05) than the average agreeableness of the population (which is 3.8). You collect data from 31 people. N = 31 X = 3.92 s = 1.52

Step 1: Write out Hypotheses Alternative hypothesis –H 1 :  sample = 3.8 Null hypothesis –H 0 :  sample = 3.8

Step 2: Calculate the Critical t N = 31 df = 30  =.05 t crit = 2.042

Step 3: Draw Critical Region t crit = 2.042t 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.27 = 1.52 / 31

Step 4: Calculate t observed t obs = (X -  ) / S x.44 = ( ) /.27

Step 5: See if t obs falls in the critical region t crit = 2.042t crit =

Step 5: See if t obs falls in the critical region t crit = 2.042t crit = t obs =.44

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 agreeableness score of Villanova students is not statistically different (  =.05) than the average agreeableness score of the population.