Example You give 100 random students a questionnaire designed to measure attitudes toward living in dormitories Scores range from 1 to 7 –(1 = unfavorable;

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 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 -  ) /.194.194 = 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 Page 747 –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 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 = -1.98 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, students tend to think favorably about living in the dorms. p <.05

Example You wonder if the average IQ score of students at Villanova significantly different (at alpha =.05)than the average IQ of the population (which is 100). You sample the students in this room. 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 = (130 - 100) / 2.5 2.5 = 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 = -2.00 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 students at Villanova is statistically different (  =.05) than the average IQ of the population.

Practice You recently finished giving 5 of your friends the MMPI paranoia measure. Is your friends average average paranoia score 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 4: Calculate t observed t obs = (X -  ) / S x -.48 = (55.2 - 56.1) / 1.88 1.88 = 4.21/ 5

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

Step 7: Put answer into words We fail to reject H 0 The average paranoia of your friends is 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 of your friends the MMPI paranoia measure. Is your friends average paranoia score 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 of your friends the MMPI paranoia measure. Is your friends average paranoia score 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 = -1.533 –If H 1 is < then t crit = negative –If H 1 is > then t crit = positive

Step 3: Draw Critical Region t crit = -1.533

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

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

Step 5: See if t obs falls in the critical region t crit = -1.533

Step 5: See if t obs falls in the critical region t crit = -1.533 t obs = -.48

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

Practice You just created a “Smart Pill” and you gave it to 150 subjects. Below are the results you found. Did your “Smart Pill” significantly (  =.05) increase the average IQ scores over the average IQ of the population (  = 100)? X = 103 s = 14.4

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

Step 2: Calculate the Critical t N = 150 df = 149  =.05 t crit = 1.645

Step 3: Draw Critical Region t crit = 1.645

Step 4: Calculate t observed t obs = (X -  ) / S x 2.54 = (103 - 100) / 1.18 1.18=14.4 / 150

Step 5: See if t obs falls in the critical region t crit = 1.645 t obs = 2.54

Step 7: Put answer into words We reject H 0 and accept H 1. The average IQ of the people who took your “Smart Pill” is statistically greater (  =.05) than the average IQ of the population.

So far... We have been doing hypothesis testing with a single sample We find the mean of a sample and determine if it is statistically different than the mean of a population

Basic logic of research

Start with two equivalent groups of subjects

Treat them alike except for one thing

See if both groups are different at the end

Notice This means that we need to see if two samples are statistically different from each other We can use the same logic we learned earlier with single sample hypothesis testing

Example You just invented a “magic math pill” that will increase test scores. You give the pill to 4 subjects and another 4 subjects get no pill You then examine their final exam grades

Hypothesis Two-tailed Alternative hypothesis –H 1 :  pill =  nopill –In other words, the means of the two groups will be significantly different Null hypothesis –H 0 :  pill =  nopill –In other words, the means of the two groups will not be significantly different

Hypothesis One-tailed Alternative hypothesis –H 1 :  pill >  nopill –In other words, the pill group will score higher than the no pill group Null hypothesis –H 0 :  pill < or =  nopill –In other words, the pill group will be lower or equal to the no pill group

For current example, lets just see if there is a difference Alternative hypothesis –H 1 :  pill =  nopill –In other words, the means of the two groups will be significantly different Null hypothesis –H 0 :  pill =  nopill –In other words, the means of the two groups will not be significantly different

Results Pill Group 5 3 4 3 No Pill Group 1 2 4 3

Remember before... Step 2: Calculate the Critical t df = N -1

Now Step 2: Calculate the Critical t df = N 1 + N 2 - 2 df = 4 + 4 - 2 = 6  =.05 t critical = 2.447

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

Now Step 4: Calculate t observed t obs = (X 1 - X 2 ) / Sx 1 - x 2

Now Step 4: Calculate t observed t obs = (X 1 - X 2 ) / Sx 1 - x 2 X 1 = 3.75 X 2 = 2.50

Standard Error of a Difference Sx 1 - x 2 When the N of both samples are equal If N 1 = N 2 : Sx 1 - x 2 = S x1 2 + S x2 2

Results Pill Group 5 3 4 3 No Pill Group 1 2 4 3

Standard Deviation S =

Standard Deviation Pill Group 5 3 4 3 No Pill Group 1 2 4 3  X 1 = 15  X 1 2 = 59  X 2 = 10  X 2 2 = 30

Standard Deviation Pill Group 5 3 4 3 No Pill Group 1 2 4 3 S =.96S = 1.29  X 1 = 15  X 1 2 = 59  X 2 = 10  X 2 2 = 30

Standard Deviation Pill Group 5 3 4 3 No Pill Group 1 2 4 3 S =.96S = 1.29 S x =.48S x =. 645  X 1 = 15  X 1 2 = 59  X 2 = 10  X 2 2 = 30

Standard Error of a Difference Sx 1 - x 2 When the N of both samples are equal If N 1 = N 2 : Sx 1 - x 2 = S x1 2 + S x2 2

Standard Error of a Difference Sx 1 - x 2 When the N of both samples are equal If N 1 = N 2 : Sx 1 - x 2 = (.48) 2 + (.645) 2

Standard Error of a Difference Sx 1 - x 2 When the N of both samples are equal If N 1 = N 2 : Sx 1 - x 2 = (.48) 2 + (.645) 2 =.80

Standard Error of a Difference Raw Score Formula When the N of both samples are equal If N 1 = N 2 : Sx 1 - x 2 =

 X 1 = 15  X 1 2 = 59 N 1 = 4  X 2 = 10  X 2 2 = 30 N 2 = 4

Sx 1 - x 2 =  X 1 = 15  X 1 2 = 59 N 1 = 4  X 2 = 10  X 2 2 = 30 N 2 = 4 15 10

Sx 1 - x 2 =  X 1 = 15  X 1 2 = 59 N 1 = 4  X 2 = 10  X 2 2 = 30 N 2 = 4 15 10 5930

Sx 1 - x 2 =  X 1 = 15  X 1 2 = 59 N 1 = 4  X 2 = 10  X 2 2 = 30 N 2 = 4 15 10 5930 44 4 (4 - 1)

Sx 1 - x 2 =  X 1 = 15  X 1 2 = 59 N 1 = 4  X 2 = 10  X 2 2 = 30 N 2 = 4 15 10 5930 44 12 56.2525

 X 1 = 15  X 1 2 = 59 N 1 = 4  X 2 = 10  X 2 2 = 30 N 2 = 4 15 10 5930 44 12 56.2525 7.75.80 =

Now Step 4: Calculate t observed t obs = (X 1 - X 2 ) / Sx 1 - x 2 Sx 1 - x 2 =.80 X 1 = 3.75 X 2 = 2.50

Now Step 4: Calculate t observed t obs = (3.75 - 2.50) /.80 Sx 1 - x 2 =.80 X 1 = 3.75 X 2 = 2.50

Now Step 4: Calculate t observed 1.56 = (3.75 - 2.50) /.80 Sx 1 - x 2 =.80 X 1 = 3.75 X 2 = 2.50

Step 5: See if t obs falls in the critical region t crit = 2.447t crit = -2.447

Step 5: See if t obs falls in the critical region t crit = 2.447t crit = -2.447 t obs = 1.56

Step 7: Put answer into words We fail to reject H 0. The final exam grades of the “pill group” were not statistically different (  =.05) than the final exam grades of the “no pill” group.

Practice You wonder if psychology majors have higher IQs than sociology majors (  =.05) You give an IQ test to 4 psychology majors and 4 sociology majors

Results Psychology 110 150 140 135 Sociology 90 95 80 98

Step 1: Hypotheses Alternative hypothesis –H 1 :  psychology >  sociology Null hypothesis –H 0 :  psychology = or <  sociology

Step 2: Calculate the Critical t df = N 1 + N 2 - 2 df = 4 + 4 - 2 = 6  =.05 One-tailed t critical = 1.943

Step 3: Draw Critical Region t crit = 1.943

9.38 =  X 1 = 535  X 1 2 = 72425 N 1 = 4 X 1 = 133.75  X 2 = 363  X 2 2 = 33129 N 2 = 4 X 2 = 90.75 535 363 7242533129 44 4 (4 - 1)

Step 4: Calculate t observed 4.58 = (133.75 - 90.75) / 9.38 Sx 1 - x 2 = 9.38 X 1 = 133.75 X 2 = 90.75

Step 5: See if t obs falls in the critical region t crit = 1.943 t obs = 4.58

Step 7: Put answer into words We Reject H 0, and accept H 1 Psychology majors have significantly (  =.05) higher IQs than sociology majors.

Practice

SPSS Problem #2 7.37 7.11

Example You give 100 random students a questionnaire designed to measure attitudes toward living in dormitories Scores range from 1 to 7 –(1 = unfavorable;

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Example You give 100 random students a questionnaire designed to measure attitudes toward living in dormitories Scores range from 1 to 7 –(1 = unfavorable;

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