2. Extra Brownie Points! Lottery To Win: choose the 5 winnings numbers
from 1 to 49
AND
Choose the "Powerball" number
from 1 to 42
What is the probability you will win?
Use combinations to answer this question

3. p of winning jackpot
Total number of ways to win / total number of possible outcomes

4. Total Number of Outcomes
Different 5 number combinations
Different Powerball outcomes
Thus, there are 1,906,884 * 42 = 80,089,128 ways the drawing can occur

5. Total Number of Ways to Win
Only one way to win –

8. 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

9. Hypothesis Alternative hypothesis H1: sample = 4
In other words, the population mean will be different than 4

10. Hypothesis Alternative hypothesis Null hypothesis H1: sample = 4
In other words, the population mean will not be different than 4

11. Results
N = 100
X = 4.51
s = 1.94
Notice, your sample mean is consistent with H1, but you must determine if this difference is simply due to chance

12. 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

13. Observed t-value
tobs = (X - ) / Sx

14. Observed t-value tobs = (X - ) / Sx
This will test if the null hypothesis H0:  sample = 4 is true
The bigger the tobs the more likely that H1:  sample = 4 is true

15. Observed t-value
tobs = (X - ) / Sx
Sx = S / N

16. Observed t-value
tobs = (X - ) / .194
.194 = 1.94/ 100

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

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

19. t distribution

20. t distribution
tobs = 2.63

21. t distribution
tobs = 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.

22. t critical
To find out how big the tobs must be to be significantly different than 0 you find a tcrit value.
Calculate df = N - 1
Page 747
First Column are df
Look at an alpha of .05 with two-tails

23. t distribution
tobs = 2.63

24. t distribution
tcrit = -1.98
tcrit = 1.98
tobs = 2.63

25. t distribution
tcrit = -1.98
tcrit = 1.98
tobs = 2.63

26. t distribution Reject H0:  sample = 4 tcrit = -1.98 tcrit = 1.98
tobs = 2.63
If tobs fall in critical area reject the null hypothesis
Reject H0:  sample = 4

27. t distribution Do not reject H0:  sample = 4 tcrit = -1.98
tobs = 2.63
If tobs does not fall in critical area do not reject the null hypothesis
Do not reject H0:  sample = 4

28. Decision
Since tobs falls in the critical region we reject Ho and accept H1
It is statistically significant, students ratings of the dorms is different than 4.
p < .05

29. 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

30. The Steps
Try to always follow these steps!

31. Step 1: Write out Hypotheses
Alternative hypothesis
H1: sample = 100
Null hypothesis
H0: sample = 100

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

33. Step 3: Draw Critical Region
tcrit = -2.00
tcrit = 2.00

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

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

36. Step 4: Calculate t observed
tobs = (X - ) / Sx
2.5 = 18.4 / 54

37. Step 4: Calculate t observed
tobs = (X - ) / Sx
12 = ( ) / 2.5
2.5 = 18.4 / 54

38. Step 5: See if tobs falls in the critical region
tcrit = -2.00
tcrit = 2.00

39. Step 5: See if tobs falls in the critical region
tcrit = -2.00
tcrit = 2.00
tobs = 12

40. 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

41. Step 7: Put answer into words
We reject H0 and accept H1.
The average IQ of students at Villanova is statistically different ( = .05) than the average IQ of the population.

42. 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)?

43. Scores

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

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

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

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

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

49. 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

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

51. SPSS

55. 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

56. 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)?

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

58. 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)?

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

60. 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

61. Step 3: Draw Critical Region
tcrit =

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

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

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

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

66. 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

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

68. 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

69. Step 1: Write out Hypotheses
Alternative hypothesis
H1: sample > 100
Null hypothesis
H0: sample < or = 100

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

71. Step 3: Draw Critical Region
tcrit = 1.645

72. Step 4: Calculate t observed
tobs = (X - ) / Sx
2.54 = ( ) / 1.18
1.18=14.4 / 150

73. Step 5: See if tobs falls in the critical region
tcrit = 1.645
tobs = 2.54

74. 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

75. Step 7: Put answer into words
We reject H0 and accept H1.
The average IQ of the people who took your “Smart Pill” is statistically greater ( = .05) than the average IQ of the population.

78. 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

79. Basic logic of research

80. Start with two equivalent groups of subjects

81. Treat them alike except for one thing

82. See if both groups are different at the end

83. 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

84. 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

85. Hypothesis Two-tailed
Alternative hypothesis
H1: pill = nopill
In other words, the means of the two groups will be significantly different
Null hypothesis
H0: pill = nopill
In other words, the means of the two groups will not be significantly different

86. Hypothesis One-tailed
Alternative hypothesis
H1: pill > nopill
In other words, the pill group will score higher than the no pill group
Null hypothesis
H0: pill < or = nopill
In other words, the pill group will be lower or equal to the no pill group

87. For current example, lets just see if there is a difference
Alternative hypothesis
H1: pill = nopill
In other words, the means of the two groups will be significantly different
Null hypothesis
H0: pill = nopill
In other words, the means of the two groups will not be significantly different

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

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

90. Now Step 2: Calculate the Critical t
df = N1 + N2 - 2
df = = 6
 = .05
t critical = 2.447

91. Step 3: Draw Critical Region
tcrit =
tcrit = 2.447

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

93. Now Step 4: Calculate t observed
tobs = (X1 - X2) / Sx1 - x2

94. Now Step 4: Calculate t observed
tobs = (X1 - X2) / Sx1 - x2

95. Now Step 4: Calculate t observed
tobs = (X1 - X2) / Sx1 - x2
X1 = 3.75
X2 = 2.50

96. Now Step 4: Calculate t observed
tobs = (X1 - X2) / Sx1 - x2

97. Standard Error of a Difference
Sx1 - x2
When the N of both samples are equal
If N1 = N2:
Sx1 - x2 = Sx12 + Sx22

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

99. Standard Deviation
S = -1

100. Standard Deviation Pill Group 5 3 4 No Pill Group 1 2 4 3 X2= 10

101. Standard Deviation Pill Group 5 3 4 No Pill Group 1 2 4 3 X2= 10

102. Standard Deviation Pill Group 5 3 4 No Pill Group 1 2 4 3 X2= 10
Sx= .48
Sx= . 645

103. Standard Error of a Difference
Sx1 - x2
When the N of both samples are equal
If N1 = N2:
Sx1 - x2 = Sx12 + Sx22

104. Standard Error of a Difference
Sx1 - x2
When the N of both samples are equal
If N1 = N2:
Sx1 - x2 = (.48)2 + (.645)2

105. Standard Error of a Difference
Sx1 - x2
When the N of both samples are equal
If N1 = N2:
Sx1 - x2 = (.48)2 + (.645)2
= .80

106. Standard Error of a Difference Raw Score Formula
When the N of both samples are equal
If N1 = N2:
Sx1 - x2 =

107. X1= 15
X12= 59
N1 = 4
X2= 10
X22= 30
N2 = 4
Sx1 - x2 =

108. X1= 15
X12= 59
N1 = 4
X2= 10
X22= 30
N2 = 4
Sx1 - x2 = 10 15

109. Sx1 - x2 = X1= 15 X12= 59 N1 = 4 X2= 10 X22= 30 N2 = 4 10 15

110. Sx1 - x2 = X1= 15 X12= 59 N1 = 4 X2= 10 X22= 30 N2 = 4 10 15
4 (4 - 1)

111. Sx1 - x2 = X1= 15 X12= 59 N1 = 4 X2= 10 X22= 30 N2 = 4 10 15
56.25 30 25 4 4 12

112. .80 = X1= 15 X12= 59 N1 = 4 X2= 10 X22= 30 N2 = 4 10 15 59
56.25 30 25
7.75 4 4 12

113. Now Step 4: Calculate t observed
tobs = (X1 - X2) / Sx1 - x2
Sx1 - x2 = .80
X1 = 3.75
X2 = 2.50

114. Now Step 4: Calculate t observed
Sx1 - x2 = .80
X1 = 3.75
X2 = 2.50

115. Now Step 4: Calculate t observed
1.56 = ( ) / .80
Sx1 - x2 = .80
X1 = 3.75
X2 = 2.50

116. Step 5: See if tobs falls in the critical region
tcrit =
tcrit = 2.447

117. Step 5: See if tobs falls in the critical region
tcrit =
tcrit = 2.447
tobs = 1.56

118. 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

119. Step 7: Put answer into words
We fail to reject H0.
The final exam grades of the “pill group” were not statistically different ( = .05) than the final exam grades of the “no pill” group.

120. SPSS

122. What if
The two samples have different sample sizes (n)

123. Results
Psychology
110
150
140
135
Sociology 90 95 80 98

124. Results
Psychology
110
150
140
135
Sociology 90 95 80

125. If samples have unequal n
All the steps are the same!
Only difference is in calculating the Standard Error of a Difference

126. Standard Error of a Difference
When the N of both samples is equal
If N1 = N2:
Sx1 - x2 =

127. Standard Error of a Difference
When the N of both samples is not equal
If N1 = N2:
N1 + N2 - 2

128. Results Psychology 110 150 140 135 Sociology 90 95 80 X1= 535
N1 = 4
X2= 265
X22= 23525
N2 = 3

129. X1= 535
X12= 72425
N1 = 4
X2= 265
X22= 23525
N2 = 3
N1 + N2 - 2

130. X1= 535
X12= 72425
N1 = 4
X2= 265
X22= 23525
N2 = 3
265
535
N1 + N2 - 2

131. X1= 535
X12= 72425
N1 = 4
X2= 265
X22= 23525
N2 = 3
265
535
23525
72425
N1 + N2 - 2

132. X1= 535
X12= 72425
N1 = 4
X2= 265
X22= 23525
N2 = 3
265
535
23525
72425 3 4 4 3

133. X1= 535
X12= 72425
N1 = 4
X2= 265
X22= 23525
N2 = 3
265
535
23525
72425 3 4 4 3 5

134. X1= 535
X12= 72425
N1 = 4
X2= 265
X22= 23525
N2 = 3
265
535
23525
72425
(.58) 3 4 4 3 5

135. = 10.69 X1= 535 X12= 72425 N1 = 4 X2= 265 X22= 23525 N2 = 3
72425
114.31 3 4 4 3 5
= 10.69