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t - test

Dr. Anshul Singh Thapa

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t - test

  • The t ratio was discovered by an English statistician, Willam S Gosset in 1908 under the pen name ‘student’. Therefore, it is also known as “Student’s t”.
  • The t –test is used to find out the difference between the two means. The difference is calculated by comparing the two means.

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t - test

  • t – test can be used in following three situations:
    • The t – test for single sample
    • The t – test for dependent means
    • The t – test for independent means

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The t – test for single sample

Sample

Population

Mean

Variance

SD

Mean

Variance

SD

Comparison

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The t – test for single sample

  • The situation when we know the population mean but not its variance, and where you have a single sample of scores. It turns out that in most research we do not even know the population’s mean, plus in most research situation we usually have not one set, but two sets of scores.
  • Here we want to compare the mean of a sample to a population for which you know the mean but the variance is unknown.

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Steps to be followed:

  • Sample Mean (M) = (ΣX)/N
  • Degree of Freedom (df) = N – 1
  • Population Mean (μ)
  • Variance of Population (S2 )= SS/df
  • Variance of Distribution of Means (S2M )= S2/N
  • Standard Deviation of Distribution of Means (SM )= √S2M
  • t = (M – μ)/SM

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Test Null Hypothesis

Rating

(X)

5

3

6

2

7

6

7

4

2

5

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Test Null Hypothesis

Rating

(X)

5

3

6

2

7

6

7

4

2

5

47

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Test Null Hypothesis

Rating

(X)

5

3

6

2

7

6

7

4

2

5

47

Mean = Σ X/N

= 47/10

= 4.7

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Test Null Hypothesis

Rating

(X)

Difference from mean

(X - M) M = 4.7

5

.30

3

-1.70

6

1.30

2

-2.70

7

2.30

6

1.30

7

-2.30

4

-.70

2

-2.70

5

.30

47

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Test Null Hypothesis

Rating

(X)

Difference from mean

(X - M) M = 4.7

Squared Difference from Mean

(X - M)2

5

.30

.09

3

-1.70

2.89

6

1.30

1.69

2

-2.70

7.29

7

2.30

5.29

6

1.30

1.69

7

-2.30

5.29

4

-.70

.49

2

-2.70

7.29

5

.30

.09

47

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Test Null Hypothesis

Rating

(X)

Difference from mean

(X - M) M = 4.7

Squared Difference from Mean

(X - M)2

5

.30

.09

3

-1.70

2.89

6

1.30

1.69

2

-2.70

7.29

7

2.30

5.29

6

1.30

1.69

7

-2.30

5.29

4

-.70

.49

2

-2.70

7.29

5

.30

.09

47

32.10

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Calculation of ‘t’

  • Sample Mean (M) = (ΣX)/N = 47/10 = 4.70
  • Degree of Freedom (df) = N – 1 = 10 – 1 = 9
  • Population Mean (μ) = 4.00
  • S2 = SS/df = 32.10/ (10 – 1) = 32.10/9 = 3.57
  • S2M = S2/N = 3.57/10 = 0.36
  • SM = √S2M = √.36 = .60
  • t = (M – μ)/SM = (4.70 – 4.00)/.60 = .70/.60 = 1.17
  • Calculated t value = 1.17
  • Tabulated t value = t(df,9) 0.01, two tailed = 3.250
  • Decision: Do not reject the Null Hypothesis

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The t – test for dependent means

Sample

Population

Mean

Variance

SD

Mean

Variance

SD

Comparison

Mean

Variance

SD

Mean

Variance

SD

Pre - test

Post - test

Difference Score

Difference Score

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�The t – test for dependent means (Paired t test)�

  • In this situation we have two scores from each person in our sample. This kind of research situation is called a repeated-measures design (also known as a within subjects design). For example pre test and the post test scores.
  • The hypothesis testing procedure for the situation in which each person is measured twice is a t test for dependent means.
  • It has the name “dependent means” because the mean for each group of scores are dependent on each other in that they are both from the same sample.
  • We do the t test for the dependent means with the following concepts
    • we use something called difference scores, and
    • we assume that the population mean (of the difference scores) is 0.

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Difference Scores

  • With the repeated measure design, the sample includes two scores for each person instead of just one. The way we handle this is to make the two scores per person into one score per person. This can be done by creating difference scores. For each person we subtract one score from other. If the difference is before verses after, difference scores are called change scores. When the two scores are a before score and an after score, we usually take the after score minus the before score to indicate the change.

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Population of Difference Scores with a mean of 0

  • The null hypothesis in the repeated measure design is that there is no difference between the two groups on the average scores.
  • Saying that there is no difference on the average scores is the same as saying that the means of the population of the difference scores is 0.

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Dependent t test

X (Pre test)

Y (Post test)

15

32

19

30

16

25

11

37

10

26

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Dependent t test

X (Pre test)

Y (Post test)

Difference Score

15

32

17

19

30

11

16

25

9

11

37

26

10

26

16

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Dependent t test

X (Pre test)

Y (Post test)

Difference Score

15

32

17

19

30

11

16

25

9

11

37

26

10

26

16

79

Mean = 79/5 = 15.8

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Dependent t test

X (Pre test)

Y (Post test)

Difference Score

Deviation (d)

15

32

17

1.2

19

30

11

-4.8

16

25

9

-6.8

11

37

26

10.2

10

26

16

0.2

79

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Dependent t test

X (Pre test)

Y (Post test)

Difference Score

Deviation (d)

(d2)

15

32

17

1.2

1.44

19

30

11

-4.8

23.04

16

25

9

-6.8

46.24

11

37

26

10.2

104.04

10

26

16

0.2

0.04

79

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Dependent t test

X (Pre test)

Y (Post test)

Difference Score

Deviation (d)

(d2)

15

32

17

1.2

1.44

19

30

11

-4.8

23.04

16

25

9

-6.8

46.24

11

37

26

10.2

104.04

10

26

16

0.2

0.04

79

Σd2 = 174.8

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Dependent t test

X (Pre test)

Y (Post test)

Difference Score

Deviation (d)

(d2)

15

32

17

1.2

1.44

19

30

11

-4.8

23.04

16

25

9

-6.8

46.24

11

37

26

10.2

104.04

10

26

16

0.2

0.04

79

Σd2 = 174.8

  • Sample Mean (M) = 79/5 = 15.08
  • Population mean (u) = 0 (assumed as no changed)
  • Variance of the population (S2) = 174.8/ (5 – 1) = 43.7
  • Variance of the distribution of means (S2M) = 43.7/5 = 8.74
  • Standard Error of Mean (SM) = √8.74 = 2.96
  • t = (M - u)/ SM = (15.08 - 0)/2.96 = 5.33
  • Calculated t value = 5.33
  • Tabulated t value = df = 4, level of significance 0.05 and two tailed = 2.776
  • Decision = Reject the Null hypothesis

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Dependent t test

Students

Before

After

A

126

115

B

133

125

C

126

96

D

115

115

E

108

119

F

109

82

G

124

93

H

98

109

I

95

72

J

120

104

K

118

107

L

126

118

M

121

102

N

116

115

O

94

83

P

105

87

Q

123

121

R

125

100

S

128

118

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Test Null Hypothesis t = 5.583

Students

Before

After

1

1487.8

1487.2

2

1329.4

1328.1

3

1407.9

1405.9

4

1236.1

1234.0

5

1299.8

1298.2

6

1447.2

1444.7

7

1354.1

1354.3

8

1204.6

1203.7

9

1322.3

1320.8

10

1388.5

1386.8

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Test Null Hypothesis = 4.88

Sr. No.

Pre test

Post test

1

42

40

2

50

62

3

51

61

4

26

35

5

35

30

6

42

52

7

60

68

8

41

51

9

70

84

10

38

50

11

62

72

12

55

63

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Independent Sample t - test

  • In previous section we have learned how to use the t test for dependent means to compare two groups of scores from a single group of people i.e., the same individual is measured before and after training programme.
  • Now we will see how to compare two groups of scores, one from each of two entirely separate groups of people.
  • The scores of the two groups are independent of each other, in such situation we used the t test for independent means.

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Independent sample t - test

Sample 1

Sample 2

16

15

19

14

12

17

10

19

9

11

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Independent sample t - test

Sample 1

Sample 2

16

15

19

14

12

17

10

19

9

11

M1 = 13.2

M2 = 15.2

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Independent sample t - test

Sample 1

Deviation

(M - X)

Sample 2

Deviation

(M - X)

16

2.8

15

-0.2

19

5.8

14

-1.2

12

-1.2

17

1.8

10

-3.2

19

3.8

9

-4.2

11

-4.2

M1 = 13.2

M2 = 15.2

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Independent sample t - test

Sample 1

Deviation

(M - X)

d12

Sample 2

Deviation

(M - X)

d22

16

2.8

7.84

15

-0.2

0.04

19

5.8

33.64

14

-1.2

1.44

12

-1.2

1.44

17

1.8

3.24

10

-3.2

10.24

19

3.8

14.44

9

-4.2

17.64

11

-4.2

17.64

M1 = 13.2

0

M2 = 15.2

0

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Independent sample t - test

Sample 1

Deviation

(M - X)

d12

Sample 2

Deviation

(M - X)

d22

16

2.8

7.84

15

-0.2

0.04

19

5.8

33.64

14

-1.2

1.44

12

-1.2

1.44

17

1.8

3.24

10

-3.2

10.24

19

3.8

14.44

9

-4.2

17.64

11

-4.2

17.64

M1 = 13.2

0

Σd12 = 70.8

M2 = 15.2

0

Σd22 = 36.8

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Independent sample t - test

Sample 1

Deviation

(M - X)

d12

Sample 2

Deviation

(M - X)

d22

16

2.8

7.84

15

-0.2

0.04

19

5.8

33.64

14

-1.2

1.44

12

-1.2

1.44

17

1.8

3.24

10

-3.2

10.24

19

3.8

14.44

9

-4.2

17.64

11

-4.2

17.64

M1 = 13.2

0

Σd12 = 70.8

M2 = 15.2

0

Σd22 = 36.8

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Independent sample t - test

Sample 1

Deviation

(M - X)

d12

Sample 2

Deviation

(M - X)

d22

16

2.8

7.84

15

-0.2

0.04

19

5.8

33.64

14

-1.2

1.44

12

-1.2

1.44

17

1.8

3.24

10

-3.2

10.24

19

3.8

14.44

9

-4.2

17.64

11

-4.2

17.64

M1 = 13.2

0

Σd12 = 70.8

M2 = 15.2

0

Σd22 = 36.8

t = 0.862

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When the raw scores are not given:

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Table value

Level of Significance = 0.05

Test = Two tailed test

Degree of Freedom = (N1 - 1) + (N2 - 1)

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Test The H0 = 4.46

Non Sports Person

Sports Person

6

4

7

3

8

2

10

1

15

5

16

9

10

10

9

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Test Null Hypothesis

Sample 1

Sample 2

77

87

88

77

77

71

90

70

68

63

74

50

62

58

93

63

82

76

79

65

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Test Null Hypothesis = 4.88

Sr. No.

Pre test

Post test

1

42

40

2

50

62

3

51

61

4

26

35

5

35

30

6

42

52

7

60

68

8

41

51

9

70

84

10

38

50

11

62

72

12

55

63

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How to calculate tabulated t value?

  • We need the following information to calculate the tabulated t – value:
    • Level of significance
    • Degree of freedom
    • Test used – one tailed or two tailed
    • t – table

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Level of significance

  • The two most common level of significance are 0.01 and 0.05
  • In simple terms, the level of significance and the level of confidence is a degree to which we accept or reject or predict a happening or incidence.
  • The amount 99% and 95% confidence is also termed as 0.01 and 0.05 level of significance. The 0.01 level means, if we repeatedly draw a sample or conduct an experiment 100 times, only one occasion, the obtained sample mean or results will fallout side the limit of population of mean ± 2.58 SEM (standard deviation of population)
  • Similarly 0.05 level means, if we repeatedly draw a sample or conduct an experiment 100 times, only on 5 occasion, the obtained sample mean will fall out side the limit of population of mean ± 1.96 SEM (standard deviation of population)

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Degree of freedom

  • It is the concept which is used in all statistical tests. The degree of freedom indicates the number of scores in a distribution that are free to vary.
  • For calculating degree of freedom in t test the formula is (N – 1)

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Two Tailed Test

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One Tailed Test

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