# Independent Sample Tests lecture at online course on statistics

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```					Testing Statistical Hypothesis
Independent Sample t-Test
Heibatollah Baghi, and

1
Research Design

Design         Manipulation    Random      Control
of        assignment   group
Intervention
Experimental                               

Quasi-                                     ?
experimental

Non-                                       
experimental
2
Steps in Test of Hypothesis

1. Determine the appropriate test
2. Establish the level of significance:α
3. Determine whether to use a one tail or
two tail test
4. Calculate the test statistic
5. Determine the degree of freedom
6. Compare computed test statistic
against a tabled value
3
1. Determine the Appropriate Test
 If comparing a sample to a population, use
one sample tests.
(n1- 1)S12  (in2order2 to 1390.14 1248
 If comparing two samples n  1)S2  draw
5. S2 pooled 
inferences about group  n2 - 2
n1 differences in the 18
population use two sample t-test. 1     1
the x1x2  S pooled based on
 Here 6. Stest statistic is ( n 1  n 2 ) a theoretical
2

sampling distribution known as sampling
Sx1x2  the
distribution of 5.41 difference between two
means.
Mdiff = X 1  X 2  95.5  105.0  1.85
7. tc 
Sx1x2       5.41
   The standard deviation of such a sampling
8. df  n1  n2 - 2  as
distribution is referred to18 the standard error
of the difference.
9. t  2.10
0.05
4
10. Statistical Decision : Fail to Reject H 0
1. Determine the Appropriate Test
 Assumptions and Requirements for the two
sample test (comparing groups means) are:
   Independent variable consists of two levels of a
nominal-level variable (when there are two and
only two groups).
   Dependent variable approximates interval-scale
characteristics or higher.
   Normal distribution or large enough sample size
to assume normality due to the central limit
theorem.
   Equal variance: assumption of the
homogeneity of variance
1 2 = 1 2

5
1. Determine the Appropriate Test
 If the two groups are independent of
each other uses independent group t-
test.
 If the two groups are not independent
of each other use dependent group t-
test also known as paired t-test.

This lecture focuses on
independent sample t-test
which is a parametric test
6
2. Establish Level of Significance

• α is a predetermined value
• The convention
• α = .05
• α = .01
• α = .001

7
3. Determine Whether to Use a One
or Two Tailed Test

• If testing for equality of means then
two tailed test
• If testing whether one mean
greater/smaller than the other then one
tailed test

8
Test (n2  1)S  1390.14
4. Calculating (n1-1)S Statistics  1248.03 
5. S   2
pooled
1
2
2
2

n1  n2 - 2                        18
1     1
•              6. Sx1x2  S
For the independentpooled( n1  n 2 ) t-test
groups                         the
2

formula is: Sx1x2  5.41
7. tc  X 1  X 2  95.5  105.0  1.85
Sx1x2        5.41
8. df  n1  n2 - 2  18
9. t0.05  2.10
• The numerator is the difference in
10. Statistical Decision : Fail to Reject H      0
means between the two samples, and
the denominator is the estimated
standard error of the difference.
9
4. Calculating Test Statistics

• The estimated standard error of the
difference is estimated on the basis of
variances of the two samples (Pooled
Variance t-test).              2
(n1- 1)S1  (n2  1)S2 2 1390.14 124
5. S2 pooled                          
n1  n2 - 2             18
1     1
6. Sx1x2  S2 pooled (  )
n1 n 2
• Where             Sx1x2  5.41
S21= variance of Group 1
S22 = variance of Group 2      95.5  105.0
7. cases in X 2  1
n 1= number of  tc  X 1 Group               1.85
Sx1x2      5.41
n 2= number of cases in Group 2
8. df  n1  n2 - 2  18                     10
9. t0.05  2.10
5. Determine Degrees of Freedom

• Degrees of freedom, df, is value indicating
the number of independent pieces of
information a sample can provide for
purposes of statistical inference.
• Df = Sample size – Number of parameters
estimated
• Df is n1 +n2 -2 for two sample test of means
because the population variance is estimated
from the sample

11
6. Compare the Computed Test
Statistic Against a Tabled Value

If |tc| > |tα|   Reject H0
If p value < α   Reject H0

12
Example of Independent Groups t-tests

• Suppose that we plan to conduct a
study to alleviate the distress of
preschool children who are about to
undergo the finger-stick procedure for a
hematocrit (Hct) determination.
• Note: Hct = % of volume of a blood
sample occupied by cells.

13
Example of Independent Groups t-
tests, Continued
• Twenty subjects will be used to
examine the effectiveness of the special
treatment.
• 10 subjects randomly assigned to
treatment group.
• 10 assigned to a control group that

14
1. Determine the Appropriate Test

independent means (t-test)
– Dependent variable = the child’s pulse rate
just prior to the finger-stick
– Independent variable or grouping variable
= treatment conditions (2 levels)

15
1. Determine the Appropriate Test
• Two samples are independent.
• Two populations are normally distributed.
• The assumption of homogeneity of
variance. (Examine Levene’s Test)
Ho: 1 2 = 12
Ha: 1 2  12
If sig. level or p-value is > .05, the assumption
is met.                                          Independent Samples Test

Pulse rate
Equal variances Equal variances
assumed          not assumed
Levene's Test for      F                                                   .134
Equality of Variances Sig.                                                 .719
t-test for Equality of t                                                 -1.847             -1.847   16
Means                  df                                                    18            17.948
Sig. (2-tailed)                                     .081               .081
2. Establish Level of Significance

• The convention
• α = .05
• α = .01
• α = .001
• In this example, assume α = 0.05

17
3. Determine Whether to Use a One
or Two Tailed Test

• H0 : µ1 = µ2
• Ha : µ1  µ2
– Where
– µ1 = population mean for the experimental group
– µ2 = population mean for the control group

18
4. Calculating Test Statistics
Group 1     Group 2
1       100         105
2        86          95
3       112         120
4        80          85
5       115         110
6        83         100
7        90         115
8        94          93
9        85         107       19
10      105         120
1        100        1

Rearrange the   2
3
86
112
1
1

Data            4         80
Experimental
1
5       115         1
6       Group
83        1
7         90        1
8         94        1
9         85        1
10       105        1
11       105        2
12        95        2
13       120        2
14        85        2
15
Control
110         2
16
Group
100         2
17       115        2
18        93        2
19       107        2   20
20       120        2
4. Calculating Test Statistics
(continued)

Group 1 (Experimental)                            Group 2 (Control)
--------------------------------------------------------------------------------------------------
X1                         X2
------------               --------------
1. X                 950                      1050
2. X                 95.0                     105.0
3. SS1  (X1 - X1)2  1390.14                 SS2  (X2 - X 2)2  1248.03
4. S2  154.46                                 S2  138.67

21
4. Calculating Test Statistics
(continued)

(n1- 1)S12  (n2  1)S2 2 1390.14 1248.03
5. S pooled 
2
                  146.565
n1  n2 - 2              18
1     1
6. Sx1x2  S2 pooled (  )
n1 n 2
Sx1x2  5.41

95.5  105.0
7. tc  X 1  X 2                1.85
Sx1x2        5.41
8. df  n1  n2 - 2  18
9. t0.05  2.10
10. Statistical Decision : Fail to Reject H 0
22
5 Compare the Computed Test
6. . S2pooled  (n1- 1)S12  (n2  1)S2 2 1390.14 1248.03
                  146.56
n1  n2 - 2
Statistic Against a Tabled Value                18
1     1
6. Sx1x2  S pooled (  )
2

n1 n 2
Sx1x2  5.41

7. tc  X 1  X 2  95.5  105.0  1.85
Sx1x2        5.41
8. df  n1  n2 - 2  18
9. t0.05  2.10
10. Statistical Decision : Fail to Reject H 0

23
6. Compare the Computed Test
(n1- 1)S  (n2  1)S
2
1390.14 1248.03
2
 Against a(n2  1)S 1390.14 1248.03
(n1- 1)S  Tabled Value
Statistic  n1  n2 - 2
1         2
 146.565
2                           2               2
5. S                            1            2
2
                   146.565
pooled
5. S           pooled
18
n1  n2 - 2                        18
chosen a
• If we had 2 ( 1  1 ) 1 one
6. Sx1x2     S2 pooled     1                       tail test:
6. Sx1x2  S pooled ( n )
– H0 : µ1 = µ2 1 n 1 2 n 2
n
S x2  5.41 µ2
–x1Ha :x2µ1 5.41
Sx1     <
95.5  105.0
7. tc  X 1  X X 2 95.5  105.0 1.85
2
              
7. tc  X 1
Sx1x2       5.41          1.85
Sx1x2         5.41
8. df  n1  n2 - 2  18
8. df  n1  n2 - 2  18
9. t0.05  2.10 1.73
9. t0.05  2.10
10. Statistical Decision : Fail to Reject H 0
10. Statistical Decision : Fail to Reject H 0
The null hypothesis can be rejected
24
SPSS Output for Two Sample
Independent t-test Example
Group Statistics

Std. Error
Group         N          Mean        Std. Deviation      Mean
Pulse rate   1                  10     95.00            12.428          3.930
2                  10    105.00            11.776          3.724

Independent Samples Test

Pulse rate
Equal variances Equal variances
assumed          not assumed
Levene's Test for      F                                                       .134
Equality of Variances Sig.                                                     .719
t-test for Equality of t                                                     -1.847             -1.847
Means                  df                                                        18            17.948
Sig. (2-tailed)                                         .081               .081
Mean Difference
-10.000           -10.000

Std. Error Difference
5.414          5.414

95% Confidence Interval Lower                          -21.374           -21.377
of the Difference       Upper                            1.374             1.377

25
Nature & Magnitude of
Relationship
Going Beyond Test of Significance

26
Point Biserial Correlation Measures
Strength of the relationship
•   Point biserial correlation is similar to
Pearson r and can be calculated using the
same formula or using the following
formula:
2

rpb 
rpb
t
t 2  df

1.85
2
(-1.85)2
rpb 
rpb                           .40
(-1.85) 
 1.85 2 +1818
2

0.402  .16                                 27
Measures of Practical Significance
•    Point biserial correlation also provides
information about the proportion of explained
variation in the dependent variable.
•    In our example 16 % of the variation in the
children’s pulse rates is explained by the group
membership.                      t
2

rpb 
rpb
t 2  df

1.85
2
(-1.85)2
rpb 
rpb                           .40
(-1.85) 
 1.85 2 +1818
2

0.402  .16
28
Effect Size

• Effect size, gamma () is a measure of
the strength of the relationship
between two variables in the
population and an index of how wrong
the null hypothesis is.
• The higher the effect size the greater
the power of the test.

29
Effect Size

• To evaluate the magnitude of the
difference between two means, a
mean difference is divided by a
“pooled standard deviation.”
• Since researches typically do not have
the value of the population effect size,
it is estimated from sample data.

( X1  X2 )
ES                  95 .0  105 .0 / 12 .85  .78
S (pooled)
30
Most Statistical Tests Assume
Randomness
• Perfect randomness is often impossible
and so researchers try to minimize the
different forms of bias in their selection
of subjects:
– Selection bias
– Attrition bias
– Non-response bias
– Cohort bias

31
Take Home Lesson

How to Compare Mean of Two
Independent Samples

32

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