I. t-Test Analysis
a. Introduction
1) Many research projects are designed to test the differences between
two groups. Differences that involve interval or ratio data, requires an
evaluation of means and distributions of each group.
b. Student's t-test
1) Named after William Gosset, who published under the pseudonym of
Student. Gosset invented the t-test as a more precise method of
comparing groups. He described a set of distribution of means of
randomly drawn samples from a normally distributed population. Distributions
are the t distributions.
2) All t distributions have a normal distribution with a mean
equal to the mean of the population. T distributions are used to
compare differences between two means. These distributions are described by
the sample of differences between means obtained from drawing pairs of
random samples from a population. If we drew an infinite number of pairs of
samples and plot the differences between the means of all possible pairs of
samples, we would find a particular distribution with a mean of zero and a
shape similar to the normal curve.
c. The Research Question
1) When compare two groups on a particular characteristic, asking
whether or not the groups are different.
1. The statistical question asks how different the groups are; that is,
the difference we find greater than that which could occur by chance alone.
Symbolically, the null hypothesis is
written:
Ho: µ1 = µ2
2. The null hypothesis states that any difference that occurs between
the means of two groups is a difference in the sampling distribution. The
means are different not because the groups are drawn from two theoretical
populations, but rather because of different random distributions of the
samples from such a population.
3. When we use the t-test to interpret the significance of the
difference between groups, we are asking the statistical question, what is
the probability of getting a difference of this magnitude in this size if we
were comparing random samples drawn from the same population? What is the
probability of getting a difference this large by chance alone?
4. Sample size (n), and standard deviation (variability), contributes to
significance of the t-test.
d. Type of Data Required
1) Independent Variable: Nominal-level variable, with two levels (two
groups).
2) Dependent Variable: Interval or ratio level, ordinal level data can
often be treated as interval-level data.
e. Assumptions
1) Interval-level data for the dependent measure.
2) Assumption of independence: subject can belong to one and only one of
the two groups and contributes one and only one score.
3) Distribution of the dependent measure is normal.
4) Homogeneity of variance: groups that are compared are similar in
their variances.
Formula:
larger variance
F = -----------------
smaller variance
df for
this F ratio are based on the ns for the groups and equal n-1.
*Meeting this assumption protects against Type II errors-incorrectly
accepting the null hypothesis.
f. T-Test
Formulas
1) Independent or Pooled Formula: Used to compare two groups when the
assumptions for the t-test (including homogeneity of variance) are met.
2) Correlated t-test (Dependent) or Paired Comparisons: Used if there
is correlation between the data taken from the two groups, adjustment must be
made for that relationship. Comparing a group of subjects on their pre- and
post scores is an example. Because these are not two independent groups, but
rather one group measured twice, the scores will most likely be correlated.
Another example is when the two groups consist of matched pairs.
g. Calculations
1) Basic (Pooled) t-test
t = |
(x1 - x2) - ( µ1- µ2)
-----------------------
s(x1 - x2) |
s(x1 -x2) = |
standard error |
calculated
by: |
s(x1-x2)
= |
___________________
√
(Σx21 + Σx22 ) (1 + 1)
--------------- --- ---
(n1 + n2 –2) (n1 n2) |
Where
Σx21 = sum
of squares of group 1
Σx22 = sum
of squares of group 2
n1 = the number of
scores in group 1
n2 = the number of
scores in group 2
when two groups have equal ns, the
formula is simplified to:
|
s(x1-x2)
= |
_____________
√ (Σx21 + Σx22 )
---------------
n (n – 1) |
To find the
sum of squares for each group, use: |
Group 1: |
(ΣX1)2
(Σx1)2 = ΣX21- --------
n1 |
Group 2: |
(ΣX2)2
(Σx2)2 = ΣX22- --------
n2 |
T-Test Example (Independent Groups or pooled)
Group I |
Group II |
9 |
81 |
6 |
36 |
6 |
36 |
7 |
49 |
8 |
64 |
7 |
49 |
8 |
64 |
9 |
81 |
9 |
81 |
8 |
64 |
∑X1 = 40 |
∑X12 = 326 |
∑X2 = 37 |
∑X22 = 279 |
Set
up your computation table.
Group I |
Group II |
_
X1 = 8 |
_
X2= 7.4 |
n1 = 5 |
n2 = 5 |
∑X1 = 40 |
∑X2 = 37 |
∑X12 =326 |
∑X22 = 279 |
Now calculate the sum of squares |
(40)2
∑x12 = 326 –- ---------
5 |
∑x12 = 6 |
(37)2
∑x22 = 279 – --------
5 |
∑x22 = 5.2 |
Now calculate t.
t = |
_ _
(X1 – X2) – (µ1 - µ2)
___________________________________
√(∑x12 + ∑x22 ) / (n1 + n2 – 2) (1/n1 + 1/ n2)
|
|
|
|
|
t =
|
(8 –
7.4) – (0)
___________________________
√(6 + 5.2) /
(5 + 5 – 2) (1/5 + 1/5)
|
.6
= ___
√(1.4) (.4)
|
.6
= ___
√ .56 |
.6
= __
.75 |
t = .8
(not significant
with p < .05) |
Now calculate the degrees of freedom.
df = (n1 + n2) – 2
or df = n - 2 |
df = (5 + 5) – 2 or 8 |
Use a
to determine
t-value. Use a one-tail distribution if hypothesize a difference in a particular
direction. If no directional hypothesis, base interpretation on the two-tailed
distribution.
