Lecture 10 - The t-distribution & t-test ((Student's t-test))
85
In a case when the population variance (∂
2
) is unknown &
the sample size is small (n≤30), we can use the sample
variance (S
2
) as a best point estimator for ∂
2
but in this
situation the distribution will not follow the standard
normal distribution (Z-distribution) but follow the t-
distribution.
The characters of t-distribution:-
1) It has a mean of zero.
2) Symmetrical about the mean.
3) Range between - ∞ & +∞.
4) Compared with the normal distribution, its curve is less
peaked and higher tails.
5) The quantity of (n-1) which is called degree of freedom
(df) is used in computing S
2
[There is a different
distribution for each sample value of (n-1)].
6) The t- distribution approaches normal distribution as (n-1)
value approaches infinity (increase the sample size).
How can we get the t-value?
As in standard normal distribution in which we have the
Z-table, here we have the t-table which depend on the df=
(n-1) → Row of the table &
⍺ (probability of error) = t
1 -
⍺
/2
→ column of the table.
Application of the t-test:
T-test can be applied for the following situations: (Not
much different than of Z- test)
1)
Is the
sample mean differs significantly from the
population mean
. [Small sample size (n≤30), and the
population variance (∂
2
) is a known]. We use the
following formula:
t=(x-µ) /(S/√n)
Ex: A certain breeds of rats show a mean weight gain of
65gm during the first 3 months of life, a sample 16 0f
these rats was taken and feed a new diet from birth until
the age of 3 months, the mean weight was 60.75gm with
S= 3.84gm. Is this mean differ significantly from
population mean?
t=(x-µ) /(S/√n)
= (60.75 – 65) / (3.84/√16) = -4.43
From the t-table (
⍺=0.05): t
1 -
⍺
/2
, df =15.
The difference between x & µ is statistically significant.
2) Comparing the significant difference between two
samples means
. [Small sample size (n≤30) and the
population variance (∂
2
) is a known]. We use the
following formula:
t=(x
1
- x
2
) - (µ
1
-µ
2
) /√ ([S
2
1
/n
1
] + [S
2
2
/n
2
])
df= n
1
+ n
2
-2
Ex: In a comparison between two groups of patients with
diverticulitis on two different types of treatment n
1
=15, n
2
=12. The recovery time in hours x
1
=68.4 hrs & S
1
=286
hrs and x
2
= 83.43 hrs & S
1
=290 hrs. Is the difference
between the means of hrs is statistically significant?
t=(x
1
- x
2
µ
1
) - (µ
1
-µ
2
) /√ ([S
2
1
/n
1
] + [S
2
2
/n
2
])
= (68.4 -83.43)- 0 / √ (286/15) + (290/12) = - 2.28
From the t-table: t
1 -
⍺
/2
, df= n
1
+ n
2
-2 =25. (
⍺≤0.05)
The difference between x
1
& x
2
is statistically significant.
3) Pairing.
Many studies are designed to produce observation in pair
e.g., single individual has pair of reading (before & after),
for example measurement of BP before and after
treatment Or when the same volunteers or participants
pass through 2 different situations (each one has 2
readings e.g. as for 2 drugs, 2 different doses for the same
drug, drug and placebo, or rest & exhaustion…).
To deal with such condition we do:
a) We find the difference (d).
b) We calculate the differences, and find the mean of
differences (d).
c) We calculate the Sd of the difference using the following
formula:
Sd =√ [n ∑d
2
- (∑d)
2
] / [n(n-1)]
d) The df =n-1, because we have one sample although
having two readings.
e) To calculate the value of t we use the following formula;
t= [d- µd] / [Sd/√n]
Lecture 10 - The t-distribution & t-test ((Student's t-test))
86
Ex: In pediatric clinic, a study was done to see the
effectiveness of certain antipyretic drug in 12 years old
children suffering from influenza , their temperature had
taken immediately before and 1 hr after administration of
the drug. The following results were found:
No.
Temperature(c
o
)
before –After
differences(d)
d
2
before
After
1
2
3
4
5
6
7
8
9
10
11
12
39.1
39.6
38.3
39.4
38.4
38.2
39.2
39.5
39.3
39.1
38.8
38.6
37.6
37.8
37.9
38.4
37.7
37.9
38.3
38.8
38.2
38.4
38.5
37.9
1.5
1.8
0.9
1.0
0.7
0.3
0.9
1.7
1.1
0.7
0.3
0.7
2.25
3.24
0.81
1.0
0.49
0.09
0.81
2.89
1.21
0.49
0.09
0.49
∑d=11.6
∑d
2
=13.86
d (mean difference) = ∑d / n = 11.6 / 12 = 0.97 C
o
Sd =√ [n ∑d
2
- (∑d)
2
] / [n(n-1)]
= √ [12(13.86) – (11.6)
2
]\ [12(12-1)] = 0.49
t= [d- µd] / [Sd/√n] = [0.97-0] / [0.49/√12] = 6.9
From the t-table (
⍺≤0.05): t
1 -
⍺
/2
, df =11.
The difference between before and after is statistically
significant.