10.2. Population Mean: Small Sample Case
t-distribution (“student” t distribution):
This distribution was invented by W. S. Gossett (published in the
name “student”). The t-distribution is a family of similar probability
distributions, with a specific t distribution depending on a parameter
known the degree of freedom (d.f.). Denote
T (n) to be the random
variable having t-distribution with n degree of freedom.
Example:
P (T (1) > 3.078) = y ⇔ y = 0.1 ; P(T (1) > 6.314) = y ⇔ y = 0.05
P(T (5) > x) = 0.025 ⇔ x = 2.571, P (T (11) > x) = 0.01 ⇔ x = 2.718.
P(T (∞) > 1.96) = 0.025 = P( Z > 1.96)
Note: T (∞ ) ~ N ( 0 ,1) . The t-distribution is symmetric about 0.
t n ,α
α
: a t value with an area of
2
2
in the upper tail of the
t-distribution with degree of freedom equal to n.
α
⇔ P ⎛⎜ T ( n ) > t n ,α ⎞⎟ =
.
2 ⎠
⎝
2
Example:
P(T (1) > 3.028) = 0.1 = P(T (1) > t1,0.1 ) ⇔ t1,0.1 = 3.078
P(T (5) > 2.571) = 0.025 = P(T (5) > t5,0.025) ⇔ t5,0.025 = 2.571
Example:
For a t distribution with 16 degrees of freedom, find the area of probability.
(a) To the left of -1.746.
(b) Between -1.337 and 2.120.
[solution:]
(a)
1
P(T (16) < −1.746) = P(T (16) > 1.746) = 0.05.
(b)
P(−1.337< T(16) < 2.120) =1− P(T(16) > 2.12) − P(T(16) < −1.337)
= 1− 0.025− P(T(16) > 1.337)
= 1− 0.025− 0.1
= 0.875
Important Result:
When the population has a normal probability distribution and σ is
unknown, then
X − μ
= T ( n − 1)
S
n
n
∑X
∑ (X
n
i
− X)
,
2
i
, and
X1, X2 ,K, Xn
variables with associated possible values
x1 , x2 ,K, xn .
where X =
i =1
n
, S2 =
i =1
n −1
are random
Derivation of (1 − α ) × 100% confidence interval:
Suppose the population has a normal probability distribution. Since
P⎛⎜ T (n − 1) ≤ t n−1,α ⎞⎟ = 1 − P⎛⎜ T (n − 1) > t n−1,α ⎞⎟
2⎠
2⎠
⎝
⎝
= 1 − 2P⎛⎜T (n − 1) > t n−1,α ⎞⎟ (QT (n − 1) is symmetric)
2⎠
⎝
= 1−α
thus
2
⎛
⎞
⎛
⎞
⎜ X −μ
⎟
⎜ μ−X
⎟
≤ t n−1,α ⎟ = P⎜
≤ t n−1,α ⎟
P⎛⎜ T (n −1) ≤ t n−1,α ⎞⎟ = P⎜
2⎠
2
2
⎝
⎜ S
⎟
⎜ S
⎟
n
n
⎝
⎠
⎝
⎠
⎛
⎞
⎜
⎟
⎛
u−X
S
S ⎞
= P⎜ − t n−1,α ≤
≤ t n−1,α ⎟ = P⎜⎜ − t n−1,α
≤ u − X ≤ t n−1,α
⎟⎟
S
2
2
2
2
n
n
⎝
⎠
⎜
⎟
n
⎝
⎠
⎛
S
S ⎞
= P⎜⎜ X − t n−1,α
≤ μ ≤ X + t n−1,α
⎟⎟ = 1 − α
2
2
n
n⎠
⎝
(1 − α ) × 100 % confidence interval based on t-distribution:
σ
As the population has a normal distribution and
x ± tn−1,α
2
is unknown,
s ⎡
s
s⎤
≡ ⎢x −tn−1,α
, x + tn−1,α
⎥
2 n
2 n⎦
n ⎣
is a (1 − α ) × 100 % confidence interval of
μ .
Example:
Consider the following random sample of 4 observations,
25, 47, 32, 56.
Suppose the population is normally distributed. Please construct a 95% confidence
interval for
μ .
[solution:]
25 + 47 + 32 + 56
x=
= 40, s =
4
(25 − 40)2 + (47 − 40)2 + (32 − 40)2 + (56 − 40)2
4 −1
= 14.07
In addition, α = 0.05 and t 3,0.025 = 3.182 . Thus,
x ± t 3,0.025
14.07
14.07 ⎤
⎡
= ⎢40 − 3.182
, 40 + 3.182
⎥
n
4
4
4 ⎦
⎣
= [17.613, 62.387]
s
= 40 ± 3.182
14.07
3
Example:
Suppose we have the following data from a normal population
50
48
55
52
53
46
54
50
Provide a 95% confidence interval for the population mean.
[solution:]
n = 8, α = 0.05, x = 51, s = 3.07.
x ± t n−1,α
s
2
n
= x ± t 7,0.025
3.07
8
Then, a 95% confidence interval is
= 51 ± 2.365
Online Exercise:
Exercise 10.2.1
Exercise 10.2.2
4
3.07
8
= 51 ± 2.57 = [48.43, 53.57]