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Estimation

Suppose that the population of interest has a normal distribution with unknown mean and
2
unknown variance σ . Assume that a random sample of size n, say, X 1 , X 2 , . . . , X n , is available,
¯
2
and let X and S be the sample mean and variance, respectively. We wish to construct a two-sided
√
¯
2
CI on µ. If the variance σ is known, we know that Z = (X − µ)/(σ n) has a standard normal
2
distribution. When σ is unknown, a logical procedure is to replace σ with the sample standard
√
¯
deviation S. The random variable Z now becomes T = (X − µ)/(S n). A logical question is,
what effect does replacing σ by S have on the distribution of the random variable T? If n is large,
the answer to this question is ”very little,” and we can proceed to use the confidence interval based
on the normal distribution. However, n is usually small in most engineering problems, and in this
situation a different distribution must be employed to construct the CI.
t- Distribution

Definition 2.2. Let X 1 , X 2 , . . . , X n be a random sample from a normal
2
distribution with unknown mean µ and unknown variance σ . The random
variable
¯
X − µ
T = √ (2.28)
S/ n
has a t distribution with n − 1 degrees of freedom. ✓

The t-probability density function is

k+1
( ) −
Γ[(k + 1)/2] x 2 2
f(x) = √ · 1 + (2.29)
πkΓ(k/2) k
where k is the number of degrees of freedom. The mean and variance of the t distribution are zero
and k(k − 2) (for k > 2), respectively.
The general appearance of the t distribution is similar to the standard normal distribution in
that both distributions are symmetric and unimodal, and the maximum ordinate value is reached
when the mean µ = 0. However, the t distribution has heavier tails than the normal; that is, it has
more probability in the tails than the normal distribution. As the number of degrees of freedom
k → ∞, the limiting form of the t distribution is the standard normal distribution. Generally,
the number of degrees of freedom for t is the number of degrees of freedom associated with the
estimated standard deviation.

t-Confidence Interval on mean. It is easy to find a 100(1 − α) percent confidence interval on
the mean of a normal distribution with unknown variance by proceeding essentially as we did in
√
¯
previous section. We know that the distribution of T = (X − µ)/(S/ n) is t with n − 1 degrees
of freedom. Letting t α/2,n−1 be the upper 100α/2 percentage point of the t distribution with n − 1
degrees of freedom, we may write

P(−t α/2,n−1 ≤ T ≤ t α/2,n−1 ) = 1 − α

or
( ¯ )
X − µ
P −t α/2,n−1 ≤ √ ≤ t α/2,n−1 = 1 − α.
S/ n
Rearranging this last equation yields

√ √
¯
¯
P(X − t α/2,n−1 S/ n ≤ µ ≤ X + t α/2,n−1 S/ n) = 1 − α. (2.30)
This leads to the following definition of the 100(1 − α)% two-sided confidence interval on µ.

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