Confidence Intervals
Confidence Intervals
Example (a variant of Problem 62, Ch5)
The total time for manufacturing a certain component is known to
have a normal distribution. However, the mean µ and variance σ 2
for the normal distribution are unknown. After an experiment in
which we manufactured 10 components, we recorded the sample
time which is given as follows:
1
2
3
4
5
time 63.8 60.5 65.3 65.7 61.9
6
7
8
9
10
time 68.2 68.1 64.8 65.8 65.4
Confidence Intervals
Example (a variant of Problem 62, Ch5)
The total time for manufacturing a certain component is known to
have a normal distribution. However, the mean µ and variance σ 2
for the normal distribution are unknown. After an experiment in
which we manufactured 10 components, we recorded the sample
time which is given as follows:
1
2
3
4
5
time 63.8 60.5 65.3 65.7 61.9
6
7
8
9
10
time 68.2 68.1 64.8 65.8 65.4
We know that both MME and MLE for the population mean µ is
the sample mean X , i.e. µ̂ = X = 64.95. How accurate is this
estimation?
Confidence Intervals
Confidence Intervals
• Assume the other parameter σ is known, e.g. σ = 2.7
Confidence Intervals
• Assume the other parameter σ is known, e.g. σ = 2.7
• X is normally distributed with mean µ and variance σ 2 /n.
X −µ
√ is a standard normal random variable.
Therefore, Z = σ/
n
Confidence Intervals
• Assume the other parameter σ is known, e.g. σ = 2.7
• X is normally distributed with mean µ and variance σ 2 /n.
X −µ
√ is a standard normal random variable.
Therefore, Z = σ/
n
• For the interval [−A, A], how large should A be such that with
95% confidence we are sure Z falls in that interval?
Confidence Intervals
• Assume the other parameter σ is known, e.g. σ = 2.7
• X is normally distributed with mean µ and variance σ 2 /n.
X −µ
√ is a standard normal random variable.
Therefore, Z = σ/
n
• For the interval [−A, A], how large should A be such that with
95% confidence we are sure Z falls in that interval?
P(−A < Z < A) = .95
Confidence Intervals
• Assume the other parameter σ is known, e.g. σ = 2.7
• X is normally distributed with mean µ and variance σ 2 /n.
X −µ
√ is a standard normal random variable.
Therefore, Z = σ/
n
• For the interval [−A, A], how large should A be such that with
95% confidence we are sure Z falls in that interval?
P(−A < Z < A) = .95
A is the 97.5th percentle, which is 1.96.
Confidence Intervals
• Assume the other parameter σ is known, e.g. σ = 2.7
• X is normally distributed with mean µ and variance σ 2 /n.
X −µ
√ is a standard normal random variable.
Therefore, Z = σ/
n
• For the interval [−A, A], how large should A be such that with
95% confidence we are sure Z falls in that interval?
P(−A < Z < A) = .95
A is the 97.5th percentle, which is 1.96.
X −µ
√ < 1.96 = .95
• P −1.96 < σ/
n
Confidence Intervals
• Assume the other parameter σ is known, e.g. σ = 2.7
• X is normally distributed with mean µ and variance σ 2 /n.
X −µ
√ is a standard normal random variable.
Therefore, Z = σ/
n
• For the interval [−A, A], how large should A be such that with
95% confidence we are sure Z falls in that interval?
P(−A < Z < A) = .95
A is the 97.5th percentle, which is 1.96.
X −µ
√ < 1.96 = .95
• P −1.96 < σ/
n
• P X − 1.96 · √σn < µ < X + 1.96 · √σn = .95
Confidence Intervals
• Assume the other parameter σ is known, e.g. σ = 2.7
• X is normally distributed with mean µ and variance σ 2 /n.
X −µ
√ is a standard normal random variable.
Therefore, Z = σ/
n
• For the interval [−A, A], how large should A be such that with
95% confidence we are sure Z falls in that interval?
P(−A < Z < A) = .95
A is the 97.5th percentle, which is 1.96.
X −µ
√ < 1.96 = .95
• P −1.96 < σ/
n
• P X − 1.96 · √σn < µ < X + 1.96 · √σn = .95
• The interval X − 1.96 · √σn , X + 1.96 · √σn is called the
95% confidence interval for µ.
Confidence Intervals
• Assume the other parameter σ is known, e.g. σ = 2.7
• X is normally distributed with mean µ and variance σ 2 /n.
X −µ
√ is a standard normal random variable.
Therefore, Z = σ/
n
• For the interval [−A, A], how large should A be such that with
95% confidence we are sure Z falls in that interval?
P(−A < Z < A) = .95
A is the 97.5th percentle, which is 1.96.
X −µ
√ < 1.96 = .95
• P −1.96 < σ/
n
• P X − 1.96 · √σn < µ < X + 1.96 · √σn = .95
• The interval X − 1.96 · √σn , X + 1.96 · √σn is called the
95% confidence interval for µ.
• In our case, 95% confidence interval for µ is (63.28, 66.62).
Confidence Intervals
Confidence Intervals
Interpretation of Confidence Interval
Confidence Intervals
Interpretation of Confidence Interval
• The 95% confidence interval for µ (63.28, 66.62) doesn’t
mean
P(µ falls in the interval(63.28, 66.62)) = .95
Confidence Intervals
Interpretation of Confidence Interval
• The 95% confidence interval for µ (63.28, 66.62) doesn’t
mean
P(µ falls in the interval(63.28, 66.62)) = .95
• It is a long-run effect: if we have 1000 random samples, then
for approximately 950 of them,µ falls in the interval
X − 1.96 · √σn , X + 1.96 · √σn .
Confidence Intervals
Confidence Intervals
Example (a variant of Problem 62, Ch5)
The total time for manufacturing a certain component is known to
have a normal distribution. However, the mean µ for the normal
distribution is unknown. After an experiment in which we
manufactured 10 components, we recorded the sample time which
is given as follows:
1
2
3
4
5
time 63.8 60.5 65.3 65.7 61.9
6
7
8
9
10
time 68.2 68.1 64.8 65.8 65.4
We know that both MME and MLE for the population mean µ is
the sample mean X , i.e. µ̂ = X = 64.95. We further assume the
standard deviation is known to be σ = 2.7. What is the 99%
confidence interval for µ?
Confidence Intervals
Confidence Intervals
Definition
A 100(1 − α)% confidence interval for the mean µ of a normal
population when the value of σ is known is given by
σ
σ
x − zα/2 · √ , x + zα/2 · √
n
n
or, equivalently, by x ∓ zα/2 ·
√σ
n
Confidence Intervals
Confidence Intervals
Graphically interpretation:
Confidence Intervals
Confidence Intervals
Example (a variant of Problem 62, Ch5)
The total time for manufacturing a certain component is known to
have a normal distribution. However, the mean µ for the normal
distribution is unknown. Thus we decide to do an experiment in
which we manufacture n components to estimate the population
mean µ. We know that both MME and MLE for the population
mean µ is the sample mean X , i.e. µ̂ = X . We further assume the
standard deviation is known to be σ = 2.7. If we want a 99%
confidence interval for µ with width 3.34, how large should
n be?
Confidence Intervals
Confidence Intervals
Proposition
To obtain a 100(1 − α)% confidence interval with width w for the
mean µ of a normal population when the value of σ is known, we
need a random sample of size at least
σ 2
n = 2zα/2 ·
w
Confidence Intervals
Proposition
To obtain a 100(1 − α)% confidence interval with width w for the
mean µ of a normal population when the value of σ is known, we
need a random sample of size at least
σ 2
n = 2zα/2 ·
w
Remark:
The half-width w2 of the 100(1 − α)% CI is called the bound on
the error of estimation associated with a 100(1 − α)%
confidence level.
Large-Sample Confidence Intervals
Proposition
If n is sufficiently large, the standardized variable
Z=
X −µ
√
S/ n
has approximately a standard normal distribution. This implies that
s
x̄ ± zα/2 · √
n
is a large-sample confidence interval for µ with confidence level
approximately 100(1 − α)%. This formula is valid regardless of the
shape of the population distribution.