Transcript MATH30-6 Lecture 10

Sampling Distribution and Point
Estimation of Parameters
MATH30-6
Probability and Statistics
Objectives
At the end of the lesson, the students are expected to
• Explain the general concepts of estimating the
parameters of a population or a probability
distribution;
• Explain the important role of the normal distribution as
a sampling distribution; and
• Understand the central limit theorem.
Point Estimator
• A point estimate of some population parameter 𝜃 is a
single numerical value 𝜃 of a statistic Θ. The statistic Θ
is called the point estimator.
Estimation problems occur frequently in engineering.
• The mean μ of a single population
• The variance σ2 (or standard deviation σ) of a single
population
• The proportion p of items in a population that belong
to a class of interest
Point Estimator
• The difference in means of two populations, 𝜇1 − 𝜇2
• The difference in two population proportions, 𝑝1 − 𝑝2
Reasonable point estimates:
• 𝜇=𝑥
• 𝜎 2 = 𝑠2
• 𝑝=𝑥 𝑛
• 𝜇1 − 𝜇2 = 𝑥1 − 𝑥2
• 𝑝1 − 𝑝2 = 𝑥1 𝑛1 − 𝑥2 𝑛2
Sampling Distribution
• The random variables 𝑋1 , 𝑋2 , … , 𝑋𝑛 are a random
sample of size n is (a) the Xi’s are independent random
variables, and (b) every Xi has the same probability
distribution.
• A statistic is any function of the observations in a
random sample.
• The probability distribution of a statistic is called a
sampling distribution.
- For example, the probability distribution of 𝑋 is called
the sampling distribution of the mean.
Central Limit Theorem
Consider determining the sampling distribution of the
sample mean 𝑋. The sample mean
𝑋1 + 𝑋2 + ⋯ + 𝑋𝑛
𝑋=
𝑛
has a normal distribution with mean
𝜇 + 𝜇 + ⋯+ 𝜇
𝜇𝑋 =
=𝜇
𝑛
and variance
2 + 𝜎2 + ⋯ + 𝜎2
2
𝜎
𝜎
𝜎𝑋2 =
=
𝑛
𝑛
Central Limit Theorem
• If we are sampling from a population that has an
unknown probability distribution, sampling distribution
of the sample mean will still be approximately normal
with mean μ and variance σ2/n, if the sample size n is
large.
• In Inferential Statistics, n ≥ 40 (Montgomery and
Runger, 2011) is considered a large sample. Otherwise,
it is considered small.
• n ≥ 30 is considered a large sample (Walpole, et al,
2012)
Central Limit Theorem
If 𝑋1 , 𝑋2 , … , 𝑋𝑛 is a random sample of size n taken from a
population (either finite or infinite) with mean μ and
variance σ2, and if 𝑋 is the sample mean, the limiting form
of the distribution of
𝑋−𝜇
𝑍=
𝜎 𝑛
(7-1)
as n → ∞, is the standard normal distribution.
Central Limit Theorem
Central Limit Theorem
Examples:
7-1/228 An electronics company manufactures resistors
that have a mean resistance of 100 ohms and a standard
deviation of 10 ohms. The distribution of resistance is
normal. Find the probability that a random sample of n =
25 resistors will have an average resistance less than 95
ohms.
Central Limit Theorem
Central Limit Theorem
7-2/228 Suppose that a random variable X has a
continuous uniform distribution
1 2,4 ≤ 𝑥 ≤ 6
𝑓 𝑥 =
0, otherwise
Find the distribution of the sample mean of a random
sample of size n = 40.
Central Limit Theorem
Central Limit Theorem
7-10/230 Suppose that the random variable X has the
continuous uniform distribution
1, 0 ≤ 𝑥 ≤ 1
𝑓 𝑥 =
0, otherwise
Suppose that a random sample of n = 12 observations is
selected from this distribution. What is the approximate
probability distribution of 𝑋 − 6? Find the mean and
variance of this quantity.
Central Limit Theorem
7-11/230 Suppose that X has a discrete uniform
distribution
1 3 , 𝑥 = 1,2,3
𝑓 𝑥 =
0, otherwise
A random sample of n = 36 is selected from this
population. Find the probability that the sample mean is
greater than 2.1 but less than 2.5, assuming that the
sample mean would be measured to the nearest tenth.
Central Limit Theorem
7-12/231 The amount of time that a customer spends
waiting at an airport check-in counter is a random
variable with mean 8.2 minutes and standard deviation
1.5 minutes. Suppose that a random sample of n = 49
customers is observed. Find the probability that the
average time waiting in line for these customers is
(a) Less than 10 minutes
(b) Between 5 and 10 minutes
(c) Less than 6 minutes
Difference in Sample Means
If we have two independent populations with means μ1
and μ2 and variances 𝜎12 and 𝜎22 and if 𝑋1 and 𝑋2 are the
sample means of two independent random samples of
sizes n1 and n2 from these populations, then the sampling
distribution of
𝑋1 − 𝑋2 − 𝜇1 − 𝜇2
𝑍=
𝜎12 𝑛1 + 𝜎22 𝑛2
(7-4)
Is approximately normal, if the conditions of the central
limit theorem apply. If the two populations are normal,
the sampling distribution of Z is exactly standard normal.
Difference in Sample Means
Examples:
7-3/229 Aircraft Engine Life The effective life of a
component used in a jet-turbine aircraft engine is a
random variable with mean 5000 hours and standard
deviation 40 hours. The distribution of effective life is
fairly close to a normal distribution. The engine
manufacturer introduces an improvement into the
manufacturing process for this component that increases
the mean life to 5050 hours and decreases the standard
deviation to 30 hours. Suppose that a random sample of
n1 = 16 components is selected from the “old” process
and a random sample of n2 = 25 components is selected
Difference in Sample Means
from the “improved” process. What is the probability that
the difference in the two sample means 𝑋2 − 𝑋1 is at
least 25 hours? Assume that the old and improved
processes can be regarded as independent populations.
Difference in Sample Means
7-13/231 A random sample of size n1 = 16 is selected from
a normal population with a mean of 75 and a standard
deviation of 8. A second random sample of size n2 = 9 is
taken from another normal population with mean 70 and
standard deviation 12. Let 𝑋1 and 𝑋2 be the two sample
means. Find:
(a) The probability that 𝑋1 − 𝑋2 exceeds 4
(b) The probability that 3.5 ≤ 𝑋1 − 𝑋2 ≤ 5.5
Difference in Sample Means
7-14/231 A consumer electronics company is comparing
the brightness of two different types of picture tubes for
use in its television sets. Tube type A has mean brightness
of 100 and standard deviation of 16, while tube B has
unknown mean brightness, but the standard deviation is
assumed to be identical to that for type A. A random
sample of n = 25 tubes of each type is selected, and 𝑋𝐵 −
𝑋𝐴 is computed. If μB equals or exceeds μA, the
manufacturer would like to adopt type B for use. The
observed difference is 𝑥𝐵 − 𝑥𝐴 = 3.5 . What decision
would you make, and why?
Summary
• The probability distribution of a statistic is called the
sampling distribution. For example, the sampling
distribution of the sample mean 𝑋 is the normal
distribution.
• The simplest form of the central limit theorem states
that the sum of n independently distributed random
variables tend to be normally distributed as n becomes
large. It is a necessary and sufficient condition that
none of the variances of the individual random
variables are large in comparison to their sum.
Summary
• Sampling Distribution of the Mean
𝑋−𝜇
𝑍=
𝜎 𝑛
• Approximate Sampling Distribution of a Difference in
Sample Means
𝑋1 − 𝑋2 − 𝜇1 − 𝜇2
𝑍=
𝜎12 𝑛1 + 𝜎22 𝑛2
References
• Montgomery and Runger. Applied Statistics and
Probability for Engineers, 5th Ed. © 2011
• Walpole, et al. Probability and Statistics for Engineers
and Scientists 9th Ed. © 2012, 2007, 2002