MATH 105 - Lesson Six

Lesson 6 : Distribution of the Sample Mean

Satya Mandal

Homework 20

Due Date: See the Lecture Notes Site.

6.1 Introduction

The goal of this lesson is to establish the foundation for estimation of parameters that we will do in the next two lesson. This lesson can be viewed as a bridge between probability (the math) and estimation (statistics). We will mainly do the estimation of the population mean μ.

In this lesson, our final theorem would be that the sample mean

X = (X₁+X₂+ ... +X_n)/n

has normal distribution, approximately.

Example. Suppose we want to study the height distribution of the U.S. population. Let X represent the height of the whole US population.

We may collect a data of size n = 713 as follows:

71	62	67	73	61	58	63	58	69	68
55	57	51	57	49	63	63	64	72	59
67	59	57	69	55	56	65	66	53	53
51	66	68	71	61	63	53	76	77	67

And so on

Our point of view is that the height

x₁, x₂, x₃, ... , x_n
(in our case 71, 62, 67, ... )
are, in fact, the observed values of random variables
X₁, X₂, X₃, ... , X_n,
respectively.

Here X₁ is the notation for height of the first member of the sample, which could be the height of anybody from the whole U.S. population, and in the case of our sample the value of X₁ is 71. Similarly, X₂ is the notation for height of the second member of the sample, which could be the height of anybody from the whole U.S. population, and in the case of our sample the value of X₂is 62.

In general, we collect sample of size n. The sample would be n numbers

x₁, x₂, …, x_n

representing the height of n individuals.

We shall consider the height x_i of the i^th individual as the observed value of a random variable X_i.
Here X_i is the notation for the height of the i^th member of the sample, which could be the height of anybody from population.
Therefore, these n measurements

x₁, x₂, …, x_n

are, respectively, the observed values of n random variables

X₁, X₂, …, X_n.

We (re)define the sample mean X as the random variable

X₁+X₂+…+X_n

So, the sample means that we computed before in Lesson 2 are the values of the random variable X.

We will consider the sampling distribution of the sample mean X.

Sampling Types

There are many ways to do sampling. Most commonly discussed among them are

Sampling without replacement,
Sampling with replacement.

The Sampling without replacement is the type of sampling where, whenever a sample member is selected, the member is excluded from the subsequent selections. It is analogous to selecting n balls from a box of N balls. Balls are selected one by one, without replacing them back in the box before subsequent selections. This type of sampling is meant to rule out the possibility of selecting a member more than once. For small populations, possibility of selecting a memeber twice may be significant. For such small populations, sampling without replacement would be appropriate.

The Sampling with replacement is the type of sampling where each selection is done without any regard to previous selections. In other words, each time a sample member is drawn, it is placed back to the whole population before the next selection is made. This way, each selection is done from the same whole population. A member could, therefore, be selected more than once. This may seem unnatural. But when working with large populations this is not likely to happen and is most natural from the statistical point of view. (How often would one receive calls twice for the same poll?) We will only consider sampling with replacement .

Properties

Let X (like height) be a random variable with mean μ and variance σ². Let X₁, X₂, … , X_n be a sample from the X - population. We assume that the sampling was done with replacement. Thw following are comments and properties of such a sample:

X would be called the parent population or the population random variable. Also μ and σ² are called the population mean and variance.
Each of the sample member X_i has the same distribution as X. So, mean of X_i is μ and variance of X_i is σ².
The sample members X₁,X₂, …, X_n are all mutually independent.
(In fact, one has to ensure that they are drawn independently. This is a very difficult task in practice, because of lack of accessibility of the population members, human bias and other reasons.)
The distribution of X is called the sampling distribution of X.
Theorem. The mean of the sample mean X is the population mean μ, that is

Mean(X) = Mean(X) = μ

The variance of the sample mean X is given by

Var(X) = σ²/n

So, the standard deviation of X, denoted by σ _X, is given by

σ_X = σ/√n.
Definition. The standard deviation σ_X is also called standard error.

6.2 Central Limit Theorem

The following theorem describes the sampling distribution of the Sample Mean. It is called the Central Limit Theorem (CLT)

Theorem (CLT). Suppose X₁,X₂, …,X_n is a sample from a population X with mean μ and variance σ².

Assume n is large. Then the sample mean X, approximately, has a normal distribution with

mean(X) = μ

and standard deviation

σ_X= σ/√n.

Therefore, approximately,

P(a < X <b) ≈ P(L < Z < R)

where L=(a-μ)/σ _X and R=(b-μ)/σ _X

P(a <

< b) ≈ P

a- μ

σ/√n

< Z <

b-μ

σ/√n

Further, if the parent population X is Normal, then the above equations are exact.

Standard Error and Precision

The standard error σ _X= σ /√n decreases to zero, as the sample size n increases. Because of this, while estimating the mean μ by the sample mean X, by increasing the sample size n, the sample mean X can simultaneously achieve precision and level of confidence (i.e. probability of a give precision). The Animation 6.1.1 below demonstrates the same. The following animation demonstration of the same.

Animation 6.1.1

Problem Solving: The Central Limit Theorem (CLT) would be used to compute approximate probability for the sample mean X. This would be similar to normal approximation to Binomial (Lesson 5).

Problems on 6.2: Central Limit Theorem

Exercise 6.2.1. It is known that the tuition paid per semester by students in a university has a distribution with mean $2,050 and standard deviation $310. If 64 students are interviewed, what is the approximate probability that the sample mean tuition paid will be above $2,060?

Solution by TI-84:
Here the population mean μ = 2,050 and the population standard deviation σ = 310. The sample size n = 64.
First step is to compute the mean
μ_X =μ = 2050
and the standard deviation
σ_X = σ/ √n = 310/√64=38.75
Let X = Tuition paid by the students. Then, the distribution of X is, approximately, N(2050, 38.75)

Now " X will be above 2060" means " X > 2060".
P(2060 < X)
= P([2060 - μ]/σ_X < [X - μ]/ σ_X )
≈ P([2060 - μ]/σ _X < Z )
= P([2060 - 2050]/38.75 < Z )
= P (.2580, < Z ) = normalcdf(.2580, 5)= .3982

Exercise 6.2.2.
The monthly water consumption X per household in a subdivision in Kansas City has normal distribution with mean 15000 gallons and standard deviation 3000 gallons. What is the probability that the mean consumption of the 44 households in the subdivision will exceed 16000 gallons?

Solution by TI-84:
Here the population mean μ = 15000 and the population standard deviation σ = 3000. The sample size n = 44.
First step is to compute the mean
μ_X =μ = 15000.
and the standard deviation
σ_X = σ/ √n = 3000/√44=452.2670
Let X = monthly water consumption by the households. Then, the distribution of X is, approximately, N(15000, 452.2670)

Now " X will exceed 16000" means " X > 16000".
P(16000 < X)
= P([16000 - μ]/σ _X < [X - μ]/ σ_X )
≈ P([16000 - μ]/σ _X < Z )
= P([16000-15000]/452.2670 < Z )
= P (2.2111, < Z ) = normalcdf(2.2111, 5)= .0135

Exercise 6.2.3. In a class of more than thousand students, the instructor announced after a test that the mean score was μ = 77 point and standard deviation σ = 24 points. You took a sample of 81 students. What would be the approximate probability that the sample mean would be less than 80?

Solution by TI-84:
Here the population mean μ = 77 and the population standard deviation σ = 24. The sample size n = 81.
First step is to compute the mean
μ_X =μ = 77.
and the standard deviation
σ_X = σ/ √n = 24/√81= 2.6667
Let X = Points scored by students. Then, the distribution of X is, approximately, N(77, 2.6667)

Now "the sample mean would be less than 80" means " X < 80".

P(X < 80)
= P([X - μ]/ σ_X < [80 - μ]/σ _X)
≈ P(Z < [80 - μ]/σ_X < Z )
= P(Z < [80 -77]/2.6667)
= P (Z < 1.1250) = normalcdf(-5, 1.1250)= .8697

Exercise 6.2.4. The mean salary X of the university professors in a state is μ = $65,000 and standard deviation σ = $14,000. You collect a sample of 75 professors. What is the probability that sample mean salary of these 75 professors would be above $60,000.

Solution by TI-84:
Here the population mean μ = 65000 and the population standard deviation σ = 14000. The sample size n = 75.
First step is to compute the mean
μ_X =μ = 65000.
and the standard deviation
σ_X = σ/ √n = 14000/√75=1616.5808
Let X = monthly water consumption by the households. Then, the distribution of X is, approximately, N(65000, 1616.5808)

Now " X would be above 60,000" means " X > 60000".

P(16000 < X)
= P([60000 - μ]/σ_X < [X - μ]/ σ_X )
≈ P([16000 - μ]/σ_X ; < Z )
= P([60000-65000]/1616.5808 < Z )
= P (-3.0929, < Z ) = normalcdf(-3.0929, 5)= .9990

Exercise 6.2.5.The time X that a child spends watching TV on week- ends has a normal distribution with mean μ = 330 minutes and standard deviation σ = 95 minutes. You sample 50 kids in a school. What is the probability that the sample time X that these kids watch TV on a weekend will be less than 300 minutes.

Exercise 6.2.6. The weight X of fish in a lake has mean μ = 12 pounds and standard deviation σ = 4.5 pounds. Suppose you catch 150 fish. What is the probability that the mean weight of of the catch will be less than 1900/150 pounds?

Solution by TI-84:
Here the population mean μ = 12 and the population standard deviation σ = 4.5. The sample size n = 150.
First step is to compute the mean
μ_X =μ = 12.
and the standard deviation
σ_X = σ/ √n = 4.5/√150= .3674
Let X = Weight of the fish population. Then, the distribution of X is, approximately, normally distributed.

P(X < 1900/150) = P(X < 12.6667)
= P([X - μ]/ σ_X < [12.6667 - μ]/σ _X)
≈ P(Z < [12.6667 - μ]/σ_X)
= P(Z < [12.6667 - 12]/.3674)
= P (Z < 1.8146) = normalcdf(-5, 1.8146)= .9652

Exercise 6.2.7. The amount X of water used when a person takes a shower has a mean μ = 30 gallons and standard deviation σ = 16 gallons. Suppose 36 people take a shower in a swimming pool facility. What is the probability that average of more than 25 gallons of water will be used by these 36 people.

Solution by TI-84:
Here the population mean μ = 30 and the population standard deviation σ = 16. The sample size n = 36.
First step is to compute the mean
μ_X =μ = 30.
and the standard deviation
σ_X = σ/ √n = 16/√36=2.6667
Let X = water used when a person takes a shower.
Then, the distribution of X is, approximately, normally distributed.

P(25 < X)
= P([25 - μ]/σ_X < [X - μ]/ σ_X )
≈ P([25 - μ]/σ _X ; < Z )
= P([25 - 30]/2.6667 < Z )
= P (-1.8750, < Z ) = normalcdf(-1.8750, 5)= .9696

Exercise 6.2.8. The waiting time for the campus bus has a mean μ= 7 minutes and the standard deviation σ = 2 minutes. A student used the bus 120 times in a month. What is the probability that the student would have waited more than 7.5 minutes, on an average, during the whole month?

Solution by TI-84:
Here the population mean μ =7 and the population standard deviation σ = 2. The sample size n = 120.
First step is to compute the mean
μ_X =μ =7.
and the standard deviation
σ_X = σ/ √n = 2/√120= .1826
Let X = waiting time for the bus.
Then, the distribution of X is, approximately, Normal.

P(7.5 < X)
= P([7.5 - μ]/σ_X < [X - μ]/ σ_X )
≈ P([7.5 - μ]/σ _X ; < Z )
= P([7.5 - 7]/.1826 < Z )
= P (2.7382, < Z ) = normalcdf(2.7382, 5)= .0031

Exercise 6.2.9. According to some data, the annual Kansas wheat export X has a mean 733 million dollars and standard deviation 163 million dollars. What is the probability that over the next 10 years mean wheat exports will exceed 804 million dollars?