Sampling Distributions

xx Sampling Distribution ofSampling Distribution of

Point EstimationPoint Estimation

IntroductionIntroduction

pp Sampling Distribution ofSampling Distribution of

Sampling Distribution for the Difference between Two Means

Sampling Distribution for the Difference between Two Proportions

A sampling distribution is a distribution of all of the possible values of a sample statistic for a given size sample selected from a population.

A sample is a portion of population.For example, suppose you sample 50 students

from your college regarding their mean GPA. If you obtained many different samples of 50, you will compute a different mean for each sample. We are interested in the distribution of all potential mean GPA we might calculate for any given sample of 50 students.

IntroductionIntroduction

The sample results provide only The sample results provide only estimatesestimates of the of the values of the population characteristics.values of the population characteristics. The sample results provide only The sample results provide only estimatesestimates of the of the values of the population characteristics.values of the population characteristics.

With With proper sampling methodsproper sampling methods, the sample results, the sample results can provide “good” estimates of the populationcan provide “good” estimates of the population characteristics.characteristics.

With With proper sampling methodsproper sampling methods, the sample results, the sample results can provide “good” estimates of the populationcan provide “good” estimates of the population characteristics.characteristics.

The reason is simply that the sample contains only aThe reason is simply that the sample contains only a portion of the population.portion of the population. The reason is simply that the sample contains only aThe reason is simply that the sample contains only a portion of the population.portion of the population.

The reason we select a sample is to collect data toThe reason we select a sample is to collect data to answer a research question about a population.answer a research question about a population. The reason we select a sample is to collect data toThe reason we select a sample is to collect data to answer a research question about a population.answer a research question about a population.

ss is the is the point estimatorpoint estimator of the population standard of the population standard deviation deviation .. ss is the is the point estimatorpoint estimator of the population standard of the population standard deviation deviation ..

In In point estimationpoint estimation we use the data from the sample we use the data from the sample to compute a value of a to compute a value of a sample statistic sample statistic that servesthat serves as an estimate of a as an estimate of a population parameterpopulation parameter..

In In point estimationpoint estimation we use the data from the sample we use the data from the sample to compute a value of a to compute a value of a sample statistic sample statistic that servesthat serves as an estimate of a as an estimate of a population parameterpopulation parameter..

Point EstimationPoint Estimation

xx

pp

Point estimationPoint estimation is a form of statistical inference. is a form of statistical inference. Point estimationPoint estimation is a form of statistical inference. is a form of statistical inference.

Relationships between the population distribution and the sampling distribution of the sample mean:

The mean of the sample means is exactly equal to the population mean

The dispersion of the sampling distribution of sample means is narrower than the population distribution.

The sampling distribution of sample means tends to become a bell-shaped and to approximate

Sampling Distribution of the Sample Mean

The probability distribution of is called its sampling distribution. It list the various values that can assume and the probability of each value of . In general, the probability distribution of a sample statistic is called its sampling distribution.

If a population is normal with mean μ and standard deviation σ, the sampling distribution of is also normally distributed with

andZ-value for the sampling distribution of

XX

X

X

X X n

( )XZ

n

X

The Central Limit Theorem

If all samples of a particular size are selected from any population, the sampling distribution of the sample mean is approximately a normal distribution.

This approximation improves with larger samples.

Sample Mean Sampling Distribution: If the Population is not Normal

We can apply the Central Limit Theorem:

Even if the population is not normal, sample means

from the population will be approximately normal

as long as the sample size is large enough.n↑

As the sample size gets large enough…

the sampling distribution becomes almost normal regardless of shape of population

Properties and Shape of the Sampling Distribution of the Sample Mean.

If n≥30, is normally distributed, where

Note: If the unknown then it is estimated by .

If n<30 and variance is known. is normally distributed

  If n<30 and variance is unknown. t distribution with

n-1 degree of freedom is use

2

~ ,X Nn

X

2s2

X

X2

~ ,X Nn

12~ n

XT t

sn

Example 1

Suppose a population has mean μ = 8 and standard deviation σ = 3. Suppose a random sample of size n = 36 is selected. What is the probability that the sample mean is between 7.8 and 8.2?Solution:

Even if the population is not normally distributed, the central limit theorem can be used (n > 30)

… so the sampling distribution of is approximately normal

with mean = 8 and standard deviation

x

xμ 0.536

3

n

σσx

0.31080.4)ZP(-0.4

363

8-8.2

nσ

μ- X

363

8-7.8P 8.2) X P(7.8

Example 2:

The amount of time required to change the oil and filter of any vehicles is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected.

What is the standard error of the sample mean to be?

What is the probability of the sample mean between 45 and 52 minutes?

What is the probability of the sample mean between 39 and 48 minutes?

Find the two values between the middle 95% of all sample means.

Solution:

X: the amount of time required to change the oil and filter of any vehicles

: the mean amount of time required to change the oil and filter of any vehicles

a) Standard error = standard deviation,

2~ 45,10 16X N n

X210

~ 45,16

X N

102.5

16

45 45 52 45b) 45 52

2.5 2.5

0 2.8

0.4974

P X P Z

P Z

39 45 48 45c) 39 48

2.5 2.5

2.4 1.2

0.4918 0.3849

0.8767

P X P Z

P Z

d) 0.95

45 45 0.95

2.5 2.5

0.95

from table:

1.96

1.96

45 1.96 40.1

2.545

1.96 49.92.5

a b

a

b

P a X b

a bP Z

P z Z z

z

z

aa

bb

Sampling Distribution for the Difference between Two Means

Suppose we have two populations, and

which are normally distributed. has mean

and variance while has mean and

variance . These two distributions can be

written as:

and

Now we are interested in finding out what is the

sampling distribution of the difference between two

sample means, the distribution of

1X 2X

1X 1 21 2X 2

22

21 1 1~ ,X N 2

2 2 2~ ,X N

1 2X X

2 21 2

1 2 1 21 2

~ ,X X Nn n

Example 3:

A taxi company purchased two brands of tires, brand A and brand B. It is known that the mean distance travelled before the tires wear out is 36300 km for brand A with standard deviation of 200 km while the mean distance travelled before the tires wear out is 36100 km for brand B with standard deviation of 300 km. A random sample of 36 tires of brand A and 49 tires of brand B are taken. What is the probability that thea) difference between the mean distance travelled before the tires of brand A and brand B wear out is at most 300 km?b) mean distance travelled by tires with brand A is larger than the mean distance travelled by tires with brand B before the tires wear out?

: the mean distance travelled before the tires of brand A wear out: the mean distance travelled before the tires of brand B wear out

Solution:

2X

1X

2 2

1 2

1 2

200 300~ 36300 36100,

36 49

~ 200,2947.846

X X N

X X N

1 2 1 2a) | | 300 300 300

300 200 300 200

2947.846 2947.846

9.21 1.84 0.9671

P X X P X X

P Z

P Z

1 2 1 2b) 0

0 200

2947.846

3.68 0.9999

P X X P X X

P Z

P Z

The population and sample proportion are denoted by p and , respectively, are calculated as,

and

where N = total number of elements in the population; X = number of elements in the population that

possess a specific characteristic;n = total number of elements in the sample; andx = number of elements in the sample that possess a

specific characteristic.

Sampling Distribution of the Sample Proportion p

Xp

N ˆ

xp

n

For the large values of n (n ≥ 30), the sampling distribution is very closely normally distributed.

Mean and Standard Deviation of Sample Proportion

Pp

P

pq

n

ˆ ,pq

p N pn

Example 4:If the true proportion of voters who support Proposition A is

what is the probability that a sample of size 200 yields a sample

proportion between 0.40 and 0.45?

0.40p

ˆ

(1 ) 0.4(1 0.4)σ 0.03464

200p

p p

n

0.40 0.40 0.45 0.40ˆ(0.40 0.45)

0.03464 0.03464

(0 1.44) 0.4251

P p P Z

P Z

Example 5:

The National Survey of Engagement shows about 87% of freshmen and seniors rate their college experience as “good” or “excellent”. Assume this result is true for the current population of freshmen and seniors. Let be the proportion of freshmen and seniors in a random sample of 900 who hold this view. Find the mean and standard deviation.

Let p the proportion of all freshmen and seniors who rate their college experience as “good” or “excellent”. Then,

p = 0.87 and q = 1 – p = 1 – 0.87 = 0.13

The mean of the sample distribution of is:

The standard deviation of is:

p

Solution:

ˆ 0.87p p

p

p

ˆ

0.87(0.13)0.011

900p

pq

n

Sampling Distribution for the Difference between Two Proportions

Now say we have two binomial populations with proportion of successes and respectively. Samples of size are taken from population 1 and samples of size are taken from population 2. Then and are the proportions from those samples.

1p

2p

2n1n

1p 2p

1 11 1

1

1ˆ ~ ,p p

P N pn

2 22 2

2

1ˆ ~ ,p p

P N pn

1 1 2 21 2 1 2

1 2

1 1ˆ ˆ ~ ,p p p p

P P N p pn n

Example 6:

A certain change in a process for manufacture of component parts was considered. It was found that 75 out of 1500 items from the existing procedure were found to be defective and 80 of 2000 items from the new procedure were found to be defective. If one random sample of size 49 items were taken from the existing procedure and a random sample of 64 items were taken from the new procedure, what is the probability that

a) the proportion of the defective items from the new procedure exceeds the proportion of the defective items from the existing procedure?

b) proportions differ by at most 0.015?

c) the proportion of the defective items from the new procedure exceeds proportion of the defective items from the existing procedure by at least 0.02?

:The proportion of defective items from the new procedure:The proportion of defective items from the existing procedure

Solution:

NP

EP

80 750.04 0.05

2000 15000.04(0.96) 0.05(0.95)ˆ ˆ~ 0.04, ~ 0.05,

64 49

N E

N E

p p

P N P N

0.05(0.95) 0.04(0.96)ˆ ˆ ~ 0.04 0.05,49 64

ˆ ˆ ~ 0.01,0.0016

N E

N E

P P N

P P N

ˆ ˆ ˆ ˆa) 0

0 0.01

0.0016

0.25

0.4013

N E N EP P P P P P

P Z

P Z

ˆ ˆ ˆ ˆb) | | 0.015 0.015 0.015

0.015 0.01 0.015 0.01

0.0016 0.0016

0.125 0.625

N E N EP P P P P P

P Z

P Z

0.2838

ˆ ˆ ˆ ˆc) 0.02 0.02

0.02 0.01

0.0016

0.75

0.2266

N E N EP P P P P P

P Z

P Z

Exercises:

BQT 173Institut Matematik Kejuruteraan, UniMAP25

1. Assume that the weights of all packages of a certain brand of cookies are normally distributed with a mean of 32 ounces and a standard deviation of 0.3 ounce. Find the probability that the mean weight of a random sample of 20 packages of this brand of cookies will be between 31.8 and 31.9 ounces.

Answer: 0.0667

2. According to the BBMG Conscious Consumer Report, 51% of the adults surveyed said that they are willing to pay more for products with social and environmental benefits despite the current tough economic times (USA TODAY, June 8, 2009). Suppose this result is true for the current population of adult Americans. There is a sample 1050 adult Americans who will hold the said opinion. Find the probability that the value of proportion is between 0.53 and 0.55.

Answer: 0.0921

3. It is known that 30% and 35% of the residents in Taman Sutera and Bandar Mas subscribe to New Straits Times newspaper respectively. If a random sample of 50 newspaper readers from Taman Sutera and 50 readers from Taman Mas were taken randomly, what is the probability that the proportion of New Straits Times subscribes in Taman Sutera is larger than Bandar Mas?

Answer: 0.2981

End of Chapter 1