International University IU

Powered by statisticsforbusinessiuba.blogspot.com

Stat

istic

s for

Bus

ines

s | C

hapt

er 0

5: S

ampl

ing

and

Sam

plin

g Di

sitr

ibut

ion

1

STATISTICS FOR BUSINESS [IUBA]

CHAPTER 05

SAMPLING AND SAMPLING DISTRIBUTION

The sampling distribution of a statistic is the probability distribution of all possible values the statistic may take when computed from random samples of the sample size, drawn from a specified population.

Sample Population Sampling Distribution Mean 푥̅ 휇 푋 Variance 푠 휎 푆 표푟푆 Standard Deviation 푠 휎 푆표푟푆 Proportion 푝̂ 푝 푃

PART I

SAMPLING DISTRIBUTION OF THE SAMPLE MEAN 퐗

The sampling distribution of 푋 is the probability distribution of all possible values the random variable 푋 may take when a sample of size 풏 is taken from a specified population.

The expected value of the sample mean 푿 is equal to the population mean 흁 and the standard deviation of 푿 is equal to the population standard deviation divided by the square root of the sample size.

The expected value of the sample mean is

푬(푿) = 흁

The standard deviation of the sample mean is

푺푫(푿) = 흈푿 = 흈/√풏

When sampling is done from a normal distribution with mean 흁 and standard deviation 흈, the sample mean 푿 has a normal sampling distribution:

푿~푵(흁,흈ퟐ/풏)

International University IU

Powered by statisticsforbusinessiuba.blogspot.com

Stat

istic

s for

Bus

ines

s | C

hapt

er 0

5: S

ampl

ing

and

Sam

plin

g Di

sitr

ibut

ion

2

Case 1: If the following two conditions are fulfilled:

The population is normally distributed. The population standard deviation 휎 is known (where that the

sample size either small, 푛 < 30, or large,푛 ≥ 30).

then we use:

푧푑푖푠푡푟푖푏푢푡푖표푛 (standard normal distribution) population mean 휇 standard deviation 휎/√푛

To find the probability with given values, we can use the following the formula:

풁 =푿 − 흁흈/√풏

To fine the values with given the probability, we can use the following the formula:

푿 = 흁 ± 풛흈√풏

Case 2: If the following two conditions are fulfilled:

The population is normally distributed. The population standard deviation 휎 is unknown. The sample size either small, 푛 < 30.

then we use: 푡푑푖푠푡푟푖푏푢푡푖표푛 (student distribution)

(The t distribution and its uses will be discussed in detail in Chapter 06. In addition, it would be impossible to use the calculator in order to find the probability of t distribution with given values. For these reasons, this kind of problems would not be mentioned in this chapter).

International University IU

Powered by statisticsforbusinessiuba.blogspot.com

Stat

istic

s for

Bus

ines

s | C

hapt

er 0

5: S

ampl

ing

and

Sam

plin

g Di

sitr

ibut

ion

3

Case 3: If the following two conditions are fulfilled:

The population is normally distributed. The population standard deviation 휎 is unknown. The sample size is large, 푛 ≥ 30.

then we use:

푧푑푖푠푡푟푖푏푢푡푖표푛 (standard normal distribution) population mean 휇 standard deviation 푠/√푛 where 푠 is sample standard deviation

To find the probability with given values, we can use the following the formula

풁 =푿 − 흁풔/√풏

To fine the values with given the probability, we can use the following the formula

푿 = 흁 ± 풛풔√풏

Example 1.1: (Case 1) The population is normally distributed. The population standard deviation 흈 is known. PROBLEM: Japan’s birthrate is believed to be 1.57 per woman. Assume that the population

standard deviation is 0.4. If a random sample of 200 women is selected, what is the probability that the sample mean will fall between 1.52 and 1.62?

SOLUTION: Assuming that the population is normally distributed.

휇 = 1.57 휎 = 0.4 푛 = 200

푃(1.52 ≤ 푋 ≤ 1.62) = 푃1.52 − 휇

휎√푛

≤푋 − 휇휎√푛

≤1.62 − 휇

휎√푛

International University IU

Powered by statisticsforbusinessiuba.blogspot.com

Stat

istic

s for

Bus

ines

s | C

hapt

er 0

5: S

ampl

ing

and

Sam

plin

g Di

sitr

ibut

ion

4

= 푃1.52− 1.570.4/√200

≤ 푍 ≤1.62 − 1.570.4/√200

= 푃 −5√24

≤ 푍 ≤5√24

≈ 0.9229

Example 1.2: (Case 3) The population is normally distributed. The population standard deviation 흈 is known. The sample size is large, 풏 ≥ ퟑퟎ. PROBLEM: A random sample with size 36 is taken from a normal population having mean µ = 25,

standard deviation is unknown. Sample standard deviation is s = 4.35. Find the probability that sample mean is at least 19.5.

SOLUTION: Assuming that the population is normally distributed.

휇 = 25 푠 = 4.35 푛 = 36

푃(푋 ≥ 19.5) = 푃푋 − 휇푠√푛

≥19.5 − 휇

푠√푛

= 푃 푍 ≥19.5 − 254.35/√36

= 푃(푍 ≥ −7.5862)

≈ 1 PART II

SAMPLING DISTRIBUTION OF THE SAMPLE PROPORTION 푷

The sampling distribution of the sample proportion 푃 is based on the binomial distribution with parameters 푛 and 푝, where 푛 is the sample size and 푝 is population proportion.

Condition: 풏풑 > 5 and 풏(ퟏ − 풑) > 5

As the sample size 푛 increases, the sampling distribution of 푃 approaches a

normal distribution with mean 푝 and standard deviation 푝(1 − 푝) 푛⁄

For all instances, we use 푧푑푖푠푡푟푖푏푢푡푖표푛 (standard normal distribution)

population proportion 푝

standard deviation 푝(1 − 푝) 푛⁄

International University IU

Powered by statisticsforbusinessiuba.blogspot.com

Stat

istic

s for

Bus

ines

s | C

hapt

er 0

5: S

ampl

ing

and

Sam

plin

g Di

sitr

ibut

ion

5

To find the probability with given values, we can use the following the formula

풁 =푷 − 풑

풑(ퟏ − 풑)풏

To fine the values with given the probability, we can use the following the formula

푷 = 풑 ± 풛풑(ퟏ − 풑)

풏

Example 2 PROBLEM: When sampling is done for the proportion of defective items in a large shipment,

where the population proportion is 0.18 and the sample size is 200, what is the probability that the sample proportion will be at least 0.20?

SOLUTION 푝 = 0.18

푛 = 200 푛푝 = 200× 0.18 = 36 > 5

푛(1 − 푝) = 200(1 − 0.18) = 164 > 5

푃 푃 ≥ 0.20 = 푃

⎝

⎛ 푃 − 푝

푝(1 − 푝)푛

≥0.20 − 0.18

0.18(1 − 0.18)200 ⎠

⎞ = 푃(푍 ≥ 0.7362) = 0.2308