Sampling Distribution
Sampling Distribution
• A sampling distribution is the distribution results if you actually
selected all possible samples.
• Sampling could be simple random sampling or restricted
random sampling (i.e. (i) Stratified sampling, (ii) Cluster
sampling, (iii) Systematic sampling, (iv) Multistage sampling).
Sampling Distribution
• Stratified sampling is based on the concept of homogeneity and
heterogeneity.
• In Cluster sampling, we divide the whole population into cluster or
area.
• In Systematic sampling, sample elements are selected from the
population at uniform intervals in terms of time order, or space.
• Multistage sampling involves the selection of sample in more than
one units.
Example 01: The population has 5 units (3,5,7,9,11) from
which a sample of 2 units is selected. Write down samples.
Find sample mean and show that the sample mean is equal
to population mean.
Solution: Population: (3,5,7,9,11), N=5
Sample size, n=2
So number of sample= 𝑁𝐶𝑛 = 5𝐶2 = 10
Samples are: 𝑆1 = (3,5), 𝑆2 = (3,7), 𝑆3 = (3,9), 𝑆4 = (3,11),
𝑆5 = (5,7), 𝑆6 = (5,9), 𝑆7 = (5,11), 𝑆8 = (7,9), 𝑆9 = (7,11), 𝑆10 = (9,11).
Example 01: cont…
Calculating sample mean:
3+5 3+7 3+9 3+11 5+7
𝑥1ҧ = = 4, 𝑥ҧ2 = = 5, 𝑥ҧ3 = = 6, 𝑥ҧ4 = = 7, 𝑥ҧ5 = =6
2 2 2 2 2
5+9 5+11 7+9 7+11 9+11
𝑥ҧ6 = = 7, 𝑥ҧ7 = = 8, 𝑥ҧ8 = = 8, 𝑥ҧ9 = = 9, 𝑥10
ҧ = = 10
2 2 2 2 2
4+5+6+7+6+7+8+8+9+10
sample mean, 𝑥ҧ = =7
10
3+5+7+9+11
Population mean, 𝜇 = =7
5
Example 02: select a sample of size 2 from a
population (3,5,6,8) (i) with replacement, (ii) without
replacement.
Solution: Population: (3,5,6,8), N=4, Sample size, n=2
(i) On basis of replacement, number of sample=𝑁 𝑛 = 42 = 16.
Samples are: 𝑆1 = (3,5), 𝑆2 = (3,6), 𝑆3 = (3,8), 𝑆4 = (5,3), 𝑆5 = (5,6), 𝑆6 = (5,8), 𝑆7 = (6,3),
𝑆8 = (6,5), 𝑆9 = (6,8), 𝑆10 = (8,3), 𝑆11 = (8,5), 𝑆12 = (8,6), 𝑆13 = (3,3), 𝑆14 = (5,5), 𝑆15 = 6,6
and 𝑆16 = (8,8).
(i) On basis of without replacement, number of sample= 𝑁𝐶𝑛 = 4𝐶2 = 6
Samples are: 𝑆1 = (3,5), 𝑆2 = (3,6), 𝑆3 = (3,8), 𝑆4 = (5,6), 𝑆5 = (5,8), 𝑆6 = (6,8).
Sampling distribution of mean when population is
normally distributed.
ҧ 𝑥ഥ
𝑥−𝜇
• Formula for z-test, 𝑧 =
𝜎𝑥ഥ
• 𝜇𝑥ҧ = sample mean= 𝜇
𝜎
• 𝜎𝑥ҧ =sample S.D. =
𝑛
ҧ
𝑥−𝜇 σ(𝑥−𝑥)ҧ 2
• If 𝜎 is unknown we use t-test instead of z-test, t = 𝑠 , 𝑠=
𝑛−1
𝑛
Example 03: the time between two arrivals in a queuing model is
normally distributed with a mean 2 minutes and SD 0.25 minute. If a
random sample of size 36 is drawn, what is the probability that the
sample mean will be greater than 2.1 minutes?
Solution: Since the population is normally distributed 𝜇𝑥ҧ = 𝜇 = 2
𝜎 0.25
and S.D. 𝜎𝑥ҧ = 𝑛
= 36
= 0.042
ҧ 𝑥ഥ
𝑥−𝜇
We will use, z = 𝜎𝑥ഥ
Therefore the probability that the sample mean will be greater than 2.1 minutes is given by,
𝑥ҧ − 𝜇𝑥ҧ 2.1 − 𝜇𝑥ҧ 2.1 − 2
𝑃 ≥ =𝑃 𝑧≥ = 𝑃[𝑧 ≥ 2.38]
𝜎𝑥ҧ 𝜎𝑥ҧ 0.042
From the table we get the area for 𝑧 = 2.38 is 0.99134.
Thus 𝑃 𝑥ҧ ≥ 2.1 = 1 − 0.99134=0.0087=0.87%
