When variance is unknown (if the population is finite with mean µ and standard deviation o or the sampling is without When variance is unknown (if the population is finite with mean u and standard deviation o or the sampling is without replacement) Example 1. Random samples with size 4 are drawn from a population having the values 14,19,26,20,44 and 47. Find the mean and the standard error of the sample means. replacement) Example 2. Consider a group of N = 4 people with the following ages: 16,18, 20 and 22. Consider samples of size n = 2 from the group. Find the mean and variance of the sampling distribution. Solution: Solution: A. Solving the population mean: A. Solving the population mean: 14+19+26+ 20+44+47 _ 170 16+18+20+22_ 76 =19 28.33 6 B. Solving for sample mean: x = µ = 28.33 B. Solving for sample mean: * = 4 x = µ = 19 C. To compute for the population standard deviation: (16 – 19)} + (18 – 19)* + (20 – 19) +(22 – 19) _ 20 C. To get the standard error of the sample means, compute first the population standard deviation: 14-28 33) - (19 - 28.33)* + (26 - 28.33)* + (20 - 28. 33) + (44 – 28.33)° + (47 –- 28.33 D. Solving for variance of the sample: o N-n 5 4-2 = =2.5•.67=1.675 N-1 2 4-1 205.35 87.05 5.43 + 69 39 + 245.55 + 348.5 - Te0. 22 - 12 66 of = D. Solving for standard of error of the mean: N-n 12.66 6-4 12.66 = 3.165..6325 =2.00 N-1 АCTIVITY: 1. Ifa population consist of the values (3,5,8,9). Consider possible samples of size 2. Find the mean and variance of the sample mean. 2. A population consists of three numbers (3,4,7). Consider all possible samples of size 2 which can be drawn from the population. Find the mean and variance of the sampling distribution of the sample means. 3. A population consist of five numbers 6, 9, 12, 15, and 18. Find the mean and variance of the sample means when sample size is 3.

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When variance is unknown (if the population is finite with mean
When variance is unknown (if the population is finite with mean
u and standard deviation o or the sampling is without
u and standard deviation o or the sampling is without
replacement)
Example 1. Random samples with size 4 are drawn from a
population having the values 14,19,26,20,44 and 47. Find the
mean and the standard error of the sample means.
replacement)
Example 2. Consider a group of N = 4 people with the following
ages: 16,18, 20 and 22. Consider samples of size n = 2 from the
group. Find the mean and variance of the sampling distribution.
Solution:
Solution:
A. Solving the population mean:
16+18+20+22_ 76
A. Solving the population mean:
14+19+26+20+44+47 _ 170
6.
= 28.33
=19
4
6
B. Solving for sample mean:
B. Solving for sample mean: * = 4
X = u = 28.33
I=μ= 19
C. To compute for the population standard deviation:
(16 – 19)* + (18 – 19)° +(20 – 19) +(22 – 19) _ 20
C. To get the standard error of the sample means, compute
first the population standard deviation:
(14- 28 33) + (19– 28.33)° + (26 – 28.33)* + (20 – 28. 33)° + (44 –- 28.33)° + (47 – 28.33F
4
D. Solving for variance of the sample:
o N-n 5 4–2
N-1 2 4-1
205.35 + 87.05 + S.43 + 69 39 + 245.53 + 348.57
961.33
- JT0. 22 - 12.66
of =
:= 2.5.67 =1.675
6
=-..
D. Solving for standard of error of the mean:
N-n
12.66
6-4
12.66
= 3.165 ..6325 = 2.00
N-1
4
6-1
4
АСTIVITY:
1. If a population consist of the values (3,5,8,9). Consider possible samples of size 2. Find the mean and variance of
the sample mean.
2. A population consists of three numbers (3,4,7). Consider all possible samples of size 2 which can be drawn from
the population. Find the mean and variance of the sampling distribution of the sample means.
3. A population consist of five numbers 6, 9, 12, 15, and 18. Find the mean and variance of the sample means when
sample size is 3.
Transcribed Image Text:When variance is unknown (if the population is finite with mean When variance is unknown (if the population is finite with mean u and standard deviation o or the sampling is without u and standard deviation o or the sampling is without replacement) Example 1. Random samples with size 4 are drawn from a population having the values 14,19,26,20,44 and 47. Find the mean and the standard error of the sample means. replacement) Example 2. Consider a group of N = 4 people with the following ages: 16,18, 20 and 22. Consider samples of size n = 2 from the group. Find the mean and variance of the sampling distribution. Solution: Solution: A. Solving the population mean: 16+18+20+22_ 76 A. Solving the population mean: 14+19+26+20+44+47 _ 170 6. = 28.33 =19 4 6 B. Solving for sample mean: B. Solving for sample mean: * = 4 X = u = 28.33 I=μ= 19 C. To compute for the population standard deviation: (16 – 19)* + (18 – 19)° +(20 – 19) +(22 – 19) _ 20 C. To get the standard error of the sample means, compute first the population standard deviation: (14- 28 33) + (19– 28.33)° + (26 – 28.33)* + (20 – 28. 33)° + (44 –- 28.33)° + (47 – 28.33F 4 D. Solving for variance of the sample: o N-n 5 4–2 N-1 2 4-1 205.35 + 87.05 + S.43 + 69 39 + 245.53 + 348.57 961.33 - JT0. 22 - 12.66 of = := 2.5.67 =1.675 6 =-.. D. Solving for standard of error of the mean: N-n 12.66 6-4 12.66 = 3.165 ..6325 = 2.00 N-1 4 6-1 4 АСTIVITY: 1. If a population consist of the values (3,5,8,9). Consider possible samples of size 2. Find the mean and variance of the sample mean. 2. A population consists of three numbers (3,4,7). Consider all possible samples of size 2 which can be drawn from the population. Find the mean and variance of the sampling distribution of the sample means. 3. A population consist of five numbers 6, 9, 12, 15, and 18. Find the mean and variance of the sample means when sample size is 3.
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