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1. VTU Edusat Programme – 16
Subject : Engineering Mathematics
Sub Code: 10MAT41
UNIT – 8: Sampling Theory
Dr. K.S.Basavarajappa
Professor & Head
Department of Mathematics
Bapuji Institute of Engineering and of Technology
Davangere-577004
Email: ksbraju@hotmail.com
2. Statistical Inference:
It is necessary to draw some valid and reasonable conclusions
concerning a large mass of individuals or things. Every individual
or the entire group is known as population. Small part of this
population is known as a sample. The process of drawing some
valid and reasonable conclusion about the entire population is
Statistical Inference.
Random sampling:
A large collection of individuals or attributed or numerical
data can be understood as population or universe.
A finite subset of the universe is called a sample. The
number of individuals in a sample is called a Sample Size (n).
Sampling distribution:
For every sample size (n) we can compute quantities like
mean, median, standard deviation etc., obviously these will not be
the same.
Suppose we group these characteristics according to their
frequencies, the frequency distributions so generated are called
Sampling Distributions.
The sampling distribution of large samples are assumed to be
a normal distribution. The standard deviation of a sampling
distribution is also called as the standard error (SE).
Testing of Hypothesis
3. Making certain assumption to arrive at a decision regarding
the population a sample population will be referred to as
hypothesis
The hypothesis formulated for the purpose of its rejection
under the assumption that the true is called as the null hypothesis
denoted as H0 .
Errors
In a test process there can be four possible situations lead to
the two types of errors and same is tabulated as follows:
Accepting the
hypothesis
Rejecting the
hypothesis
Hypothesis is true Correct decision Wrong decision
Type I error
Hypothesis is false Wrong decision
Type II error
Correct decision
In order to minimize both these types of errors we need to increase
the sample size.
Significance level:
The probability level below which leads to the hypothesis is
known as the significance level. This probability is conventionally
fixed at 0.05 or 0.01 i.e., 5% or 1%
Therefore rejecting hypothesis at 1% level of significance,
implies that at 5% level of significance, there may be errors of
either types (Type I or II) is 0.05.
4. TESTS OF SIGNIFICANCE AND CONFIDENCE
INTERVALS
The process which helps us to decide about the acceptance or
rejection of the hypothesis is called as the test of significance.
Suppose that we have a normal population with mean μ and S
D as ߪ. If xത is the sample mean of a random sample size (n), the
quantity “t” defined by
ܜ =
ሺܠതି ࣆሻ√
࣌
(1)
is called as the standard normal variate (SNV) whose xത = 0 , σ =
1
From the table of the normal areas, we find that 95% of the
area lies between
t = -1.96 and t = 1.96
Further 5% level of significance is denoted by t0.05, therefore,
−1.96 ≤
ሺ୶തି ఓሻ√
ఙ
≤ 1.96
ఙ
√
൫– 1.96൯ ≤ xത − ߤ ≤
ఙ
√
1.96 (2)
ߤ ≤ xത +
ߪ
√݊
ሺ1.96ሻ and xത −
ߪ
√݊
ሺ1.96ሻ ≤ ߤ
∴ ܠത −
࣌
√
ሺ. ૢሻ ≤ ࣆ ≤ ܠത +
࣌
√
ሺ. ૢሻ
(3)
Similarly from the table of the normal areas 99% of the area
lies between
5. -2.58 and 2.58. This is equivalent to the form,
∴ ܠത −
࣌
√
ሺ. ૡሻ ≤ ૄ ≤ ܠത +
࣌
√
ሺ. ૡሻ
(4)
Therefore representation (3) is that 95% confidence interval
and Representation (3) is the 99% confidence level.
Graph:
Tests of significance for large samples:
Let N be the large sample having n members. Let p and q
denote number of success and failure respectively, then p+ q = 1.
By binomial distribution, N (p + q) n
denotes the frequencies of
samples. Therefore N (p + q) n
denotes the sampling distribution of
the number of successes in the sample.
We know that by binomial distribution ܠത = ܘܖ and ો = ඥܙܘܖ
then,
6. • Mean proportion of success =
୬୮
୬
= p
• S.D.(or S.E) proportion of success =
ඥܙܘܖ
ܖ
= ටቀ
ܙܘ
ܖ
ቁ
Let ‘x’ be the observed number of successes in a sample size
(n) and ૄ = .ܘܖ
The standard normal variate Z is defined as,
Z =
୶ିஜ
=
୶ି୬୮
ඥ୬୮୯
If Z ≤ 2.58, we conclude that the differences is highly significant
and reject the hypothesis. Then p ± 2.58ටቀ
ܙܘ
ܖ
ቁ be the probable
limits of Z.
p − 2.58ටቀ
ܙܘ
ܖ
ቁ ≤ ܈ ≤ p + 2.58ටቀ
ܙܘ
ܖ
ቁ
For a normal distribution, only 5% of members lie outside
μ ± 1.96 σ while only 1% of the members lie outside μ ± 2.58 σ
If x be the observed number of successes in the sample and Z is the
standard normal variate the Z =
୶ିஜ
=
୶ି୬୮
ඥ୬୮୯
We have the following test of significance
• If Z < 1.96, difference between the observed and expected
number of successes is not significant.
• If Z > 1.96 difference is significant at 5% level of
significance.
• If Z > 2.58, difference is significant at 1% level of
significance.
7. Example:
A coin is tossed 1000 times and it turns up head 540 times , decide
on the hypothesis is un biased .
Solution: Let us suppose that the coin is unbiased
P = probability of getting a head in one toss = 1/2
Since p + q = 1, q =
ଵ
ଶ
Expected number of heads in 1000 tosses ൟ = np
= 1000 ×
ଵ
ଶ
= 500
݂ ݎܾ݁݉ݑܰ ݈ܽݑݐܿܣℎ݁ܽ݀ݏ = 540 = ݔ
ݐℎ݁݊ ݔ − ݊ = 540 − 500 = 40
ܿݖ ݎ݁݀݅ݏ݊ =
ݔ − ݊
ඥ݊ݍ
=
40
ට1000 ×
ଵ
ଶ
×
ଵ
ଶ
= 2.53 < 2.58
∴ ݖ = 2.53 < 2.58 ⇉ 99% ሺܷ݊݀݁ݎሻ
⇉ ܶℎ݁ ݀݁ݏܾܽ݅݊ݑ ݏ݅ ݊݅ܥ
Example:
A survey was conducted in one locality of 2000 families by
selecting a sample size 800. It was revealed that 180 families were
illiterates. Find the probable limits of the literate families in a
population of 2000.
8. Solution: Probability of illiterate families = ܲ =
ଵ଼
଼
= 0.225
Also ݍ = 1 − ܲ = 1 − 0.225 = 0.775
Probability limits of illiterate families = ܲ ± 2.58ට
= 0.225 ± 2.58ඨ
ሺ0.225ሻሺ0.775ሻ
800
= 0.187 ܽ݊݀ 0.263
Therefore Probable limits of illiterate families in a sample of 2000
is
= 0.187ሺ2000ሻ ܽ݊݀ 0.263ሺ2000ሻ
= 374 and 526
Example:
A die was thrown 9000 times and a throw of 5 or 6 was
obtained 3240 times. On the assumption of random throwing, do
the data indicate an unbiased die.
Solution:
Suppose ‘the die is unbiased’
then Probability of throwing 5 or 6 with one die
= p(5) or p(6) = p(5) + p(6) = (1/6 ) + (1/6) = 1/3
q = 1-p = 1- (1/3) = 2/3
9. Then expected number of successes ሺnpሻ =
1
3
× 9000 = 3000
= μ ሺsayሻ
But the observed value of successes = 3240
Excess of observed value of successes = x − np = 3240 − 3000
= 240
Here n = 9000, p =
1
3
, q =, np = 3000
∴ S. D = ඥnpq = ඨ9000 × ൬
1
3
൰ × ൬
2
3
൰ = 44.72
∴ Z ሺSNVሻ =
x − np
ඥnpq
=
240
44.72
= 5.36 ≈ 5.4 > 2.58
⇉ Highly Signiϐicant
⇉ hypotheses is to be rejected at 1% level of Signiϐicance
∴ die is biased.
Example:
A biased dice is tossed 500 times a particular appears120 times.
Find the 95% confidence limit of obtaining the value. Also find the
standard error of proportion of success (Use binomial distribution).
Solution:
Let p =
ଵଶ
ହ
= 0.24
then q = 0.76, n = 500.
Standard error = 9.55
Then mean proportion of success = np/n = p = 0.24 and
10. mean proportion of S. E = ඥnpq /n= 0.019
then 95% confidence interval for proportion of success is
nሺ0.203ሻ ≤ np ≤ nሺ0.277ሻ
⇉ 500ሺ0.203ሻ ≤ np ≤ 500ሺ0.277ሻ
101 ≤ np ≤ 138
The interval is [101 , 138 ].
We say that with 95% confidence that out of 500 times always we
get particular number between 101 and 138 times.
Degrees of freedom (d.f )
It is the number of values in a set which may be set
arbitrarily.
d.f = n -1 for n number of observations
d.f = n -2 for n -1 number of observations
d.f = n -3 for n - 2 number of observations etc.
Ex: for 25 observations we have 24 d.f
Student’s t distribution
It is to test the significance of a sample mean for a normal
population where the population S is not known.
It is given by ܜ =
ሺܠതି ࣆሻ√
࢙
where
̅ݔ = ݉݁ܽ݊ =
∑ ௫
, ߤ = , ݊ܽ݁݉ ݊݅ݐ݈ܽݑ
11. ݏଶ
=
1
ሺ݊ − 1ሻ
ሺݔ − ̅ݔሻଶ
We need to test the hypothesis, whether the sample mean ̅ݔ differs
significantly from the population mean ߤ.
If the calculated value of t i.e. |ݐ| is greater than the table value of t
say t 0.05, we say that the difference between ̅ݔ and ߤ is significant
at 5% level.
If |ݐ| > t 0.01, the difference is significant at 1% level.
Note: 95% confidence limits for the population mean ߤ. Is
̅ݔ ± ሺ
௦
√
ሻ
Example:
A machine is expected to produce nails of length 3 inches. A
random sample of 25 nails gave an average length of 3.1 inches
with standard deviation 0.3 can it be said that the machine is
producing nail as per the specification.(value of students t 0.05 for
24 d.f is 2.064 )
Solution:
Given ߤ = 3 , ̅ݔ = 3.1 , n = 25 , s = 0.3
ܜ =
ሺܠതି ࣆሻ√
࢙
= 1.67 < 2.064
∴ The hypothesis that the machine is producing nails as per
speciϐication is accepted at 5% level of signiϐicance .
12. Example:
Ten individuals are chosen at random from a population and
their heights in inches are found to be 63, 63, 66, 67, 68, 69, 70,
71,71, test the hypothesis that the mean height of the universe is 66
inches (value of t 0.05 = 2.262 for 9 d.f).
Solution:
We have ߤ = 66 , n = 10, ∴ d.f = 9
̅ݔ =
∑ ௫
=
଼
ଵ
= 67.8
ݏଶ
=
ଵ
ିଵ
∑ሺݔ − ̅ݔሻଶ
=
ଵ
ଽ
ሾሺ63 − 67.8ሻଶ
+ … … +
ሺ71 − 67.8ሻଶሿ =9.067
S = 3.011
We have ܜ =
ሺܠതି ࣆሻ√
࢙
=
ሺૠ.ૡି ሻ
.
√ = 1.89 < 2.262 (given in
the problem)
⇉ The hypothesis is accepted at 5% level of significance.
Example:
Eleven school boys were given a test in drawing. They were
given a month’s further tution and a second test of equal difficulty
was healed at the end of it do the marks give evidence that the
students have benefitted by extra coaching (t 0.05 for d.f = 10) =
2.228
13. Boys 1 2 3 4 5 6 7 8 9 10 11
Marks
test 1
23 20 19 21 18 20 18 17 23 16 19
Marks
test 2
24 19 22 18 20 22 20 20 23 20 17
Chi-Square distribution: (࣑
)
It provides a measure of correspondence between the
Theoretical frequencies and observed frequencies
Let Oi ( i = 1 , 2 , ….. n ) – observed frequencies
Ei ( i = 1 , 2 , ….. n ) – estimated frequencies
The quantity ࣑
(chi square) distribution is defined as
࣑
= ∑
ሺି ሻమ
ୀଵ ; degrees of freedom = n-1
Chi – square test as a test of goodness of fit:
࣑
test helps us to test the goodness of fit of the distributions
such as Binomial, Poisson and Normal distributions.
If the calculated value of ࣑
is less than the table value of ࣑
at a specified level of significance, the hypothesis is accepted.
Otherwise the hypothesis is rejected.
14. Example:
A die is thrown 264 times and the number appearing on the
face (x) follows the following frequency distribution
x 1 2 3 4 5 6
f 40 32 28 58 54 60
Calculate the value of ࣑
Solution:
Frequencies in the given table are the observed frequencies.
Assuming that the die is unbiased the expected number of
frequencies for the numbers 1, 2, 3,4,5,6 to appear on the face is
264/6 = 44 each
Then the data is as follows
No. on the
die
1 2 3 4 5 6
Observed
frequency(Oi)
40 32 28 58 54 60
Expected
frequency(Ei)
44 44 44 44 44 44
࣑
= ∑
ሺି ሻమ
ୀଵ
࣑
= 22
15. Example:
Five dice were thrown 96 times and the numbers 1 or 2 or 3
appearing on the face of the die follows the following frequency
distribution
No. of
dice
showing 1
or 2 or 3
5 4 3 2 1 0
Frequency 7 19 35 24 8 3
Test the hypothesis that the data follows a binomial
distribution.
Solution:
Probability of a single die throwing 1 or 2 or 3 is
P = 1/6+1/6+1/6 = ½
q = ½
Binomial distribution to fit the data
Nሺq + pሻ୬
= 96
1
2
+
1
2
൨
ହ
=96 ቀ
ଵ
ଶ
ቁ
ହ
,96 × 5ܿଵ ×
ቀ
ଵ
ଶ
ቁ
ହ
, … … 96 ቀ
ଵ
ଶ
ቁ
ହ
∴ New table of values are
16. No. on the
die 12 or 3
5 4 3 2 1 0
Observed
frequency(Oi)
7 19 35 24 8 3
Expected
frequency(Ei)
3 15 30 30 15 3
࣑
= ∑
ሺି ሻమ
ହ
ୀ
࣑
0.05 = 11.7 > 11.07ሺ table valueሻ
Hence the hypothesis that the data follows a binomial
distribution is rejected.
Example:
Fit the Poisson distribution for the following data and test the
goodness of fit given that X2
0.05 = 7.815 for degrees of freedom =
4
Solution:
Poisson distribution to fit the data = Np(x) = Ne-m
mx
/x!
m = np =
∑ ௫
ே
=
ଵ
ଶ
x 0 1 2 3 4
f 122 60 15 2 1
17. Ne-m
mx
/x! = 200 ቂ
షభ/మሺଵ/ଶሻೣ
௫!
ቃ where x = 0, 1, 2, 3, 4
= 121, 61, 15, 3, 0
Therefore new table is
࣑
= ∑
ሺି ሻమ
ସ
ୀ
࣑
= 0.025 < ࣑
0.05 = 7.815
Therefore the fitness is considered good.
∴ The hypothesis that the fitness is good can be accepted.
Example:
The number of accidents per day (x) as recorded in a textile
industry over a period of 400 is given below. Test the
goodness of fit in respect of Poisson distribution of fit to the
given data
x 0 1 2 3 4
f(oi) 122 60 15 2 1
Ei 121 61 15 3 0
x 0 1 2 3 4 5
f 173 168 37 18 3 1
18. Solution:
Poisson distribution to fit the data = Np(x) = Ne-m
mx
/x!
m = np =
∑ ௫
ே
= 0.7825
Therefore new table is
࣑
= ∑
ሺି ሻమ
ହ
ୀ
࣑
= 12.297 ≈ 12.3 > ࣑
0.05 = 9.49
Therefore the fitness is not good
∴ The hypothesis that the fitness is good is rejected.
Example:
In experiments of pea breeding, the following frequencies of seeds
were obtained
Round &
yellow
Wrinkled &
yellow
Round &
green
Wrinkled &
green
total
315 101 108 32 556
Theory predicts that the frequency should be in proportion 9:3:3:1.
Examine the correspondence between theory and experiment.
x 0 1 2 3 4 5
f(oi) 173 168 37 18 3 1
Ei 183 143 56 15 3 0
19. Solution:
Corresponding frequencies are 313, 104, 104, 35.
࣑
= 0.51 < ࣑
0.05 = 7.815
⟹ The calculated value of ࣑
is much less than ࣑
0.05
⟹ There exists agreement between theory and experiment.
