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Module-V

Statistical Techniques-III
Application in Engineering:

1
...


3
...


The application of the t distribution to the following four types of problem will now be
considered
...

The mean and standard deviation of a sample are calculated and a value is postulated for the
mean of the population
...
Could both samples have
been taken from the same population?
Paired observations are made on two samples (or in succession on one sample)
...
In the medical literature, the Chi-square
is used most commonly to compare the incidence (or proportion) of a characteristic in one group
to the incidence (or proportion) of a characteristic in other group(s)
...

The application data were analysed using computer program MATLAB that performs these
calculations
...


BL-2,3

Syllabus
Sampling, Testing of Hypothesis and Statistical Quality Control: Introduction, Sampling Theory (Small and Large),
Hypothesis, Null Hypothesis, Alternative Hypothesis, Testing a Hypothesis, Level of Significance, Confidence Limits,
Test of Significance of Difference Means, T-test, F-test, and Chi-square test, One way Analysis of Variance (ANOVA),
Statistical Quality Control SQC, Control Charts, Control Chart for Variables (𝑋̅ and 𝑅 Charts) Control Charts for
Variables (𝑝, 𝑛𝑝 and 𝐶 Charts)
...

No
...
1

Topic

Page no
...
2

Test of significance

3

5
...
4

Snedecor’s Variance Ratio Test or F-test

9

5
...
6

Z test

23

5
...
8

Control Charts

31

5
...
Thus it is a collection of individuals or of their
attributes or of results of operations which can be numerically specified
...

Example: Universe of weights of students in a particular class, books in
college library
...

Example: Universe of pressures of various points in the atmosphere or stars
...

Example: Employ of central govt
...

(4) Hypothetical population- The collection of all possible ways in which
specified event can happen is called hypothetical universe
...
The number of individuals in a sample is called
sample size
...

Example: In a shop, we assess the quality of sugar, rice or any other commodity by taking
only a handful of it from the bag and then decide whether to purchase it or not
...
2 Test of Significance
Population mean and variance are denoted by 𝜇 and 𝜎 2 while sample mean and variance are
represented by 𝑥̅ and 𝑠 2
...

Null Hypothesis:-It is a definite statement about population parameter and denoted by 𝐻0
...

Alternative hypothesis:- Any hypothesis which complimentary to the null hypothesis (𝐻0 ) is
called an alternative hypothesis
...

For example, If we want to test the null hypothesis that the population has a specified mean
𝜇0 then we have
𝐻0 : 𝜇 = 𝜇0
Then alternative hypothesis will be
(i)
(ii)
(iii)

𝐻1 : 𝜇 ≠ 𝜇0 (two tailed alternative hypothesis)
𝐻1 : 𝜇 > 𝜇0 (right tailed alternative hypothesis)
𝐻1 : 𝜇 < 𝜇0 (left tailed alternative hypothesis)
Level of Significance
The probability level below which we reject the
known as level of significance
...
The
significance usually employed in testing of hypothesis are

hypothesis is
level of
5% and 1%
...
01)
5% (0
...
58
|𝑧𝛼 | = 1
...
33
𝑧𝛼 = 1
...
33
𝑧𝛼 = −1
...
1)
|𝑧𝛼 | = 1
...
28
𝑧𝛼 = −1
...
3 Student’s t-Distribution (t- Test)
t- distribution is used when sample size ≤ 30
...

If the standard deviation of the sample ′𝑠′ is given then t-statics is defined as
𝑥̅ − 𝜇
𝑡=
𝑠/√(𝑛 − 1)
Application of t-distribution:
1
...

2
...

3
...


•
•

Test I: To test whether the mean of a sample drawn from normal population deviates
significantly from a stated value when variance of population is unknown
...

Calculate t-statics:
𝑡=

𝑥̅ −𝜇
,
𝑆/√𝑛

where 𝑆 = √

∑(𝑥−𝑥̅ )2
𝑛−1

with degree of freedom (𝑛 − 1)
...

Note:
1
...
05 𝑆/√𝑛
2
...
01 𝑆/√𝑛
Example 1: A random sample size 16 and 53 as mean
...
Can this sample be regarded as taken from the population
having 56 as mean? Obtain 95% and 99% confidence limits of the mean of the population
...
e
...

Alternative hypothesis 𝐻1 : 𝜇 ≠ 56 (two tailed test)
...

•

∑(𝑥−𝑥̅ )2

Now we have 𝑆 = √

𝑛−1

135

= √ 15 = 3
⟹𝑡=

𝑥̅ − 𝜇
53 − 56
=
= −4
𝑆
3
√𝑛
√16
⟹ |𝑡| = 4
4

𝑑
...
05 = 2
...
05 = 2
...
Hence the sample mean has not
come from a population having 56 as mean
...
05 = 53 ±

3
√16

(2
...
4025, 54
...
95) = 50
...
2125
= 𝑥̅ ±
𝑡0
...
2

2
4
...
9

4
4
...
2

6
3
...
9

8
4
...
4

10
5
...
e
...

Alternative hypothesis 𝐻1 : 𝜇 ≠ 4000 ℎ𝑟𝑠 (two tailed test)
Now we have
𝑥̅ − 𝜇
𝑡=
𝑆/√𝑛
The calculation table is
𝑥
𝑥 − 𝑥̅
(𝑥
− 𝑥̅ )2

4
...
2
0
...
6
0
...
04

3
...
5
0
...
1
-0
...
09

5
...
8
0
...
4,
𝑛
10

3
...
6
0
...
9
-0
...
25

4
...
1
0
...
4
0
0

5
...
2
1
...
12

= 0
...
4 − 4
= 2
...
589
(
)
√10

Degree of freedom = 𝑛 − 1 = 10 − 1 = 9, 𝑡0
...
26
Since the calculated value of 𝑡 is less than the tabulated value of 𝑡 at 5% level of significance
...
i
...
the average lifetime of bulbs could be 4000 hrs
...
Before a heavy advertisement
...

5

After the campaign a sample of 26 shampoo was taken and the mean sales figure was found to
be 147 dozens with standard dozens with standard deviation 16
...
05,25 = 1
...
e
...
e
...

Given 𝑛 = 26, mean 𝑥̅ = 147, 𝑆
...

We know that
𝑡=

𝑥̅ −𝜇

,

(S
...
is known)

𝑠/√(𝑛−1)

=

(147 − 140)√25
= 2
...

16

The tabulated value of 𝑡 at 5% level of significance for 𝑑
...
708
...
e
...
05,25 =
1
...

Hence calculated value of |𝑡| = 2
...
05,25 = 1
...

∴ Null hypothesis is rejected
...


Test II : 𝑡 −test for difference of means of two small samples (from a normal population)
Let two samples 𝑥1 , 𝑥2 , … , 𝑥𝑛1 , 𝑦1 , 𝑦2 , … , 𝑦𝑛 2 of size 𝑛1 , 𝑛2 have been drawn from two normal
populations with mean 𝜇1 and 𝜇2 respectively under the assumption that the population
variance are equal (𝜎1 = 𝜎2 = 𝜎)
...
= 𝑛1 + 𝑛2 − 2
•

If two sample’s standard deviations 𝑠1 , 𝑠2 are given then we have 𝑆 2 =

•

If 𝑠1 , 𝑠2 are not given then 𝑆 2 =

2
2
∑(𝑥1 −𝑥
̅̅̅̅)
̅̅̅̅)
1 +∑(𝑥2 −𝑥
2

𝑛1 +𝑛2 −2

𝑛1 𝑠12 +𝑛2 𝑠22

...


Example 4: Two samples of sodium vapour bulbs were tested for length of life and the
following results were got:
Type I
Type II

Size
8
7

Sample mean
1234 hrs
1036 hrs

Sample S
...

36 hrs
40 hrs

Is the difference in the means significance to generalize that Type I is superior to Type II
regarding length of life?
Solution: Null hypothesis: 𝐻0 : 𝜇1 = 𝜇2 i
...
two types of bulbs have same lifetime
...
e
...

Hence we use right tailed test
...
076

∴ 𝑆 = 40
...
(1)
𝑡=

1234 − 1036
1 1
40
...
1480

𝑑
...
05 at 13 𝑑
...
77
...

∴ 𝐻0 is rejected
...

⟹ There are discriminate between two horses at 5% level of significance
...

Those of 9 randomly chosen soldiers are 61, 62, 65, 66, 69, 70, 71, 72 and 73
...

Solutions: Let 𝑥1 and 𝑥2 be the two samples denoting the heights of sailors and soldiers
...
e
...

Alternative hypothesis 𝐻1 : 𝜇1 > 𝜇2 (one tailed test)
We know that

̅̅̅̅
𝑥̅1 −𝑥
2

𝑡=
∑ 𝑥1
𝑛1

2
2
∑(𝑥1 −𝑥
̅̅̅̅)
̅̅̅̅)
1 +∑(𝑥2 −𝑥
2

𝑛1 +𝑛2 −2

1
+
𝑛1 𝑛2

𝑆√

𝑥1 =
̅̅̅

, where 𝑆 2 =
1

= 68,

𝑥2 =
̅̅̅

∑ 𝑥2
𝑛2

= 67
...
66
-5
...
66
1
...
34

(𝑥2 − ̅̅̅)
𝑥2 2
44
...
035
7
...
7556
1
...
34
3
...
34
5
...
4756
11
...
8356
28
...
0002

From (1) eq
...
038
⟹ 𝑡 = 0
...

The value of 𝑡 at 5% level of significance for 13 𝑑
...
77
...
05 = 1
...
The sailors are not on the average
taller than the soldier
...
The manufacturer of certain make of LED bulb claims that his bulbs have a mean life of 20
months
...
Can you regard
the producer’s claims to be validate 1% level of significance?
2
...
D
...
Test the hypothesis that it is a random
sample from a normal population with mean 45 units
...
The 9 items of a sample have the following values:
45, 47, 50, 52, 48, 47, 49, 53, 51
...
5 ?
4
...
In the light of these data, discuss the suggestion
that mean marks of population of students is 66
...
Samples of sizes 10 and 14 were taken from two normal populations with S
...
3
...
2 the
sample mean were found to be 20
...
6
...

6
...
D
...
A second
sample of 17 motors chosen from a different batch showed a mean life of 1280 hrs with a S
...

of 398 hrs
...
The marks obtained by a group of 9 regular course students and another group of 11 part time
course students in the a test are given bellow:
Regular 56
62
63
54
60
51
67
69
58
Part time 62
70
71
62
60
56
75
64
72
68
66
Examine whether the marks obtained by regular students and part time students differ
significantly at 5% and 1% level of significance
...
The average number of articles produced by two machines per day are 200 and 250 with
standard deviation 20 and 25 respectively on the basis of records of 25 days production
...
4 Snedecor’s Variance Ratio Test or F-test
This test is known as Fisher’s F-test or simply F-test
...

Let 𝑛1 and 𝑛2 be sizes of two samples with variances 𝑠12 and 𝑠22
...

3
...

2
...

AssumptionsThe populations for each sample must be normally distributed
...
The sample must be random and independent
...
That is why we take the larger
variance in the numerator of ratio
...

Whether the two independent estimates of the population variance are homogeneous or not
...
f
...

DecisionIf 𝐹𝑐𝑎𝑙 < 𝐹𝑡𝑎𝑏 , accept 𝐻0
...

Example 1: The random samples are drawn from two populations and the following results
we obtained:
Sample
20
16 26
27
23
22
18
24
25
19
𝑥
Sample
27
33 42
35
32
34
38
28
41
43
39
37
𝑦
Find variance of two populations and test whether two samples have same variance (Given
𝐹0
...
f
...
112)
Solution: Null hypothesis 𝐻0 : Let 𝜎12 = 𝜎22 i
...
two sample have same variance
...

Calculation of 𝐹 −statitistic:
𝑥
20

𝑥 − 𝑥̅
-2

(𝑥 − 𝑥̅ )2
4

𝑦
27
9

𝑦 − 𝑦̅
-8

(𝑦 − 𝑦̅)2
64

16
26
27
23
22
18
24
25
19

-6
4
5
1
0
-4
2
3
-3

36
16
25
1
0
16
4
9
9

220

0

120

33
42
35
32
34
38
28
41
43
30
37
420

-2
7
0
-3
-1
3
-7
6
8
-5
2
0

4
49
0
9
1
9
49
36
64
25
4
314

𝑛1 = 10, 𝑛2 = 12,
Degree of freedom 𝑣1 = 𝑛1 − 1 = 9, 𝑣2 = 𝑛2 − 1 = 11
𝑥̅ =

∑𝑥
𝑛1

,

𝑦̅ =

∑𝑦
𝑛2

𝑠12

∑(𝑥 − 𝑥̅ )2 120
=
=
= 13
...
5
𝑛2 − 1
11

Hence,
𝑠2

28
...
3 = 2
...
f
...
112 i
...
𝐹0
...
112
...
14 < tabulated value of 𝐹0
...
112
∴ The null hypothesis is accepted
...

Example 2: Two random samples drawn from 2 normal populations are as follows:
A
B

17
16

27
16

18
20

25
27

27
26

29
25

Test whether the samples are drawn from the same normal population
...

F-test :
Null hypothesis 𝐻0 : 𝜎12 = 𝜎22 i
...
the population variance do not differ
significantly
...
29
28
...
14
11
...
89
54
...
39
21
...
87

𝑥1 − ̅̅̅
𝑥1
-4
...
735
-3
...
375
5
...
735
-8
...
625

𝑥2
16
16
20
27
26
25
21

𝑥2 − ̅̅̅
𝑥2
-2
...
714
1
...
286
7
...
286
2
...
859

𝑛1 = 8, 𝑛2 = 7
𝑥1 = 21
...
714
̅̅̅
𝑠12 =

∑(𝑥1 − ̅̅̅)
𝑥1 2 253
...
267
𝑛1 − 1
7

𝑠22 =

∑(𝑥2 − ̅̅̅)
𝑥2 2 182
...
47
𝑛2 − 1
6
𝐹=

𝑠12
= 1
...

𝑠22

The table value of 𝐹 for 𝑣1 = 7 and 𝑣2 = 6 d
...
at 5% level is 4
...

Since 𝐹𝑐𝑎𝑙 < 𝐹𝑡𝑎𝑏 , ∴ null hypothesis is accepted
...

t-test:
Null hypothesis 𝐻0 : 𝜇1 = 𝜇2 i
...
the population means are equal
...
87+182
...
365
7
...
653
68
...
085
39
...
226

= 33
...
796
∴ 𝑡 = 0
...

11

The tabulated value of 𝑡 at 5% level of significance for 13 d
...
is 2
...
The
calculated value of 𝑡 less than tabulated value
...
There is no
significant difference between population mean i
...
𝜇1 = 𝜇2
...

Practice Questions
1
...


Sum of squares of
deviation from mean
90
108

2
...
D
...

3
...
The standard deviation calculated from two random samples of sizes 9 and 13 are 2:1 and 1:8
respectively
...
5 Chi-Square 𝝌𝟐 Test
The Chi-square test is very powerful test for testing the significance of the
discrepancy between actual (or observed) frequencies and theoretical (or
expected) frequencies
...
𝑛) is the set of observed frequencies
and 𝐸𝑖 (𝑖 = 1,2, … 𝑛) is the corresponding set of expected frequencies, then 𝜒 2
is defined as
𝑛

(𝑂𝑖 − 𝐸𝑖 )2
𝜒 = ∑[
],
𝐸𝑖
2

𝑖=1

where ∑ 𝑂𝑖 = ∑ 𝐸𝑖 = 𝑁 (total frequency)
Application of Chi-square test
(1) Test of independents of attributes – With the help of chi-square test we can
find out whether two or more attributes associated or not
...
Such as poison, binomial or normal
...

(4) Test for specified standard deviation i
...
it may be used to test of population
variance
...

The members of sample should be independent
...

Total frequencies 𝑁 should be reasonably large, say greater than 50
...

Step 2- Calculate the expected frequency 𝐸𝑖 corresponding to each cell
...

Step 4- See the value of 𝜒 2 from the table i
...
value of 𝜒 2 at 𝛼% level of
significance and for dof 𝑣, as calculated in step 3
...
Then accept the null
hypothesis
...
Then reject the null
hypothesis i
...
accept the alternative hypothesis
...
appeared on the
die
Frequency

1

2

3

4

5

6

40

32

29

59

57

59

29
46
289

59
46
169

57
46
121

59
46
169

Test whether the die is biased or not
...

The expected frequency for each digit is
𝑂𝑖
𝐸𝑖
(𝑂𝑖 − 𝐸𝑖 )2

40
46
36

276
6

= 46
...
30
𝐸𝑖
46

The tabulated value of 𝜒 2 at 5% level of significance for (6 − 1) = 5 dof is 11
...

Since calculated value of 𝜒 2 > the tabulated value
...
That is, die is not unbiased or die is biased
...

Examine the correspondence between theory and experiment
...

Calculation of expected frequenciesThe given frequencies in proportions 9:3:3:1
Total sum of proportions =9+3+3+1=16
...
75
556 = 104
...
25
1
×
16

556 = 34
...
75

104
...
25

34
...
016187

0
...
134892

0
...
470024
𝐸𝑖

The tabular value of 𝜒 2 at 5% level of significance for 𝑛 − 1 = 3 𝑑
...
815 i
...
𝜒0
...
815
Since calculated value of 𝜒 2 < the tabulated value
...
i
...
experimental result support the theory
...
of mistakes in a
0
page (𝑥)
No
...
1
116
...
1
4
...
5
0
...
41
1998
...
61

4
...
121
0
...
01

19
...
937

Fit a poisson distribution to the above data and test the goodness of fit
...
4821

The frequency of 𝑥 mistakes per page is given by the poisson law as follow:
𝑁(𝑥) = 𝑁𝑃(𝑥)
=

392[𝑒 −0
...
4821)𝑥 ]
𝑥!
=

242
...
4821)𝑥
𝑥!

Expected frequencies are
𝑁(0) = 242
...
69, 𝑁(2) = 28
...
52, 𝑁(4) = 0
...
052, 𝑁(6) = 0
...
937
𝐸𝑖

𝑑
...
05
= 5
...


∴ Poisson distribution is not a good fit to the given data
...
In a sample study, the following information was obtained
Days
Mon
Tue
Wed
Thurs
Fri
Sat
No
...
of parts demanded does not depend on the day of the week
...
of parts demanded are uniformly distributed i
...
no
...

Expected no
...
of parts

𝐸𝑖

(𝑂𝑖 − 𝐸𝑖 )2

Mon
Tue
Wed
Thurs
Fri
Sat

1124
1125
1110
1120
1126
1115

1120
1120
1120
1120
1120
1120

16
25
100
0
36
25

2
⟹ 𝜒cal
=∑

(𝑂𝑖 − 𝐸𝑖 )2
𝐸𝑖
0
...
22321
0
...
03214
0
...
180346

(𝑂𝑖 − 𝐸𝑖 )2
= 0
...
= 6 − 1 = 5
Tabulated value of 𝜒 2 for 5 𝑑
...
07
...


⟹ 𝐻0 , the null hypothesis is accepted
...
We take the null hypothesis that there is no association between
the attributes under study, i
...

𝐻0 : Given attributes are independent
...
, 𝐴𝑟 and 𝐵 divided into

---