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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 𝑠 classes 𝐵1 , 𝐵2 , …
...
𝐴𝑟

𝐵1

(𝐴1 𝐵1 )

(𝐴2 𝐵1 )

(𝐴3 𝐵1 )

…
...
(𝐴𝑟 𝐵2 )

(𝐵2 )

𝐵3

(𝐴1 𝐵3 )

(𝐴2 𝐵3 )

(𝐴3 𝐵3 )

…
...

…
...

…
...

…
...

…
...

…
...
(𝐴𝑟 𝐵𝑠 )

…
...

(𝐵𝑠 )

Total

(𝐴1 )

(𝐴1 )

(𝐴1 )

…
...
e
...

Eye color of son
Not Light
Not Light
230
Light
151

Eye color of
father

Light
148
471

Solution: Null Hypothesis 𝐻0 - The color of son’s eyes is not associated with the color of
father’s eyes, i
...
they are independent
...
29

Total
378

622

1000

Tabulated value of 𝜒 2 at 5% level, for 1 d
...
is 3
...
i
...
they are not independent, the color eye of son’s eye is
associated with that of father
...
of persons)

Not attacked
160
460

Total
190
600

620

790

Use 𝜒 2 -test to defend or refute the statement that the inoculation prevents
attack from choler
Solutions: Null hypothesis 𝐻0 −The inoculation does not prevent attack from
cholera
...
11
600 × 170
600 × 620
Not
=
129
...
89
(30 − 40
...
11)2 (140 − 129
...
886
149
...
11
(460 − 470
...
863
470
...
886
790

Tabulated value of 𝜒 2 at 5% level of significance for 1 d
...
is 3
...

2
2
Since 𝜒cal
> 𝜒tab
at 5% level of significance
...
Hence inoculation prevents attack from cholera
...
of good and bad parts
produced by each of three shift in a factory
...

950

45

995

Solution: Null hypothesis 𝐻0 : Production of bad part is independent of the
shift
...
77
𝑁
2985

(𝐴1 𝐵2 )0 =

(𝐴1 )(𝐵2 ) 2850 × 990
=
= 945
...
27
𝑁
2985

(𝐴2 𝐵2 )0 =

(𝐴2 )(𝐵2 ) 135 × 990
=
= 44
...

𝑁
2985

To calculate the value of 𝜒 2
Class

𝑂𝑖

𝐸𝑖

(𝑂𝑖 − 𝐸𝑖 )2

(𝐴1 𝐵1 )
(𝐴1 𝐵2 )
(𝐴1 𝐵3 )
(𝐴2 𝐵1 )
(𝐴2 𝐵2 )
(𝐴2 𝐵3 )

960
940
950
40
50
45

954
...
226
950
45
...
773
45

27
...
3110
0
27
...
3215
0

(𝑂𝑖 − 𝐸𝑖 )2
𝐸𝑖
0
...
02889
0
0
...
61022
0
1
...
28126
...
= (𝑟 − 1)(𝑠 − 1) = (2 − 1)(3 − 1) = 2
∴ The tabulated value of 𝜒 2 at 5% level of significance for 2 d
...
is 5
...

19

2
2
𝜒cal
< 𝜒tab
at
Yes
No
Total
level of
Yes
56
31
87
significance
...
The
Total 74
37
111
production of
bad part is independent of the shift on which they were produced
...
841)
Solution: Null hypothesis 𝐻0 : The option on sale of liquor is not dependent
with individual drinking
...
957

Hence the value of 𝜒 2 = 0
...
f
...
841
...


20

⟹ Sale of liquor is not dependent or not associated with the individual
drinking
...
Out of 300 customers, we find 88 prefer brown color, 65 prefer grey, 52
prefer red, 40 prefer blue and 55 prefer white
...

2
...
A survey of 320 families with 5 children shows the following
distribution:
No
...
of girls
No
...
05
= 10
...

3
...
The following table shows the distribution of digit in numbers chosen at random from the
telephone directory:
Digits
0
1
2
3
4
5
6
7
8
9
Frequency 1026 1107 997 966 1075 933 1107 972 964 853
Test whether the digits may be taken to occur equally frequently in the directory
...
The following data is collected on two characters:
Smokers
Non smokers
Literate
83
57
Illiterate
45
68
Based on this information can you say that there is no relation between habit of smoking and
literacy?
6
...
School
Low
200
400
High
1000
400

21

Z test
Test of significance for single Mean:
To test whether the difference between population mean is significant or
not
...
𝑋𝑛 be a random sample of size N with mean µ and variance
𝜎 2
...
A random sample of 900 members has a mean 3
...
Can it be
reasonably regarded as a sample from a large population of mean 3
...
D 2
...

Sol: Null Hypothesis: Assume that the sample is drawn from a large
population with mean 3
...
D 2
...
4 − 3
...
261
2
...
261 < 1
...

Question 2 : Intelligence tests were given to two groups of boys and girls

Girls

75

8

60

Boys

73

10

100

Examine if the difference between means scores is significant
Solution: Null Hypothesis : There is no significant difference between mean
scores
...
3912

√ +

As the calculated value of |𝑍| < 1
...
e there is
no significant difference between mean scores
...

Let ̅̅̅
𝑥2 be the mean of an independent sample of size 𝑛2 from another
population with mean 𝜇2 and variance 𝜎22
...
e
...
If 𝜎1 , 𝜎2 are not known and 𝜎1 ≠ 𝜎2 the test statistic in this case is 𝑧 =
̅𝑥̅̅̅−
𝑥̅̅̅
1 ̅
2
2

2

𝑠
𝑠
√ 1+ 2

𝑛1 𝑛2

Note 2
...
we use 𝜎 2 =

𝑛1 𝑠12 +𝑛2 𝑠22
𝑛1 +𝑛2

to calculate 𝜎
...
The average income of persons was ₹ 210 with a S
...
of ₹ 10 in
sample 100 people of a city
...
D
...
The S
...
of incomes of the people of
the city was ₹ 11
...

Sol
...

Null hypothesis: The difference is not significant
...
e
...
𝐻0 : ̅̅̅
𝑥1 = ̅̅̅
𝑥2
Alternative hypothesis
𝐻1 : ̅̅̅
𝑥1 = ̅̅̅
𝑥2 (two tailed test)
Test statistic:
Under 𝐻0 , 𝑧 =

̅𝑥̅̅̅−
𝑥̅̅̅
1 ̅
2
𝑠2 𝑠2
√ 1+ 2
𝑛1 𝑛2

=

210− 220
2

2

√10 +12

= −7
...
1428

100 150

Conclusion: As the calculated value of |𝑧| > 1
...
e
...

Exercise:
1
...
The variance of wheat farms in the country
may be taken as 100 kg
...

Answer Highly Significantly
...


The means of two large samples of 1000 and 2000 members are 168
...
Can the samples be regarded as drawn from the same
population of standard deviation 6
...

3
...
Their average weekly food expenditure is Rs
...
D
of Rs 40
...
220 with a S
...
45
...

Answer: Highly Significant
4
...
D of 192 kg
...
D of 224 kg
...
D’s significantly different?
Answer: z=4
...
The S
...
Let us suppose we want to test the
effectiveness of various doses of a drug on blood pressure
...
For example if we want to test four different doses we
make four groups (one control group and four experimental group)
...

One way ANOVA is used to see the effects of an independent variable on a
given dependent Variable
...
We calculate the variance of the
total population, variance between the groups and the variance within the
groups
...
A table showing the source of variation , the sum of squares, degree
of freedom, mean square and the formula for F ratio is known as ANOVA
table
...

Population, from which the samples are selected are normally distributed
(iii) Each of the sample is independent of the other sample
...

(i)

The various sum of squares involved in the computation of F statistic are:
SUM of the squares of variations amongst the columns(SSC):It is the sum
of squares of deviation between the columns
...
(ii) Sum of squares of Variations within columns (SSE): It is the
sum of the square of variation between individual items and the columns
means
...
(iii) Total sum of squares
of variation (SST): SST=SSC+SSE
𝑆𝑆𝑇

Total Variance= 𝑛−1
𝑀𝑆𝐶

Test Statistic: F statistic= 𝑀𝑆𝐸 with ( c − 1)and c(r − 1)degree of freedom
...
of deaths per
month
...
From these data suggest a difference in the no
...
of deaths
...
F
...
2
𝑛

=

(60)2
15

= 240

Sum of square of sample
∑ 𝑦𝐴 2 = 9 + 16 + 9 + 25 + 0 = 59
∑ 𝑦𝐵 2 = 36 + 9 + 9 + 16 + 16 = 86
∑ 𝑦𝐶 2 = 49 + 9 + 16 + 36 + 25 + 25 = 135
Total sum: ∑ 𝑦𝐴 2 + ∑ 𝑦𝐵 2 + ∑ 𝑦𝐶 2 − 240 = 40
(∑ 𝑦𝐴 )2

Sum of squares b/w samples=
=

I
...

III
...
5=2
samples
Within
30
12
2
...
89
Conclusion: 𝐹𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 < 𝐹𝑡𝑎𝑏𝑢𝑙𝑎𝑟 𝑠𝑜 𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠 𝑖𝑠 𝑎𝑐𝑐𝑒𝑝𝑡𝑒𝑑
Question
...
F
...
2
𝑛

=

(480)2
15

= 15360

𝑛 = 𝑛𝑜
...
= 15810 −
15360 = 450
27

2

(∑ 𝑦𝐴1 )

Sum of squares b/w samples=

𝑛1

2

2

+

(∑ 𝑦𝐴2 )
𝑛2

+

(∑ 𝑦𝐴3 )
𝑛3

− 𝐶
...

II
...


ANOVA Table
Note: ANOVA table: A table showing the source of variance , the sum of
squares , degree of freedom , mean square (variance) & the formula for F-ratio
is known as ANOVA Table
Source
Sum of Degree of Mean of sum Variance ratio of F
of
square
Freedom square
variation
250
b/w
250
2
F(2,12)=125/16
...
49
= 125
samples
2
200
Within
200
12
samples
12
= 16
...
89
Conclusion: 𝐹𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 > 𝐹𝑡𝑎𝑏𝑢𝑙𝑎𝑟 𝑠𝑜 𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠 𝑖𝑠 𝑁𝑂𝑇 𝑎𝑐𝑐𝑒𝑝𝑡𝑒𝑑
Questions:
1
...
Prepare
ANOVA table and test that varieties differ Significantly
...
To test the significance of the variations of the retail prices in the
commodity in three principal cities: Mumbai, Bangalore and Chennai , the four
shops were chosen at random in each city and prices observed in rupees were
as follows:
Mumbai
Bangalore
Chennai

16
14
4

8
10
10

12
10
8

28

14
16
8

3
...
Building an information system to Satisfy
the concept of prevention and control and improving upon product quality
requires statistical thinking
...

Control Chart
A control chart is a graphical representation of the collected information
...

Types of Control Chart:
There are many types of control chart designed for different control situations
...
It includes 𝑋̅ and R charts for 𝑋̅ and 𝜎
...
A fraction defective
control chart discloses erratic fluctuations in the quality of inspection which , may result in
improvement in inspection practice and inspection standards
...
𝑥𝑛 are has been observed
...
Then we find 𝑥̿ and 𝑅̅ as
𝑅 +𝑅 +⋯……
...
𝑜𝑓 𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑎𝑡𝑢𝑎𝑙𝑙𝑦 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑒𝑑

Control limits on p chart The control limits for p chart will be
CL= 𝑝̅
Upper and lower limits for 𝑝 − chart are
𝑝̅ (1 − 𝑝̅ )
𝑈𝐶𝐿𝑝 = 𝑝̅ + 3𝜎𝑝 = 𝑝̅ + 3√
𝑛
𝑝̅ (1 − 𝑝̅ )
𝐿𝐶𝐿𝑝 = 𝑝̅ − 3𝜎𝑝 = 𝑝̅ − 3√
𝑛

Control limits for np chart : the actual number of defectives called np chart
is used
Upper and lower limits for 𝑛𝑝 − chart are
𝐶𝐿 = 𝑛𝑝̅ where 𝑝̅ =

∑ 𝑛𝑝
∑𝑛

𝑈𝐶𝐿𝑛𝑝 = 𝑛𝑝̅ + 3𝜎𝑛𝑝 = 𝑛𝑝̅ + 3√𝑛𝑝̅ (1 − 𝑝̅ )
𝐿𝐶𝐿𝑛𝑝 = 𝑛𝑝̅ − 3√𝑛𝑝̅ (1 − 𝑝̅ )

31

Question1 The following are the mean heights and ranges of lengths of a finished product
from 10 samples each of size 5
...
Construct 𝑋̅ and R
chart and examine whether the process is under control and state your recommendations
...
1
201
Mean (𝑋̅)
Range (𝑅) 5

2
198
0

3
202
7

4
200
3

5
203
3

6
204
7

7
199
2

8
196
8

9
199
5

10
201
6

Assume for n=5, 𝐴2 = 0
...
11 𝑎𝑛𝑑 𝐷3 = 0
Solution : CL=200 ; UCL (for 𝑋̅ -chart) =𝑥̿ + 𝐴2 𝑅̅=200 + (0
...
6) =
202
...
58 × 4
...
332
Control limit for R chart:
(CL)= 𝑅̅=4
...
11 × 4
...
706
LCL (for R chart) =𝐷3 𝑅̅=0 × 4
...
Hence the
process is under control
...

Question2
...
Draw the np chart for the following table and examine whether the process is
under control?
Date

1

2

3

4

5

6

32

7

8

9

10

11

12

13

14

15

No
...
011
∑𝑛
1000 × 15
Control limits are 𝐶𝐿 = 𝑛𝑝̅ =1000 × 0
...
011)
= 20
...
011) = 1
...

Question3
...

1
2
3
4
5
6
7
8
9
10

No
...

11
12
13
14
15
16
17
18
19
20

33

No
...

Solution :
𝑝̅ =

𝑁𝑜 𝑜𝑓 𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒𝑠 𝑓𝑜𝑢𝑛𝑑 𝑖𝑛 𝑎𝑛𝑦 𝑖𝑛𝑝𝑒𝑐𝑡𝑖𝑜𝑛
120
=20×100
𝑇𝑜𝑡𝑎𝑙 𝑛𝑜
...
06 +
𝑛

𝑈𝐶𝐿𝑝 = 𝑝̅ + 3𝜎𝑝 = 𝑝̅ + 3√

𝑝̅ (1−𝑝̅ )
=0
...
66

0
...
06)
100

0
...
06)
100

= 0
...
01095

Since the fraction defective can not be negative
...
1537,
calculate the control limits Given that sub- group size is 2000
...
1537

Sub group size is 2000
n=2000
Central line 𝑈𝐶𝐿𝑛𝑝 = 𝑛𝑝̅ + 3𝜎𝑛𝑝 = 𝑛𝑝̅ + 3√𝑛𝑝̅ (1 − 𝑝̅ )
=307
...
4(1 − 0
...
787742
𝐿𝐶𝐿𝑛𝑝 = 𝑛𝑝̅ − 3𝜎𝑛𝑝 = 𝑛𝑝̅ − 3√𝑛𝑝̅ (1 − 𝑝̅ )
=307
...
4(1 − 0
...
01225
Links
Population and sampling
https://www
...
com/watch?v=VPM84_yfx5Q
https://www
...
com/watch?v=iUutXUIwAvw&t=60s
Testing of hypothesis
https://www
...
com/watch?v=7kb-a7n2bcQ&t=16s
https://www
...
com/watch?v=1bywo64eIC0&t=344s
t-test
https://www
...
com/watch?v=UW1tUJFmmm8
https://www
...
com/watch?v=fF0tFUQv-c8
https://www
...
com/watch?v=gePqxb9Vxuo
F-test
https://www
...
com/watch?v=h5Glm738j84
https://www
...
com/watch?v=7eTO7faJqSg

34

https://www
...
com/watch?v=PmGHMzZ1u3Q
Chi Square test
https://www
...
com/watch?v=3aRIwDDMc88
https://www
...
com/watch?v=dXB3cUGnaxQ
https://www
...
com/watch?v=hpWdDmgsIRE
Z Test
https://www
...
cyom/library/view/common-statisticalmethods/9781607642282/ch06
...
au
...
mil/au/awc/awcgate/navy/bpi_manual/mod10control
...
youtube

---