Presentation

1. STATISTICS - II
STAT (401)

2. UMAR DARAZ
Entomologist
BZU Multan
Statistic
Any numerical value calculated from sample is
known as statistic.
 Sample mean ( )
 Sample variance (s2)
 Sample standard deviation (s)

3. UMAR DARAZ
Entomologist
BZU Multan
Parameter
Any numerical value calculated from population
is known as parameter.
 Population mean (µ)
 Population variance ( )
 Population standard deviation ( )

4. UMAR DARAZ
Entomologist
BZU Multan
Testing of hypothesis
1) Hypothesis (Null and alternative hypothesis)
2) Level of significance (α= 1%, 2%, 5%, 10%)
3) Test-statistic (Formula)
4) Calculation (Value)
5) Critical region (Frequency curve)
6) Conclusion (Result)
Formulas:
z-test , t-test , χ -test , F-test

5. UMAR DARAZ
Entomologist
BZU Multan
Statistical hypothesis
A statistical hypothesis is a statement about one or
more parameters of population, this statement
may or may not be true.
 Null hypothesis
 Alternative hypothesis

6. UMAR DARAZ
Entomologist
BZU Multan
Null hypothesis
 A null hypothesis is a hypothesis which is to be
tested for possible rejection under the
assumption that is true.
 It is denoted by Ho
 For example, suppose we think that the average
age of ICS student is 16, then null hypothesis is
written as
Ho : µ = 16

7. UMAR DARAZ
Entomologist
BZU Multan
Alternative hypothesis
 An alternative hypothesis is a hypothesis which
is accepted when null hypothesis is rejected.
 It is denoted by H1 .
 It is written as
H1 : µ ≠ 16

8. UMAR DARAZ
Entomologist
BZU Multan
Three hypothesis
Null hypothesis
 Ho : µ = 16
 Ho : µ = 16
 Ho : µ = 16
Alternative hypothesis
H1 : µ ≠ 16
H1 : µ > 16
H1 : µ < 16

9. UMAR DARAZ
Entomologist
BZU Multan
Simple hypothesis
 A hypothesis in which all parameters of
population are specified is known as simple
hypothesis.
 Only sign will be used = , ≠
Ho : µ = 16
H1 : µ ≠ 16

10. UMAR DARAZ
Entomologist
BZU Multan
Composite hypothesis
 A hypothesis in which all parameters of
population are not specified is known as
composite hypothesis.
 Only sign will be used > or <
Ho : µ = 16 Ho : µ = 16
H1 : µ > 16 H1 : µ <16

11. UMAR DARAZ
Entomologist
BZU Multan
Type-1-error
 If we reject the null hypothesis when it is actually
true is known as type-1-error.
 E.g. A deserving player is not selected by team.
Type-2-error
 If we accept the null hypothesis when it is falls is
known as type-2-error.
 E.g. A non-deserving player is selected by team.
Level of significance
 Probability of making type-1-error is known as
level of significance.
 It is denoted by α

12. UMAR DARAZ
Entomologist
BZU Multan
Test-statistic
 A test-statistic is a function or formula of
sample observation that provides basis for
testing null hypothesis.
 Most commonly used test-statistic are z-test, t-
test, F-test and χ -test
 z-test, t-test and F-test are used for quantitative
data
 χ -test is used for qualitative data

13. UMAR DARAZ
Entomologist
BZU Multan
One sided test/One tailed test
 If the rejection region is located in only one
side of test-statistic is known as one sided test.
 It is also called composite hypothesis
 For > sign
Ho : µ = 16
H1 : µ > 16
 For < sign
Ho : µ = 16
H1 : µ < 16
- Table
value
0
+ Table
value
0

14. UMAR DARAZ
Entomologist
BZU Multan
Two sided test / Two tailed
test
 If the rejection region is located in both side of
sampling distribution of test-statistic is known
as two sided test.
 It is also called simple hypothesis
Ho : µ = 16
H1 : µ ≠ 16
+ Table
value

15. UMAR DARAZ
Entomologist
BZU Multan
Rejection region
 Rejection region is the part of sampling
distribution which leads to the rejection of null
hypothesis.
Acceptance region
 Acceptance region is the part of sampling
distribution which leads to the acceptance of
null hypothesis.

16. UMAR DARAZ
Entomologist
BZU Multan
Write a general procedure for
hypothesis testing single population
mean(µ)
1) Hypothesis
 Ho : µ = µo
 H1 : µ ≠ µo
 Ho : µ = µo
 H1 : µ > µo
 Ho : µ = µo
 H1 : µ < µo

17. UMAR DARAZ
Entomologist
BZU Multan
CONT…
2) Level of significance
α= 1%, 2%, 5%, 10%
3) Test-statistic
Conditions…
 If “ ” is given and the value of “n” is small or
large then we apply z-test
z
µ

18. UMAR DARAZ
Entomologist
BZU Multan
CONT…
 If “ ” is not given and n ≥ 30 then we apply z-
test
z
µ
 If “ ” is not given and n ≤ 30 then we apply t-
test
t
µ

19. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
S
5) Critical region
Frequency curve

20. UMAR DARAZ
Entomologist
BZU Multan
CONT…
6) Conclusion
 If calculated value lie in rejection region, then
we reject the null hypothesis and accept the
alternative hypothesis.
 If calculated value lie in acceptance region,
then we accept the null hypothesis and reject
the alternative hypothesis.

21. UMAR DARAZ
Entomologist
BZU Multan
Example : 1
A random sample of size n=36 is taken from a population with
variance =25 , sample mean =42.6 test a null hypothesis is Ho : µ
= 45 , H1 : µ < 45 with α= 5% and table value is 1.645
1) Hypothesis
 Ho : µ = 45
 H1 : µ < 45
2) Level of significance
α= 5%
3) Test-statistic
z
µ
One sided test

22. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
 =42.6
 µ = 45
 =25 , =5
 n=36
z
45
z
z
z

23. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in rejection region so, we
reject the null hypothesis and accept the
alternative hypothesis.
-2.88 -1.645

24. UMAR DARAZ
Entomologist
BZU Multan
Example : 2
A random sample of nine values are taken from a population 66, 68,
62, 61, 59, 67, 64, 66, 63, test a hypothesis that population mean is
60, =3, α= 10% and table value is 1.645
1) Hypothesis
 Ho : µ = 60
 H1 : µ ≠ 60
2) Level of significance
α= 10%
3) Test-statistic
z
µ
Two sided test

25. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
 = 64
 µ = 60
 =3
 n=9
z
z
z

26. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in rejection region so, we
reject the null hypothesis and accept the
alternative hypothesis.
-4 -1.645 +1.645 +4

27. UMAR DARAZ
Entomologist
BZU Multan
Example : 3
A random sample of size n=10 is taken from a population with
standard deviation =1.2 , sample mean =27 test a null hypothesis
is Ho : µ = 26.3 , H1 : µ > 26.3 with α= 5% and table value is 1.645
1) Hypothesis
 Ho : µ = 26.3
 H1 : µ > 26.3
2) Level of significance
α= 5%
3) Test-statistic
z
µ
One sided test

28. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
 =27
 µ = 26.3
 =1.2
 n=10
z
z
z
z

29. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in rejection region so, we
reject the null hypothesis and accept the
alternative hypothesis.
+1.645 +1.84

30. UMAR DARAZ
Entomologist
BZU Multan
Example : 4
A random sample of size n=49 is taken from a population with
standard deviation =4.2 , sample mean =125 test a null hypothesis
is Ho : µ = 123.5 , H1 : µ > 123.5 with α= 1% and table value is 2.33
1) Hypothesis
 Ho : µ = 123.5
 H1 : µ > 123.5
2) Level of significance
α= 1%
3) Test-statistic
z
µ
One sided test

31. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
 =125
 µ = 123.5
 =4.2
 n=49
z
z
z
z

32. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in rejection region so, we
reject the null hypothesis and accept the
alternative hypothesis.
+2.33 +2.5

33. UMAR DARAZ
Entomologist
BZU Multan
Example : 5
The marks obtain by students at a large number of colleges are
known to be normally distribution of mean 25 and a random sample
of 36 student has an average marks of 27 with standard deviation of 5
at 5% level of significance, what conclusion should be taken, table
value is 1.96
1) Hypothesis
 Ho : µ = 25
 H1 : µ ≠ 25
2) Level of significance
α= 5%
3) Test-statistic
z
µ
Two sided test

34. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
 = 27
 µ = 25
 s = 5
 n= 36
z
z
z
z

35. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in rejection region so, we
reject the null hypothesis and accept the
alternative hypothesis.
-2.40 -1.96 +1.96 +2.40

36. UMAR DARAZ
Entomologist
BZU Multan
Example : 6
A random sample of 12 graduate pass students from a certain typing
distribute was taken with a average speed 72.6 words/minute and
is 17.64 words/minute use α= 1% test the claim that institute graduate
average less than 75….table value is 2.718
1) Hypothesis
 Ho : µ = 75
 H1 : µ < 75
2) Level of significance
α= 1%
3) Test-statistic
t
µ
One sided test

37. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
 =72.6
 µ = 75
 =17.64 , =4.2
 n=12
t
75
t
t
t

38. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in acceptance region so,
we accept the null hypothesis and reject the
alternative hypothesis.
-2.718 - 1.98

39. UMAR DARAZ
Entomologist
BZU Multan
Example : 7
A random sample of 100 workers with children in day care shows a
mean day care cost of Rs.2600 and standard deviation of Rs.500
verify the department claim that the mean exceed Rs.2500 at 5%
level of significance with in this information and table value is 1.645
1) Hypothesis
 Ho : µ = 2500
 H1 : µ > 2500
2) Level of significance
α= 5%
3) Test-statistic
z
µ
One sided test

40. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
 =2600
 µ = 2500
 =500
 n=100
z
z
z
z

41. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in rejection region so, we
reject the null hypothesis and accept the
alternative hypothesis.
+1.645 +2

42. UMAR DARAZ
Entomologist
BZU Multan
Example : 8
A random sample of 100 observation from a population produced
sample value =182 and =299 test the hypothesis that Ho : µ = 180,
H1 : µ > 180 and α= 5% ,table value is 1.645
1) Hypothesis
 Ho : µ = 180
 H1 : µ > 180
2) Level of significance
α= 5%
3) Test-statistic
z
µ
One sided test

43. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
 =182
 µ = 180
 =299 , =17.29
 n=100
z
z
z
z

44. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in acceptance region so,
we accept the null hypothesis and reject the
alternative hypothesis.
+1.16 +1.645

45. UMAR DARAZ
Entomologist
BZU Multan
Write a general procedure for
hypothesis testing of difference
between population mean(µ1 µ2)
1) Hypothesis
 Ho : µ1 µ2 = 0
 H1 : µ1 µ2 ≠ 0
 Ho : µ1 µ2 = 0
 H1 : µ1 µ2 > 0
 Ho : µ1 µ2 = 0
 H1 : µ1 µ2 < 0

46. UMAR DARAZ
Entomologist
BZU Multan
CONT…
2) Level of significance
α= 1%, 2%, 5%, 10%
3) Test-statistic
Conditions…
 If “ 1 , 2” is given and the value of “ 1 , 2” is
small or large then we apply z-test
( 1 2 (µ1 µ2)
1 2

47. UMAR DARAZ
Entomologist
BZU Multan
CONT…
 If “ 1 , 2” is not given and 1 , 2 ≥ 30 then we
apply z-test
( 1 2 (µ1 µ2)
1 2
 If “ 1 , 2” is not given and 1 , 2 ≤ 30 then we
apply t-test
t
( 1 2 (µ1 µ2)
SP
1 2
SP=Pooled standard deviation

48. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
SP
SP
5) Critical region
Frequency curve

49. UMAR DARAZ
Entomologist
BZU Multan
CONT…
6) Conclusion
 If calculated value lie in rejection region, then
we reject the null hypothesis and accept the
alternative hypothesis.
 If calculated value lie in acceptance region,
then we accept the null hypothesis and reject
the alternative hypothesis.

50. UMAR DARAZ
Entomologist
BZU Multan
Example : 9
A random sample of size 36 from a population with variance 24 and
1=15 , a second sample size of 28 from an other population with
variance 80 and 2=13 test Ho : µ1 µ2 = 0, H1 : µ1 µ2 ≠ 0 at 5% of
level of significance and table value is 1.96
1) Hypothesis
 Ho : µ1 µ2 = 0
 H1 : µ1 µ2 ≠ 0
2) Level of significance
α= 5%
3) Test-statistic
( 1 2 (µ1 µ2)
1 2
Two sided test

51. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
 1 = 36 2 = 28
 = 24 = 80
 1 =15 2 =13
𝑧 =
(15 − 13) − (0)
24
36
+
80
28
𝑧 =
2 − 0
0.67 + 2.86
𝑧 =
2
3.53
𝑧 =
.
= ±1.06

52. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in acceptance region so,
we accept the null hypothesis and reject the
alternative hypothesis.
-1.96 -1.06 +1.06 +1.96

53. UMAR DARAZ
Entomologist
BZU Multan
Example : 10
A random sample of size 40 from a population produce the sample
values 1=70.4 , = 31.40 and other random sample size of 50 from
a second population produce the sample values 2=65.3, = 44.82
test Ho : µ1 µ2 = 2, H1 : µ1 µ2 > 2 at 5% of level of significance and
table value is 1.645
1) Hypothesis
 Ho : µ1 µ2 = 2
 H1 : µ1 µ2 > 2
2) Level of significance
α= 5%
3) Test-statistic
( 1 2 (µ1 µ2)
1 2
One sided test

54. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
 1 = 40 2 = 50
 = 31.40 = 44.82
 1 =70.4 2 =65.3
𝑧 =
(70.4 − 65.3) − (2)
31.40
40
+
44.82
50
𝑧 =
5.1 − 2
0.79 + 0.90
𝑧 =
3.1
1.69
𝑧 =
.
.
= 2.38

55. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in rejection region so, we
reject the null hypothesis and accept the
alternative hypothesis.
+1.645 +2.38

56. UMAR DARAZ
Entomologist
BZU Multan
Example : 11
On an examination, in a statistics course the average marks of 12
boys was 78 with a standard deviation of 6 while average marks of 15
girls was 74 with a standard deviation of 8 test the hypothesis that
there is no difference between perform of boys and girls, level of
significance is 1% and table value is 2.06
1) Hypothesis
 Ho : µ1 µ2 = 0
 H1 : µ1 µ2 ≠ 0
2) Level of significance
α= 1%
3) Test-statistic
t
( 1 2 (µ1 µ2)
SP
1 2
Two sided test

57. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
 1 = 12 2 = 15
 = 6 = 36
 = 8 = 64
 1 =78 2 =74
SP
t
( ̅1 ̅2) (µ1 µ2)
SP
1 2
SP
t
( ) ( )
7.18
SP
t 7.18 . .
SP
t 7.18 .
SP
=7.18 t = ±1.44

58. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in acceptance region so,
we accept the null hypothesis and reject the
alternative hypothesis.
-2.06 -1.44 +1.44 +2.06

59. UMAR DARAZ
Entomologist
BZU Multan
Example : 12
The test was given to a group of 100 scouts and to a group of 144
guides, the mean score for scout was 27.53 and mean score for guide
was 26.81 assuming a common population standard deviation of 3.48
using 5% level of significance without scout performance was better
than that of guides, table value is 1.645
1) Hypothesis
 Ho : µ1 µ2 = 0
 H1 : µ1 µ2 > 0
2) Level of significance
α= 5%
3) Test-statistic
( 1 2 (µ1 µ2)
1 2
One sided test

60. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
 1 = 100 2 = 144
 = 3.48 = 12.11
 1 =27.53 2 =26.81
(27.53 26.81) ( )
12.11 12.11
.
. .
.
.
.
.
= 1.6

61. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in acceptance region so,
we accept the null hypothesis and reject the
alternative hypothesis.
+1.6 +1.645

62. UMAR DARAZ
Entomologist
BZU Multan
Example : 13
The two sample A and B taken from population of standard deviation
0.8 test the difference of mean are low, A= 10.5, 11.6, 12.7, 12.9,
13.5, 13.6, 14.8 and B= 11.3, 12.4, 12.4, 13.9, 14.2, 14.7, 14.9, 15.6
and level of significance is 5% and table value is 1.96
1) Hypothesis
 Ho : µ1 µ2 = 0
 H1 : µ1 µ2 < 0
2) Level of significance
α= 5%
3) Test-statistic
( 1 2 (µ1 µ2)
1 2
One sided test

63. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
 1 = 7 2 = 8
 = 0.8 = 0.64
 1 =
1
10.5+11.6+12.7+12.9+13.5+13.6+14.8
=
.
=12.8
 2 =
1
11.3+12.4+12.4+13.9+14.2+14.7+14.9+15.6
=
.
=13.7
𝑧 =
(12.8 13.7) ( )
0.64 0.64
𝑧 =
.
. .
𝑧 =
.
.
𝑧 =
.
.
= − 2.19

64. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in rejection region so, we
reject the null hypothesis and accept the
alternative hypothesis.
-2.19 -1.96

65. UMAR DARAZ
Entomologist
BZU Multan
Write a general procedure for hypothesis
testing about the difference between mean
when the observations are paired
 Sometimes in testing hypothesis about two
means the two samples are dependent such
samples are known as pair sample.
 Suppose, we take a sample of 10 patients then
we get two measurements from each patient
blood pressure before medicine and blood
pressure after medicine and hence we obtain
paired data.
 Always be used t-test

66. UMAR DARAZ
Entomologist
BZU Multan
 Hypothesis testing technique in paired observation
is as follows…
1) Hypothesis
 Ho : µd = 0
 H1 : µd ≠ 0
 Ho : µd = 0
 H1 : µd > 0
 Ho : µd = 0
 H1 : µd < 0

67. UMAR DARAZ
Entomologist
BZU Multan
CONT…
2) Level of significance
α= 1%, 2%, 5%, 10%
3) Test-statistic
t
µ
s
4) Calculation
=
s
d= difference =After Before

68. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
Frequency curve
6) Conclusion
 If calculated value lie in rejection region, then
we reject the null hypothesis and accept the
alternative hypothesis.
 If calculated value lie in acceptance region,
then we accept the null hypothesis and reject
the alternative hypothesis.

69. UMAR DARAZ
Entomologist
BZU Multan
Example : 14
The follows are the weight of persons when they started weight
reducing diet and two week later,
Before =212,193,241,218,205,216,215,298,200,233,209,198
After = 195,185,225,199,194,193,205,176,188,224,201,195 and test a
hypothesis mean weight loss of person two weeks on this diet is at most
10 pound and level of significance is 1% and table value is 2.718
1) Hypothesis
 Ho : µd = 10
 H1 : µd < 10
2) Level of significance at most – (<)
α= 1% at least – (≥)
3) Test-statistic
t
µd
sd
One sided test

70. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
= 12
µ 𝒅 = 10
Before After d=After Before 𝟐
212 195 -17 289
193 185 -8 64
241 225 -16 256
218 199 -19 361
205 194 -11 121
216 193 -23 529
215 205 -10 100
298 176 -122 14884
200 188 -12 144
233 224 -9 81
209 201 -8 64
198 195 -3 9
𝟐

71. UMAR DARAZ
Entomologist
BZU Multan
CONT…
= =
sd
t
µ
s
sd
t
sd
t
sd
t
sd
t
sd

72. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in rejection region so, we
reject the null hypothesis and accept the
alternative hypothesis.
-3.39 -2.718

73. UMAR DARAZ
Entomologist
BZU Multan
Example : 15
10 young recruits were put through physical training program by the
army their weight were recorded before and after the training with the
following results…
Before = 125,195,160,171,140,201,170,176,195,139
After=136,201,158,184,145,195,175,190,190,145 and test the
hypothesis at the level of significance 5% and table value is 2.262
1) Hypothesis
 Ho : µd = 0
 H1 : µd 0
2) Level of significance
α= 5%
3) Test-statistic
t
µd
sd
Two sided test

74. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
= 10
µ 𝒅 = 0
Before After d=After Before 𝟐
125 136 11 121
195 201 6 36
160 158 -2 4
171 184 13 169
140 145 5 25
201 195 -6 36
170 175 5 25
176 190 14 196
195 190 -5 25
139 145 6 36
𝟐

75. UMAR DARAZ
Entomologist
BZU Multan
CONT…
= =
sd
t
µ
s
sd
t
sd
t
sd
t
sd
t
sd

76. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in acceptance region so,
we accept the null hypothesis and reject the
alternative hypothesis.
-2.262 -2.18 +2.18 +2.262

77. UMAR DARAZ
Entomologist
BZU Multan
Write a general procedure for testing
of hypothesis about variance of
population
1) Hypothesis
 Ho : =
 H1 : ≠
 Ho : =
 H1 : >
 Ho : =
 H1 : <

78. UMAR DARAZ
Entomologist
BZU Multan
CONT…
2) Level of significance
α= 1%, 2%, 5%, 10%
3) Test-statistic
4) Calculation

79. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
Frequency curve
6) Conclusion
 If calculated value lie in rejection region, then
we reject the null hypothesis and accept the
alternative hypothesis.
 If calculated value lie in acceptance region,
then we accept the null hypothesis and reject
the alternative hypothesis.

80. UMAR DARAZ
Entomologist
BZU Multan
Example : 16
Given two random sample of size 1=12 and 2=10 from two
population with 1=2.3 and 2=1.5 test the hypothesis at 5% level of
significance, Ho : = H1 : > and table value is 3.10
1) Hypothesis
 Ho : =
 H1 : >
2) Level of significance
α= 5%
3) Test-statistic
One sided test

81. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
 1 = 2.3 = 5.29
 2 = 1.5 = 2.25
5.29
2.25

82. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in acceptance region so,
we accept the null hypothesis and reject the
alternative hypothesis.
+2.35 +3.10

83. UMAR DARAZ
Entomologist
BZU Multan
Example : 17
Two random sample from two population are =20, 16,26,27,23, 22,
18,24,15,19 and =27,33,42,35,32,34,38,28,41,43,30,37 obtain
estimate variance of population and test the two population have
same variance at 5% level of significance and table value is 3.92
1) Hypothesis
 Ho : =
 H1 : ≠
2) Level of significance
α= 5%
3) Test-statistic
Two sided test

84. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
= 10
= 12
𝟐 𝟐
20 400 27 729
16 256 33 1089
26 676 42 1764
27 729 35 1225
23 529 32 1024
22 484 34 1156
18 324 38 1444
24 576 28 784
15 225 41 1681
19 361 43 1849
30 900
37 1369
𝟐 𝟐

85. UMAR DARAZ
Entomologist
BZU Multan
CONT…
.
.

86. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in acceptance region so,
we accept the null hypothesis and reject the
alternative hypothesis.
-3.92 -0.58 +0.58 +3.92

87. UMAR DARAZ
Entomologist
BZU Multan
Example : 18
In an experiments on reaction times in seconds of two individuals
A&B measured under ideal conditions. The following results are
obtained…
A=0.41, 0.38, 0.37, 0.42, 0.35, 0.38 and B=0.32, 0.36, 0.38, 0.33,0.38
test the hypothesis Ho : = , H1 : ≠ at 5% level of
significance and table value is 9.36
1) Hypothesis
 Ho : =
 H1 : ≠
2) Level of significance
α= 5%
3) Test-statistic
Two sided test

88. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
= 6
= 5
0.41 0.1681 0.32 0.1024
0.38 0.1444 0.36 0.1296
0.37 0.1369 0.38 0.1444
0.42 0.1764 0.33 0.1089
0.35 0.1225 0.38 0.1444
0.38 0.1444
𝑩 𝑩

89. UMAR DARAZ
Entomologist
BZU Multan
CONT…
. .
.
.

90. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in acceptance region so,
we accept the null hypothesis and reject the
alternative hypothesis.
-9.36 -0.86 +0.86 +9.36

91. UMAR DARAZ
Entomologist
BZU Multan
Example : 19
Two independent random sample of size = 10 and = 7 were
observed two have sample variance and using 10%
level of significance and test the hypothesis Ho : = ,H1 : ≠
and table value is 4.10
1) Hypothesis
 Ho : =
 H1 : ≠
2) Level of significance
α=10 %
3) Test-statistic
Two sided test

92. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
 = 16
 = 3
 = 10
 = 7
16
3

93. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in rejection region so, we
reject the null hypothesis and accept the
alternative hypothesis.
-5.33 -4.10 +4.10 +5.33

94. UMAR DARAZ
Entomologist
BZU Multan
Write a general procedure for testing
of hypothesis of checking association
between two attributes
 Always be used χ -test
 Always be used two sided test
1) Hypothesis
 Ho : There is no association between two attributes
 H1 : There is association between two attributes

95. UMAR DARAZ
Entomologist
BZU Multan
CONT…
2) Level of significance
α= 1%, 2%, 5%, 10%
3) Test-statistic
χ It will be used for more than two rows or two column
χ
| | .
It will be used for two rows and two column
Observed frequency
Expected frequency
4) Calculation

96. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
Frequency curve
6) Conclusion
 If calculated value lie in rejection region, then
we reject the null hypothesis and accept the
alternative hypothesis.
 If calculated value lie in acceptance region,
then we accept the null hypothesis and reject
the alternative hypothesis.

97. UMAR DARAZ
Entomologist
BZU Multan
Example : 20
In an investigation into eye color and left or right handed person, the
following results were obtained.
Do these results indicate at 5% level of significance and association
between eye color and left or right handed person and table value is
3.84
Left
handed
Right
handed
Blue 15 85 100
Brown 20 80 100
35 165 200 = n

98. UMAR DARAZ
Entomologist
BZU Multan
CONT…
1) Hypothesis
 Ho : There is no association between eye color and left
or right handed person
 H1 : There is association between eye color and left or
right handed person
2) Level of significance
α= 5%
3) Test-statistic
χ

99. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
Expected frequency of 15 =
×
and so on…
χ
15 17.5 2 4 0.23
85 82.5 2 4 0.05
20 17.5 2 4 0.23
80 82.5 2 4 0.05
= 0.56

100. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in acceptance region so,
we accept the null hypothesis and reject the
alternative hypothesis.
-3.84 -0.56 +0.56 +3.84

101. UMAR DARAZ
Entomologist
BZU Multan
Example : 21
Give the following table of hair color and eye color. Test the
hypothesis at 5% level of significance that there is no association
between hair color and eye color and calculate the co-efficient of
contingency.
Table value is 9.49
ontingency
χ
χ
Black Brown Grey
Black 15 5 20 40
Grey 20 10 20 50
Blue 25 15 20 60
60 30 60 150 = n
Hair color
Eye
color

102. UMAR DARAZ
Entomologist
BZU Multan
CONT…
1) Hypothesis
 Ho : There is no association between eye color and
hair color
 H1 : There is association between eye color and hair
color
2) Level of significance
α= 5%
3) Test-statistic
χ

103. UMAR DARAZ
Entomologist
BZU Multan
CONT…
4) Calculation
Expected frequency of 15 =
×
and so on…
χ
𝒐 𝒆 𝒐 𝒆 𝒐 𝒆
𝟐
𝒐 𝒆
𝟐
𝒆
15 16 -1 1 0.0625
20 20 0 0 0
25 24 1 1 0.0417
5 8 -3 9 1.125
10 10 0 0 0
15 12 3 9 0.75
20 16 4 16 1
20 20 0 0 0
20 24 -4 16 0.667
= 3.65

104. UMAR DARAZ
Entomologist
BZU Multan
CONT…
5) Critical region
 Frequency curve
6) Conclusion
 Calculated value lie in acceptance region so,
we accept the null hypothesis and reject the
alternative hypothesis.
-9.49 -3.65 +3.65 +9.49

105. UMAR DARAZ
Entomologist
BZU Multan
CONT…
Contingency:
χ
χ