This powerpoint has ten problems and solutions for test of number of success to test of significance for attributes.

1. Test for number of success
(Problems and solutions)
K.Sudha Rameshwari
Assistant Professor, Department of biochemistry,
V.V.Vanniaperumal college for women, Virudhunagar,
Tamilnadu, India

2. Tests of significance for attributes
• Attributes, we can find out the pressure or
absence of a particular characteristics.
• for example, in the study of attribute : literacy” a
sample may be taken and people classified as
literates and illiterates.
• while testing the significance for attributes,
binomial type of problem may be formed.
• The presence of an attribute is taken as success ie
P and its non-appearance as failure ie q.

3. Three tests
1. Tests for number of success
2. Test for proportion of success
3. Test for difference between proportion

5. Problems
1. In a sample of 400 population from a
village, 230 are found to be eaters of
vegetarian items and the rest non-
vegetarian items. Can we assume that
both vegetarian and non vegetarian food
are equally popular?

7. Since the difference between observed and
excepted number of vegetable eaters is more
than 1.96 SE at 5% level of significance. So, the
hypothesis is rejected or not supported.
We concluded that both vegetarian and non -
vegetarian

8. Problem-2:
In a hospital, 620 female and 600 male babies are born.
Do these figures confirm the hypothesis that males and
females are born in equal number?
Solution:
Let us take hypothesis that male and female babies
are born in equal number
ie p=q=1/2 ;n=1220
Therefore, the excepted frequencies of male and
female babies -1220/2
=610
Observed number of female babies =620

9. S.E of female babies =17.46
Difference between observed and excepted female babies=620-610
=10
Difference 10
--------------- = ------- =0.57
S.E 17.46
Since the difference between excepted and observed female babies
is less than 1.96 SE at 5% level of significance, it can be
concluded that the male and female babies are born in equal
number.

10. Problem 3:

11. Problem 4:

12. Problem 5:
A coin was tossed 400 times and the head
turned up 216 times. Test the hypothesis that
the coin is unbiased
Solution : Let us take hypothesis that the coin is
unbiased. On the basis of this hypothesis the
probability of getting head or tail would be equal
ie p=q=1/2 ; n=400
Observed number of heads =216
Expected frequency of head =400/2
=200

13. n=400; p=q=1/2
=10
Difference between observed of heads and expected number of
heads =216-200
=16
Difference 16
--------------- = ------- =1.6
S.E 10
The difference between observed and expected number of heads is
less than 1.96 S.E (5% level of significance). Hence the
hypothesis is accepted and we conclude that the coin is unbiased

14. Problem 6:
In 324 throws of a six-faced dice, odd points appeared 180 times. Would you
say that the dice is fair at 5 percent level of significance
Solution: let us take hypothesis that the dice is fair
ie p=q=1/2 ; n=324
Observed number of odd points =180
Expected frequency of odd points=324/2
=162

15. n = 324 ; p=q=1/2
=9
Difference between observed number of odd points and
expected number of odd points = 180-162
= 18
Difference 18
--------------- = ------- =2
S.E 9
The difference between observed and expected number of
heads is more than 1.96 S.E (5% level of significance),
the hypothesis is rejected . Hence we cannot say that the
die is fair at 5% level of significance

16. Problem 7:
A person throws 10 dice 5000 times and obtains 2560 times 4,5
or 6. Can this be attributed to fluctuations of sampling?
Solution : Let us take the hypothesis that there is no significant
difference in the sample results and the expected results
Since the difference is less than 1.96 S.E at 5% level of
significance the hypothesis holds true. Hence the
difference could be attributed to fluctuations of
sampling.

17. Problem 8:
A coin is tosses 10000 times and the head appeared 5195 times.
Would you consider the coin biased.
Solution : Let us take hypothesis that the coin is
unbiased. On the basis of this hypothesis the
probability of getting head or tail would be
equal ie p=q=1/2 ; n=10000
Observed number of heads =5195
Expected frequency of head =10000/2
=5000

18. n=10000; p=q=1/2
=50
Difference between observed of heads and expected
number of heads = 5195-5000
= 195
Difference 195
--------------- = ------- =3.9
S.E 50
The difference between observed and expected number
of heads is more than 1.96 S.E (5% level of
significance). Hence the hypothesis is rejected and we
conclude that the coin is biased

19. Problem 9:
In a hospital 480 female and 520 male babies are born in a week.
Do these figures confirm the hypothesis that males and females
are born equal number
Solution :Let us take hypothesis that the male
and female babies are born in equal number
ie p=q=1/2 ; n=1000
Observed number of male babies =520
Expected frequency of male babies =1000/2
=500

20. =15.81
Difference between observed and excepted number
of male babies are born=520 -500
=20
Difference 20
--------------- = ------- =1.265
S.E 15.81
The difference between observed and expected
number of male babies is less than 1.96 S.E (5%
level of significance). Hence the hypothesis is
accepted and we conclude that the both male and
female babies are born in equal number

21. Problem 10:
In a 600 throws of a six-faced die, odd points appeared 360
times. Would you say that the die is fair at 5% level of
significance?
Since the difference is more than 1.96 at 5%
level of significance , the hypothesis is
rejected . Hence we cannot say that the die is
fair at 5% level of significance

22. References
• Statistical methods –Dr.S.P.Gupta
• Statistical methods for biologists
(Biostatistics)- S.Palanichamy & M.Manoharan