This document presents hypotheses concerning proportions. It introduces proportions and the binomial distribution. It discusses hypotheses for one proportion using a z-test and provides an example. It also discusses hypotheses for two proportions using a z-test and provides an example. Finally, it discusses analyzing contingency tables using a chi-squared test and provides an example. References for further information are also included.

1. HYPOTHESES CONCERNING
PROPORTIONS
Submitted to:
Prof. Rakesh Jain
Dated: 20-Nov-2013
Presented By:
• Kapil Jain
• Charan SIngh
2013PIE5199
2012PMM5004
1

2. Overview
 Introduction of Proportions
 Hypotheses Concerning One Proportion with
Example
 Hypotheses Concerning Two Proportions with
Example
 Analysis of Contingency Table with Example
 References
2

3. Introduction of Proportions
 For estimating a proportion of occurrences in a
population, Statisticians use Sample.
 Sample Discrete-Binomial Distribution.
Population, p =
X
x
n
N
Sample
Population
 Sample
size
Distribution.
x
n
increases,
use
Continuous
Normal
3

4.  Binomial Distribution:
 Mean, µ=n p
 Standard Deviation, σ= √(n p q)
n= Number of trails
p= probability of success
q= 1-p= probability of a
failure
 Mean of sampling Distribution of the proportion
 µ = (n p)/ n = p
 Standard Deviation, σ=
npq
n
pq
n
 n is large, for large sample Confidence Interval for p :
x
n
Z
x
n
2
1
n
x
n
 p 
x
n
Z
x
n
2
1
x
n
n
 Where α = level of significance
4

5.  Now, Error, E= p- x
n
 So, Maximum Error, E max= p-
x
n
Z
/2
x
x
1
n
n
n
Z
/2
p(1 p)
n
Physical Significance
 U.S.A. Government estimates proportion of unemployed
people by sampling procedure.
 It is quite beneficial in Acceptance Sampling.
5

6. Hypotheses Concerning One Proportion
In many methods, NULL Hypotheses Proportion equals some
specified constant(po)
 Statistic for large Sample Test Concerning p
Z
X npo
npo 1 po
 Critical Region for testing p= po (Large Sample)
Alternative
Hypothesis
Reject Null
Hypothesis if
p < p0
p > p0
Z < -zα
Z > zα
p ≠ p0
Z < -zα/2
P ≠ p0
Z > zα/2
6

7. Hypotheses Concerning One Proportion
Example-A manufacturer of submersible pumps claims that at
most 30% of the pumps require repairs within the first 5 years of
operation. If a random sample of 120 of these pumps includes 47
which required repairs within the first 5 years, test the NULL
hypotheses p=0.30 against the alternative hypotheses p > 0.30 at
the 0.05 level of significance.
Solution- 1. NULL Hypotheses: p=0.30
Alternative Hypotheses: p > 0.30
2. Level of Significance: α = 0.05
3. Criterion: Reject the NULL hypotheses, If Z > 1.645 (from table)
Where
Z
X npo
npo 1 po
7

8. 4. Calculations: Substituting x=47, n=120 and po=0.30 into the
formula, we get
Z
47 120 (0.30 )
120 0.30 1 .0.30
2.19
5. Decision: Since Z=2.19 is greater than 1.645, We reject the NULL
hypotheses at level 0.05. In other words, there is sufficient
evidence to conclude that the proportion of repairable pumps are
greater than 0.30 or 30%.
-4
-2
0
1.645
2
2.19
4
8

9. Hypotheses Concerning Two Proportion
When we compare Two different products, whether proportion of a
given process remains constant from day by day.
 Statistic for test concerning difference between Two Proportions:
x1 x 2
n1 n 2
Z
1
1
ˆ
ˆ
p1 p
n1 n 2
 Where Pooled Estimator
ˆ
p
x1
n1
x2
n2
 Large Sample Confidence Interval for the difference of two
Proportions:
x1
n1
x2
n2
Z
/ 2
x1
x1
1
n1
n1
n1
x2
x2
1
n2
n2
n2
9

10. Hypotheses Concerning Two Proportions
Example-A Study showed that 64 of 180 persons who saw a
photocopying machine advertised during the telecast of a baseball
game & 75 of 180 other persons who saw it advertised on a variety
show remembered the brand name 2 hours later. Use the Statistic
to test at the 0.05 level of significance whether the difference
between the corresponding sample proportions is significant.
Solution- 1. NULL Hypotheses: p1=p2
Alternative Hypotheses: p1 ≠ p2
2. Level of Significance: α = 0.05
3. Criterion: Reject the NULL hypotheses, If Z > 1.96 & Z <-1.96(from table)
x1
n1
Where Z
ˆ
p1
x2
n2
1
ˆ
p
n1
1
n2
10

11. 4. Calculations: Substituting x1=64, n1=180, x2= 75and n2=180 into
the formula, we get
64 75
ˆ
p
Z
0.386
180 180
64
180
75
180
1
0.386 0.614
180
1.191
1
180
5. Decision: Since Z=1.191 is lesser than 1.96, We are failed to reject the NULL
hypotheses at level 0.05. In other words, there is sufficient evidence to
conclude that the difference between the corresponding sample proportions is
not significant.
1.191
-4
-2
0
-1.96
2
1.96
4
11

12. Analysis of Contingency Table
 Practically, We have to compare more than two sample
proportions, so we arrange the statistic in the form of Contingency
Observed
Table
Frequency,
fo/Oij
Success
Sample 1
Sample 2
..........
Sample k
Total
x2
..........
xk
x
Failure
x1
x1
n1-x1
n2-x2
..........
nk-xk
n-x
Total
n1
n2
..........
nk
n
Row.Total . X .Column.Total
 Expected Frequency fe/eij=
Grand .Total , n
A Chi Square Statistic,
2
fo
fe
fe
2
o
e
e
2
12

13. Analysis of Contingency Table
Example-A large electronics firm that hires many workers
with disabilities wants to determine whether their
disabilities affect such workers performance. Use the level of
significance α=0.05 to decide on the basis of the sample data
shown in the following table whether it is reasonable to
maintain that the disabilities have no effect on the worker’s
performance:
Performance
Above average
Average
Below average
Blind
21
64
17
Deaf
16
49
14
No disability
29
93
28
13

14. Solution- 1. NULL Hypotheses: p1=p2=p3
Alternative Hypotheses: p1 ≠ p2 ≠ p3
2. Level of Significance: α = 0.05
3. Criterion: Reject the NULL Hypotheses, If χ2 > 11.143 & χ2<11.143, the value of for (3-1)(3-1)=4 degree of freedom, where χ2
is given by the formula.
4. Calculation: Calculating first the expected frequency for all the
cells.
Total
21
64
17
102
16
49
14
79
29
93
28
150
66
206
59
331
e11=(102x66)/331= 20.33
e12=(102x206)/331=63.48
e13=(102x59)/331=18.18
e21=(79x66)/331= 15.75
e22=(79x206)/331=49.166
e23=(79x59)/331=14.08
e31=(150x66)/331= 29.90
e32=(150x206)/331=93.35
e33=(150x59)/331=26.73
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15. 2
21 20 .33
20 .33
14 14 .08
14 .08
2
2
64 63 .48
63 .48
2
17 18 .18
18 .18
2
29 29 .90
29 .90
2
93 93 .35
93 .35
2
16 15 .75
15 .75
28 26 .73
26 .73
2
49 49 .16
49 .16
2
2
0.1958
5. Decision: Since χ2=0.1958 before 11.143, we are failed to reject
NULL hypotheses. We conclude that the disabilities have no
effect on the worker’s performance.
15

16. references
 Probability & Statistics for Engineers, Eighth
Edition: Richard A. Johnson
Chapter-10, Inferences Concerning Proportions.
 Statistics
for Management, Seventh Edition:
Richard I. Levin, David S. Rubin
Chapter-8,9-Testing Hypotheses: One Sample &
Two Sample Test.
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