CHAPTER 2

Hypothesis Testing
-Test for one and two means
-Test for one and two proportions
Definitions
Hypothesis Test:
It is a process of using sample data and statistical procedures to
decide whether to reject or not to reject the hypothesis (statement)
about a population parameter value (or about its distribution
characteristics).
Null Hypothesis,
:
Generally this is a statement that a population has a specific value.
The null hypothesis is initially assumed to be true. Therefore, it is
the hypothesis to be tested.
Alternative Hypothesis,
:
It is a statement about the same population parameter that is used
in the null hypothesis and generally this is a statement that
specifies that the population parameter has a value different in
some way, from the value given in the null hypothesis. The
rejection of the null hypothesis will imply the acceptance of this
alternative hypothesis.
Test Statistic:
It is a function of the sample data on which the decision is to be
based.
Critical/ Rejection region:
It is a set of values of the test statistics for which the null
hypothesis will be rejected.
Critical point:
It is the first (or boundary) value in the critical region.
P-value:
The probability calculated using the test statistic. The smaller the
p-value is, the more contradictory is the data to
.
Procedure for hypothesis testing
1. Define the question to be tested and formulate a hypothesis for a stating
the problem.
H o :  a
H1:  a or   a or  > a
2. Choose the appropriate test statistic and calculate the sample statistic
value. The choice of test statistics is dependent upon the probability
distribution of the random variable involved in the hypothesis.
3. Establish the test criterion by determining the critical value and critical
region.
4. Draw conclusions, whether to accept or to reject the null hypothesis.
Example 2.1:
The average monthly earnings for women in managerial and
professional positions is RM 2400. Do men in the same positions
have
average monthly earnings that are higher than those for women? A
random sample of n = 40 men in managerial and professional
positions showed X = RM3600 and s = RM 400. Test the appropriate
hypothesis using  = 0.01.
Solution:
The hypothesis to be tested are:
H 0 :  2400
H1:  2400
We use normal distribution n > 30
Z  z
Rejection region:
z  z0.01  2.33 (from normal distribution table)
Test statistic:
Z
x   3600  2400

 18.97
s
400
n
40
Since 18.97 > 2.33, falls in the rejection region, we reject H 0 and conclude
that average monthly earnings for men in managerial and professional
positions are significantly higher than those for women.
Example 2.2:
Aisyah makes “kerepek ubi” and sell them in packets of 100g each. 12
randomly selected packets of “kerepek ubi” are taken and their weights in g
are recorded as follows:
98
102
98
100
96
91
97
97
100
94
101
97
Perform the required hypothesis test at 5% significance level to check
whether the mean weight per packet if “kerepek ubi” is not equal to 100g.
Solution:
The hypothesis to be tested are:
H 0 :   100
H1:   100
We use t distribution,
Two-tailed test
 unknown, n = 12 < 30

 0.025  t0.025 ,11  2.201 and t0.025,11  2.201
2
Test Statistic:
Z
x   97.5833  100

 2.737
s
3.0588
n
12
ttest  2.737  2.201  t0.025,11
Since – 2.737 < -2.201, falls in the rejection region, we reject H 0 and
conclude that weight per packet of “kerepek ubi” is not equal to 100g.
Exercise 2.1:
A teacher claims that the student in Class A put in more hours studying
compared to other students. The mean numbers of hours spent studying per
week is 25hours with a standard deviation of 3 hours per week. A sample of
27 Class A students was selected at random and the mean number of hours
spent studying per week was found to be 26hours. Can the teacher’s claim
be accepted at 5% significance level? Answer: Z = 1.7321, Do not reject H 0
Hypothesis testing for the differences between two population
mean,  1  2 
Test hypothesis
H 0 : 1  2  0
H1: 1  2  0 Reject H 0 when Z test  Z
H1: 1  2  0 Reject H 0 when Z test   Z
H1: 1  2  0 Reject H 0 whenZ   z 2 or Z  z 2
Test statistics
2
i) Variance  1 and  2 2 are known, and both n1 and n2 are samples of any
sizes.
X  X 2  0
Z test  1
 12  2 2
n1

n2
ii) If the population variances,  1 and  2 are unknown, then the following
tables shows the different formulas that may be used depending on the sample
sizes and the assumption on the population variances.
2
2
Equality of
variances,
 12 , 2 2 when
are unknown
Sample size
n1  30, n2  30
Z test 
  2
2
1
 12   22
X 1  X 2  0
s12 s2 2

n1 n2
2
Z test
ttest 
Sg 
 n1  1 s
  n2  1 s2
n1  n2  2
s12 s2 2

n1
n2
2
 s12 s2 2 



n
n
2 
v   12
2
 s12 
 s2 2 




 n1    n2 
n1  1
n2  1
X1  X 2

1
1
Sg

n1 n2
2
1
n1  30, n2  30
X 1  X 2  0
ttest 
2
Sg 
X 1  X 2  0
1
1
Sg

n1 n2
 n1  1 s12   n2  1 s2 2
n1  n2  2
v  n1  n2  2
Example 2.3:
The mean lifetime of 30 bateries produced by company A is 50 hours and the
mean lifetime of 35 bulbs produced by company B is 48 hours. If the standard
deviation of all bulbs produced by company A is 3 hour and the standard
deviation of all bulbs produced by company B is 3.5 hours, test at 1 %
significance level that the mean lifetime of bulbs produced by Company A is
better than that of company B. ( Variances are known)
Solution:
H 0:  A  B  0
H 1:  A   B  0
Z test 
 50  48  0  2.4807
32 3.52

30
35
Ztest  2.4807  2.3263  Z 0.01
We reject H 0. The mean lifetime of bulbs produced by company A is better
than that of company B at 1% significance level.
Example 2.4:
A mathematic placement test was given to two classes of 45 and 55 student
respectively . In the first class the mean grade was 75 with a standard
deviation of 8, while in the second class the mean grade was 80 with a
standard deviation of 7. Is there a significant difference between the
Performances of the two classes at 5% level of significance? Assume the
population variances are equal.
Solution:
H 0 : 1  2  0
H1: 1  2  0
Z test 
X 1  X 2  0
Sg
Sg 
1
1

n1 n2
 75  80   0

7.4656
 44  82   54  7 2
44  55  2
1
1

45 55
 74656
 3.3319
Since
Ztest  3.3319  1.96  Z0.025 , so we reject H 0
So there is a significant difference between the perforance of the two classes
at 5% level of significance.
Exercise 2.2:
A sample of 60 maids from country A earn an average of RM300 per week
with a standard deviation of RM16, while a sample of 60 maids from country B
earn an average of RM250 per week with a standard deviation of RM18. Test
at 5% significance level that country A maids average earning exceed country
B maids average earning more than RM40 per week. Answer : Z = 16.0817,
Reject H 0
Example 2.5:
When working properly, a machine that is used to make chips for calculators
produce 4% defective chips. Whenever the machine produces more
than 4% defective chips it needs an adjustment.To check if the
machine is working properly, the quality control department at the
company often takes sample of chips and inspects them to determine if they
are good or defective. One such random sample of 200 chips taken recently
from the production line contained 14 defective chips. Test at the 5%
significance level whether or not the machine needs an adjustment.
Exercise 2.3:
A manufacturer of a detergent claimed that his detergent is least 95%
effective is removing though stains. In a sample of 300 people who had used the
Detergent and 279 people claimed that they were satisfied with the result.
Determine whether the manufacturer’s claim is true at 1% significance level.
Answer: Do not Reject H 0
Example 2.6:
A researcher wanted to estimate the difference between the percentages of
two toothpaste users who will never switch to other toothpaste. In a sample
of 500 users of toothpaste A taken by the researcher, 100 said that they will
never switched to another toothpaste. In another sample of 400 users of
toothpaste B taken by the same researcher, 68 said that they will never
switched to other toothpaste. At the significance level 1%, can we conclude
that the proportion of users of toothpaste A who will never switch to
other toothpaste is higher than the proportion of users of toothpaste
B who will never switch to other toothpaste?
Exercise 2.4:
In a process to reduce the number of death due the dengue fever, two district,
district A and district B each consists of 150 people who have developed
symptoms of the fever were taken as samples. The people in district A is given
a new medication in addition to the usual ones but the people in district B is
given only the usual medication. It was found that, from district A and from
district B, 120 and 90 people respectively recover from the fever. Test the
hypothesis that the new medication better to cure the fever than the using the
usual ones only using a level of significance of 5%.
Answer: reject H 0
SOLVE USING P-VALUE
Seven people who have a problem with obesity
were placed on a diet for one month. Their weight
at the beginning and the end of the month were
recorded as follows: (Assume variance are equal)
Subject
Begin (in kg)
End (in kg)
1
105
85
2
120
105
3
90
75
4
110
95
5
100
85
6
104
88
7
98
72
Can we conclude that there is a difference in the
mean for two populations at significance level 95%.
1.
Construct hypothesis
H 0 : 1  2  0
H1 : 1  2  0
Test Statistic
p-value (get from output using Excel)
2.
3.
Rejection Region
p  value  0.00874 

2
 0.025
Reject H 0
3. Conclusion

Since p  value 
, we reject H 0 . We can
conclude that there2is a difference in the means of
two populations.
EXERCISES
Exercise
1.A new concrete mix being designed to provide
adequate compressive strength for concrete blocks. The
specification for a particular application calls for the
blocks to have a mean compressive strength 
greater than 1350kPa. A sample of 100 blocks is
produced and tested. Their mean compressive strength
is 1366 kPa and their standard deviation is 70 kPa.
Test hypothesis using  = 0.05. Answer : Do not reject
2. A comparing properties of welds made using carbon
dioxide as a shielding gas with those of welds made
using a mixture of a argon and carbon dioxide. One
property studied was the diameter of inclusions,
which are particles embedded in the weld. A sample
of 544 inclusions in welds made using argon shielding
averaged 0.37  m
in diameter, with a standard
deviation 0f 0.25  m . A sample of 581 inclusions in
welds made using carbon dioxide shielding average
0.40
in diameter, with a standard deviation of 0.26  m
m
. Can you conclude that the mean diameters of
inclusions differ between the two shielding gases?
Both variances for population are not equal.
Answer : Reject
3. A method for measuring orthometric heights
above the sea level is presented. For a sample of
1225 baselines, 926 gave results that were within
the class C spirit leveling tolerance limits. Can
we conclude that this method procedures results
within the tolerance limits more than 75% of the
time?
4. A survey asked which methods played a major
role in the risk management strategy of their
firms. In a sample of 43 oil companies, 22
indicated that risk transfer played a major role,
while in a sample of 93 construction companies,
55 reported that risk transfer played a major
role. Can we conclude at level 5% that the
proportion of oil companies that employ the
method of that risk transfer played is less than
the proportion of construction companies that do?
5.The nicotine content in miligrams of two samples
of tobacco were found to be as follows:
Sample A
24
27
26
21
25
Sample B
27
30
28
31
22
36
Can we conclude that the mean between these two
samples have difference at significance level 95%?
Use p-value.
6.A researcher wants to compare two companies
which use to appraise the value of residential
homes. He selected a sample of 10 residential
properties and scheduled both firms for an
appraisal. He get some results. Then the data are
being analyzed and transform into an output as
follows: t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Variable 1
226.8
208.8444
Variable 2
222.2
204.1778
10
10
Pooled Variance
206.5111
Hypothesized Mean
Difference
df
t Stat
0
18
0.715766
P(T<=t) one-tail
0.241659
t Critical one-tail
1.734064
P(T<=t) two-tail
0.483319
t Critical two-tail
2.100922
From the output, can we conclude that there is a
difference mean between these two populations at
significance level 95%?Use p-value.