Hypothesis Testing
Dhon G. Dungca, PIE, M.Eng’g.
Hypothesis Testing
• Hypothesis – is an assumption about a
population or an assertion about the
possible value of a population parameter.
• Hypothesis Testing – is a statistical process
of determining whether a hypothesis made
is reasonable or not, based upon sample
evidence.
Hypothesis Testing
• Example:
– Hypothesis : “The mean of a population is 50”
– Test: We decide to take 40 samples and
determine x to test if our hypothesis is valid.
Steps in Hypothesis Testing
• Step 1: Formulate the Null Hypothesis
– Denoted by H0
– Is the hypothesis we hope to reject
– This is the hypothesis used for testing and is the starting
point of the testing process
– Must always express the idea of a no significant
difference or relationship
– It is a precise statement of equality, such as =10.
– We always first consider that the null hypothesis is true
unless proven otherwise by sample evidence.
Steps in Hypothesis Testing
• Assuming k to be any constant, we have the
following possible forms of the null hypothesis:
– H0 :  = K
– H0 :  = K
• As a precaution, we never make hypothesis tests
regarding the statistic. Hence, the null hypothesis
H0 : x = K is not acceptable. We only hypothesize
about the value of the population.
Steps in Hypothesis Testing
• Step 2: Formulate the Alternative
Hypothesis
– Denoted by H1
– Is opposite the null hypothesis
– It specifies existence of a difference or a
relationship
– The acceptance of H1 would mean that the
difference between the statistic and population
parameter hypothesized is too great for us to
allow the acceptance of H0.
Steps in Hypothesis Testing
• Assuming k to be any constant, we have the
following possible forms of the alternative
hypothesis:
–
–
–
–
H1 :  = K
H1 :  = K
H1 :  < K or H1 :  > K
H1 :  < K or H1 :  > K
• If H1 is in the form of H1 :  = K, we have a
two-tailed test.
Steps in Hypothesis Testing
H0:  = K
H1:  = K
Two-tail Test
H0 is rejected

H0 is not rejected
H0 is rejected
Steps in Hypothesis Testing
H0:  = K
H1:  < K
One-tail Test

H0 is rejected
H0 is not rejected
Steps in Hypothesis Testing
H0:  = K
H1:  > K
One-tail Test

H0 is not rejected
H0 is rejected
Steps in Hypothesis Testing
• Step 3: Specify the level of significance, 
–
–
–
–
Specifies the area within H1 is accepted.
1- is the Level of confidence
 is the Level of significance
 divides the graph into 2 regions, the region of
the acceptance of H0 and the region of the
acceptance of H1 (critical region).
– It is customary to use an  of 0.05 or 0.01.
Level of Significance
H0:  = K
H1:  = K
Two-tail Test
1-
/2
/2
H0 is rejected
H0 is not rejected
H0 is rejected
Level of Significance
H0:  = K
H1:  < K
One-tail Test
1-

H0 is rejected
H0 is not rejected
Level of Significance
H0:  = K
H1:  > K
One-tail Test
1-

H0 is not rejected
H0 is rejected
Steps in Hypothesis Testing
• Step 4: Decide which sampling distribution
to choose
– Determine the appropriate test statistic whose
sampling distribution is known under the
assumption that H0 is true.
– Parameter Tested
Test Statistic Used
•
z or t
•
x2
Steps in Hypothesis Testing
• Step 5: Determine the critical value and the
critical region
– The critical value is the dividing point between
the acceptance and rejection points.
– The critical region is the region within which
H0 is rejected.
– The critical region (CR) depends upon the
value of the level of significance selected in
Step 3.
Critical Regions
CR: Z < -Z /2
H0:  = K
Z > Z 1-/2
H1:  = K
Two-tail Test
1-
/2
/2
H0 is rejected
-Z /2
H0 is not rejected
Z 1-/2
H0 is rejected
Critical Regions
CR: Z < -Z 
H0:  = K
H1:  < K
One-tail Test
1-

H0 is rejected
-Z 
H0 is not rejected
Critical Regions
CR: Z > Z 1-
H0:  = K
H1:  > K
One-tail Test
1-

H0 is not rejected
Z 1-
H0 is rejected
Steps in Hypothesis Testing
• Step 6: Compute the value of the statistic
– Using the data from a sample of size n,
compute the value of the test statistic z, t or x2
whichever was selected in Step 4 using the
appropriate formula for each case.
• Step 7: State the conclusion and make the
decision
– Determine whether the test statistic is in the
critical region or not. If the test statistic is in
the CR, we accept H1. Otherwise, H0 is
selected.
Test of Means
Gives us an indication of the true
average of a population.
Test of Means (Summary)
H0:  = K
H1:  = K,  < K or  > K
 : 0
CR: Z < -Z /2 or
Z < -Z 
or
Z > Z 1-
or
t> t
Z > Z 1-/2
t < -t /2
t > t /2
or
t < -t 
Test of Means (Summary)
Computations:
Used if n  30
z=
x-
/ n
Used if n < 30
t=
x-
s/ n
Conclusion:
or
z= x-
s/ n
Example 1:
• A new process will be installed if its mean
processing time is at most 20 minutes. The
new procedure was tried. In a random
sample of 50 trials, an average processing
time of 22.2 minutes with a standard
deviation of 4.3 minutes was obtained. At a
level of significance =0.05, should the
new process be installed?
Answer:
H0:  = 20 (install the new process)
H1:  > 20 (do not install)
: 0.05
CR: z > 1.645
Computations:
z = 22.2 – 20
H0 is not rejected
H0 is rejected
4.3 / 50
1.645
3.62
= 3.62
Conclusion:
Since z is greater than 1.645, we reject H0. We
should not install the new process.
Example 2:
• A company that makes chocolates claims
that the mean weight of a bag of chocolates
is 240 grams with a standard deviation of
20.5 grams. Using a 0.05 significance level,
would you agree with the company if a
random sample of 50 bags of chocolates
was found to have a mean weight of 230
grams?
Answer:
H0:  = 240
H1:   240
: 0.05
CR: z < -1.96
H1 is accepted
H1 is accepted
z > 1.96
H0 is accepted
Computations:
z = 230-240
-1.96
1.96
20.5 / 50
-3.45
= -3.45
Conclusion:
Reject Ho. The claim of the manufacturer is not true
Example 3:
• The Edison Electric Institute has published figures
on the annual number of kilowatt-hours expended
by various home appliances. It is claimed that a
vacuum cleaner expends an average of 46
kilowatt-hours per year. If a random sample of 12
homes included in a planned study indicates that
vacuum cleaners expend an average of 42
kilowatt-hours with a standard deviation of 11.9
kilowatt-hours, does this suggest at the 0.05 level
of significance that vacuum cleaners expend, on
the average, less than 46 kilowatt-hours annually?
Answer:
H0:  = 46
H1:  < 46
: 0.05
CR: t < -1.796 w/ v = n-1
degrees of freedom
Computations:
t = 42 - 46
11.9 / 12 H1 is accepted
H0 is accepted
= -1.16
Conclusion:
-1.796
-1.16
Do not reject Ho.
Example 4:
• A new process for producing synthetic diamonds
can be operated at a profitable level only if the
average weight of the diamonds is greater than
0.5K. To evaluate the profitability of the process,
6 diamonds are generated with a mean and a
standard deviation of 0.53 and 0.0559
respectively. Do the 6 diamonds’ measurements
present sufficient evidence to indicate that the
average weight of the diamond produced by the
process is in excess of 0.5K?
Answer:
H0:  = 0.5
H1:  > 0.5
: 0.05
CR: t > 2.015 w/ v = 5
degrees of freedom
Computations:
t = 0.53 – 0.5
.0559 / 6
= 1.32
Conclusion:
Do not reject Ho.
H1 is accepted
H0 is accepted
1.32
2.015
Test of Variance
Gives us the degree of dispersion of
the population in consideration
Test of Variance (Summary)
H 0 : 2 = K
H1: 2 = K, 2 < K or 2 > K
 : 0
CR: x2 < x2 1-/2 or
x2 < x21-
or
x2 > x2
x2 > x2/2
Computations:
x2 =
(n-1) S2
2
Conclusion:
Accept Ho
Test of Variance (Summary)
x2 > x2 
x2 < x21-
Accept H1
Accept Ho
Accept Ho
x21-
Accept H1
x2
x2 < x2 1-/2
x2 > x2/2
Accept H1
Accept Ho
x2 1-/2
x2 /2
Accept H1
Example 5:
• In the past, scores in DIFFCAL test have been
found to be normally distributed with a variance of
100. A class of 29 students now taught by an
outstanding teacher whose method of teaching is
expected to have a wider dispersion of test scores.
Suppose that the class can be considered to be a
random sample and suppose that the variance of
student scores in the final examinations is found to
be 150; can we conclude that this instructor’s
method of teaching results in a wider dispersion at
the 5% level of significance?
Answer:
H0: 2 = 100 (no wider dispersion)
H1: 2 > 100 (there is wider dispersion)
: 0.05
CR: x2 > 41.337 (x20.05 at v = 29-1 = 28)
degrees of freedom
Computations:
x2 = (29-1)(150)
Accept H1
100
Accept Ho
= 42
Conclusion:
41.337
42
Reject Ho. Teaching method results in a wider
dispersion
Example 6:
• The volume of containers of a particular
lubricant is known to be normally
distributed with a variance of 0.03 liter.
Test the hypothesis that 2 = 0.03 against
the alternative that 2  0.03 for the random
sample of 10 containers if their variance is
0.06. Use 0.01 level of significance.
Answer:
H0: 2 = 0.03
H1: 2  0.03
: 0.01
CR: x2 < 1.735 (x20.995 at v = 10-1 = 9)
x2 > 23.589 (x20.005 at v = 10-1 = 9)
Computations:
x2 = (10-1)(0.06)
0.03
Accept Ho
= 18
Accept H1
Accept H1
Conclusion:
1.735
23.589
Do not reject Ho.
18
Example 7:
• Past experience indicates that the time for
high school seniors to complete a
standardized test is normal random variable
with a standard deviation of 6 minutes. Test
the hypothesis that  = 6 against the
alternative that  < 6 if a random sample of
20 high school seniors has a standard
deviation s = 4.51. Use a 0.05 level of
significance.
Answer:
H0: 2 = 36
H1: 2 < 36
: 0.05
CR: x2 < 10.117 (x20.95 at v = 20-1 = 19)
Computations:
x2 = (20-1)(20.34)
36
Accept Ho
= 10.74
Accept H1
Conclusion:
10.117
Do not reject Ho.
10.74
Extra Challenge 1:
• For some years now, a certain hospital has been using
Formula  for patients known to have a certain type of
disease. It has been proven that the mean recovery time of
patients under Formula  is equal to 20 days with a
standard deviation of 4 days. A new type of treatment is
now available. This new type of treatment is branded as
Formula . A random sample of 25 patients was created
with this new formula and showed a mean recovery time of
18 days with a standard deviation of 3.5 days. The hospital
is not willing to gamble on new formula unless it is proven
to be better than the old formula as far as mean recovery
time and variability are concerned. Decide whether the
hospital should employ the new formula at the 5% level of
significance. (Hint: 2 tests)
Extra Challenge 2:
• The specifications for a component of a particular
item provide that the mean length should not be
less than 4 cm with a standard deviation 0.45 cm.
A random sample of 50 components drawn from a
very large shipment showed a mean length of 3.95
cm and a standard deviation of 0.45 cm. If you
were the manufacturer of this item who had
received this shipment, would you accept it?
Assume 5% level of significance. What if the
standard deviation is 0.55 cm?