Hypothesis testing

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11 Hypothesis Testing
Null Hypothesis, Alternative Hypothesis, Type I and Type II
Errors, One-Tailed Test, Two-Tailed Test
11.1 HYPOTHESIS TESTING
We define hypothesis testing as a procedure to verify claims about population
parameters. The tentative assumption is referred to as the null hypothesis and is denoted
by 0H . But another hypothesis is developed to replace the initial tentative assumption
once the null hypothesis has been rejected. This other hypothesis is referred to as the
alternative hypothesis and is denoted by aH . The procedure to test both claims will
require using data obtained from a sample of the population.
In the following examples we shall show how null and alternative hypotheses are
developed.
Example 11.1
Consider a car model that currently is able to cover 8.5 km per litre of gasoline.
A product-research group has developed a new carburettor designed to increase
the kilometre-per-litre performance. This group will look for evidence that will
enable them to conclude that the new design increases the mean  number of
kilometre per litre. Develop the null and alternative hypothesis for this test.
Solution
58580 .:.:   aHH
Hypothesis Testing is a statistical procedure to determine which hypothesis is more
acceptable as true or which hypothesis is more likely to be false.
Null Hypothesis 0H is the hypothesis that is being tested; it represents the assertion
that the experimenter doubts to be true.
Alternative Hypothesis aH is the statement the experimenter believes to be true and
wishes to prove. The rejection of the null hypothesis leads to the acceptance of the
alternative hypothesis.

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In this situation, the researchers wish to see their claims about the new
carburettor validated. Hence, their claim about the new carburettor should
be the alternative hypothesis. And the existing kilometre-per-litre
performance is the null hypothesis.
:0H The mean kilometre-per-litre performance of the new
carburettor is less than or equal to 8.5.
:aH The mean kilometre-per-litre performance of the new
carburettor is greater than 8.5.
Example 11.2
A manufacturer of soft drinks states that the 2-litre containers of its products have
an average content of at least 67.6 fluid ounces. A sample of 2-litre containers
will be obtained, and their contents will be measured to test the manufacturer’s
statement. Develop the null and alternative hypotheses for this test.
Solution
6676670 .;.:   aHH
The null hypothesis is the manufacturer’s claims about their product, that
is, the mean content of the containers is at least 67.6 fluid ounces. The
alternative hypothesis is: the mean content is less than 67.6.
0H : The mean content of the containers for soft drink is greater
than or equal to 67.7 fluid ounces.
:aH The mean content of the containers for soft drink is less than
67.7 fluid ounces.
Example 11.3
The manager of Bantugon Resort & Hotel stated that their guests spend on
average 1,500Php or less during their stay. One member of the hotel accounting
staff has noticed that the total charges for guest bills have been increasing in
recent months. If the accounting staff wishes to test the manager’s claims by
collecting samples from their guest bills how should he formulate the hypotheses?
Solution
150015000   :: aHH

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:0H The mean amount guests spend in the resort is less than or
equal to 1500Php.
:aH The mean amount guests spend in the resort is greater than
1500Php
11.2 TYPE I ERROR AND TYPE II ERROR
The null and alternative hypotheses are competing statements about the true state of a
situation. They are mutually exclusive statements: either 0H is true or aH is true, but not
both. It would be nice if the process of hypothesis testing always leads us to accept a
statement which is in fact true. But this result is not always possible. Bear in mind,
hypothesis testing is based on information drawn from samples, therefore it makes sense
to allow for the possibility of errors. Table 11.1 gives a summary of the correct decisions
and the errors in hypothesis testing.
The first column shows what happens when we accept 0H . If 0H is true, and we accept
it, then our decision is correct. But if we accept 0H when in fact, aH is true then we
make a Type II error.
The second column shows what happens when we reject 0H . If we reject 0H when in
fact, 0H is true then , we commit a Type I error. If we reject 0H and aH is true then, we
make the correct decision.
Since we cannot eliminate the possibility of errors in hypothesis testing, we must
consider the probability of their occurrence. The probabilities of making the two errors
are defined as follows:
Correct
Decision
Type I
Error
Type II
Error
Correct
Decision
Accept H0 Reject H0
H0 is true
Ha is true
Table 11.1 Errors and Correct Decisions in
Hypothesis Testing

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The probability of making a Type I error  is also referred to as the level of significance.
Example 11.4
The label on a 3 quart container of orange juice claims that the orange juice
contains on average, 1 gram of fat or less.
a) Construct the appropriate null and alternative hypothesis.
Solution
10 :H 1:aH
:0H The mean amount of fat in the orange juice in a 3 quart container is
less than or equal to 1 gram.
aH : The mean amount of fat in the orange juice in a 3 quart container is
greater than 1 gram
b) What is Type I error in this situation?
Answer
Type I error involves rejecting the claim that the mean amount of fat in the
orange juice in a 3 quart container is less than or equal to 1 gram, when in
fact, it is true. (or claiming 1:aH when it is not true)
c) What is Type II error in this situation?
Answer
Type II error involves accepting the claim that the mean amount of fat in
the orange juice in a 3 quart container is less than or equal to 1 gram,
when in fact, it is greater than 1 gram. (or claiming 10 :H is true when
it is not true)
Example 11.5
Mumbai Carpet House has been selling an average of 100,000Php of carpeting a
week. Sofia Visco, the firm’s marketing manager, has proposed a compensation
 the probability of making a Type I error
 the probability of making a Type II error

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plan with a new selling incentive scheme. Mrs. Visco plans to use a trial selling
period to demonstrate that the new incentive scheme will increase the average
sales per salesperson.
a) Develop an appropriate null and alternative hypothesis.
Solution
1000000 H 100000aH
:0H The mean sales of Mumbai Carpet House is less than or equal to
100000Php even with the new incentive scheme.
:aH The new incentive scheme will produce a new mean sales which is
greater than 100000Php.
b) What is Type I error in this situation?
Answer
Type I error in this case involves rejecting the null hypothesis when in
fact, the mean sales remains less than or equal to 100000Php. As a
consequence, the company might adapt the new incentive scheme when in
fact it will not help increase sales.
c) What is the Type II error in this situation?
Answer
Type II error involves accepting the null hypothesis, when in fact, it is not
true. That is, the new incentive scheme does help increase sales. The
consequence of Type II error is the company will miss an opportunity to
improve their sales by failing to implement the new incentive scheme.
STEPS IN HYPOTHESIS TESTING
1. State the null hypothesis 0H and the alternative hypothesis aH
2. Choose the level of significance  .
3. Choose the test statistic and set the critical region.
4. Collect data and compute the test statistic based on the sample data.
5. Decide: Reject 0H if the value of the test statistic falls in the critical region.
If it falls outside, then accept 0H .

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11.3 One-Tailed Hypothesis Test for Population Mean
A one – tailed test of hypothesis for the population mean is a test which specifies only
one tail of the distribution as the rejection region, either to the left tail or right tail.
Setting up the null and alternative hypotheses can take on the following forms:
yxayx
yxayx
a
a
uHuuH
uHuuH
cHcH
cHcH








::
::
::
::
0
0
0
0
11.4 Two-Tailed Hypothesis Test for Population Mean
A two – tailed test of hypothesis for the population mean is a test which specifies both
tails of the distribution as the rejection region. Setting up the null and alternative
hypothesis can take on the following forms:
yxayx
a
HH
cHcH




::
::
0
0
A test statistic is a statistic whose value is calculated from sample measurements and
on which the statistical decision will be based.
The critical region is the set of values of the test statistic for which the null hypothesis
will be rejected. The acceptance region is the set of values of the test statistic for
which the null hypothesis will be accepted.
The critical value of the test statistic is the demarcation line between the critical and
acceptance regions.
Test Statistic
a. ( Large Sample ) If  is known and 30n , the test statistic is
x
x
z

0
 where
nx

 
b) ( Small Sample ) If  is unknown and 30n , the test statistic is
x
s
x
t 0
 where
n
sx


With 1 nv degrees of freedom (t – statistic)

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The z – statistic is to be applied for a large sample size, that is, 30n . When the sample
size less than 30, use the statistict .
Example 11.6
A random sample of 500 car owners shows that the average mileage of a car is
23,500 km a year with a standard deviation of 3,900 km. Test the hypothesis that
the average mileage of a car is 23,400 km a year at 5% level of significance.
Solution
a) Set the null and alternative hypotheses.
40023400230 ,:,:   aHH
b) The population standard deviation  is not known but we can replace
it with the sample standard deviation s. The sample size is also large.
We shall use the z – statistic.
Critical Region
a) For one – tailed hypothesis test:
z – test
0
00




:
:
aH
H
Rejection Rule at a Level of Significance of  : Reject zzH if0
z – test
0
00




:
:
aH
H
Rejection Rule at a Level of Significance of  : Reject zzH if0
b) For two – tailed test:
z – test
0
00




:
:
aH
H
Rejection Rule at a Level of Significance of  :
Reject 220  zzzzH  andif

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x
s
x
z 0
 where
n
s
sx

Since this is a two – tailed test, the critical points are 2z . At
961050 0250 .,. .  z
c) Computation
570
5003900
23400235000
.




x
s
x
z

d) Since 570.z is less than 1.96 we do not reject the null hypothesis,
that is, the average mileage of all cars is not significantly different
from 23,400 km.
Example 11.7
Conrad Dotong Real Estate Co. advertises that the mean selling time of a
residential home is 40 days or less after it is listed in their company. A sample of
50 recently sold homes shows a sample mean selling time of 45 days and a
standard deviation of 20 days. Using a 0.02 level of significance, test the validity
of the company’s claim.
Solution
a) Set the null and alternative hypothesis:
40400   :: aHH
b) The population standard deviation  is not known but we shall replace
it with the sample standard deviation 20s . Sample size is large with
50n . Use the z – statistic.
Critical Value 961.z
Acceptance Region

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This is a one – tailed test with 020. . The critical value is
052020 .. z .
c) Computation
771
5020
40450
.




x
s
x
z

d) Since 771.z is less than 2.05, we do not reject the null hypothesis.
A sample selling time of 45 days is not significantly greater than the
population mean of 40 days at 0.02 level of confidence.
Example 11.8
The height of adults in a certain town is found to have a mean of 166.17 cm with
standard deviation of 5.89 cm. A random sample of 144 adults in the slum
section of the town is discovered to have a mean height of 164.65 cm. Does this
height indicate that the residents of the slum area are significantly shorter in
height at 0.05 level of significance?
Solution
a) Set the null and alternative hypothesis.
cm.:cm.: 17166171660   aHH
b) The population standard deviation  is cm.895 , and the sample size
is large at 144n . Use the z – statistic.
This is also a one-tailed test with 050. . The critical point is
651050 .. z . We shall reject 0H if 651.z .
Critical Value 052.z
Acceptance Region

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c) Computation
933
114895
17166651640
.
.
..





x
x
z


d) Since 933.z is less than 651. , we reject the null hypothesis.
This means the height of adults in the slum section of the town is
significantly lower than 166.17 cm at 0.05 level of significance.
Example 11.9
The average aluminium concentration recovered from samples of aluminium
measurements from 20 different locations was discovered to be 2.8 grams/ml with
standard deviation of 0.3 gram/ml. Does this suggest that the average amount of
aluminium in the river is significantly more than 2.5 grams/ml at 0.05 level of
significance?
Solution
a) Set the null and alternative hypotheses.
grams/ml.:grams/ml.: 52520   aHH
b) We shall use the sample standard deviation 30.s . Since the sample
size is small 20n , we shall use the t – statistic.
x
s
x
t 0
 where
n
sx


With 1 nv degrees of freedom.
Critical Value 651.z
Acceptance Region

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This is a one – tailed test, and the critical point at 19 degrees of freedom is
7291050 .. t (see t-distribution table)
c) Computations
474
2030
52820
.
.
..





x
s
x
t

d) Since 474.t is greater than the critical point of 1.729, the null
hypothesis has to be rejected. This means the average amount of
aluminium in the river is significantly greater than 2.5 grams/ml at
0.05 level of significance.
Critical Value 7291.t
Acceptance Region

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Exercise 11.1 Hypothesis Testing: Null and Alternative Hypotheses
Complete the following exercises neatly and orderly.
1. The coffee dispenser of LPU canteen was readjusted. The canteen manager
wished to know if the dispenser is dispensing the right quantity of coffee. He
took a sample of 50 cups filled by the dispenser. The dispenser can be
deemed in good condition only if the average fill per cup is 8 ounces.
Define the appropriate null and alternative hypotheses for this situation.
0H :
aH :
2. A triathlete plans to reduce his weight to improve his racing performance in
triathlon races. He heard about a food supplement which is marketed to be an
effective supplement for weight loss. The food supplement however, is very
expensive. He talked to people who had tried the supplement themselves and
decided that he will use the supplement only if the proportion of people who
would claim it to be effective is greater than 60%.
Define the appropriate null and alternative hypotheses for this situation.
0H :
aH :
Name Date
Course-Section Score

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3. Bantugon Pharmaceuticals claims that its new medicine is at least 90%
effective in relieving an allergy for a period of 8 hours. Students in the
medical technology program of LPU Batangas wish to test this claim.
Define the appropriate null and alternative hypotheses.
0H :
aH :