This document discusses hypothesis testing and outlines the five-step procedure: 1) State the null and alternative hypotheses 2) Select the level of significance 3) Identify the appropriate test statistic 4) Formulate the decision rule 5) Make a decision about whether to reject the null hypothesis Examples are provided to illustrate one-tailed and two-tailed hypothesis tests using z-tests when the population standard deviation is known. Critical value and decision rule approaches are demonstrated.

1. HYPOTHESIS TESTING

2.  We develop a procedure to test the
validity of a statement about a population
parameter.
Examples:
 The mean starting salary for graduates of
four year business schools is Rs. 32,000
per month.
 Eighty percent of those who play the state
lottery regularly never win more than $
100 in any one play.

3. What is Hypothesis?
 Is a statement about a population
developed for the purpose of testing.
 In most cases the population is too
large that it is not feasible to study
all items in the population.
 We can, therefore, test a statement
to determine whether the sample
does or does not support the
statement concerning the population.

4. HYPOTHESIS TESTING
 A procedure based on sample
evidence and probability theory to
determine whether the hypothesis is
a reasonable statement.

5. Five-Step Procedure for Testing a
Hypothesis
 State the Null Hypothesis (Ho) and
the Alternate Hypothesis (Hı)
 Select a level of significance.
 Identify the test statistic.
 Formulate a decision rule.
 Make a decision

6. Step 1: State the Null Hypothesis (Ho)
and the Alternate Hypothesis (Hı)
 The first step is to state the
hypothesis being tested. It is called
the null hypothesis, designated (Ho).
 The capital H stands for hypothesis,
and the subscript zero implies “no
difference”.
 There is usually a “not” or a “no”
term in the null hypothesis, meaning
that there is “no change”.

7.  For example, the null hypothesis is that
the number of miles driven on the steel-
belted tire is not different from 60,000.
 Therefore, Ho: μ = 60,000.
 We either reject or fail to reject the null
hypothesis.
 The null hypothesis is a statement that is
not rejected unless our sample data provide
convincing evidence that it is false.

8. ALTERNATE HYPOTHESIS
 The alternate hypothesis describes
what you will conclude if you reject
the null hypothesis. It is written as
Hı.
 It is also called the research
hypothesis.
 The alternate hypothesis is accepted
if the sample data provide us with
enough statistical evidence that the
null hypothesis is false.

9. Example:
 A recent article indicated that the
mean age of U.S. commercial
aircraft is 15 years.
 The null hypothesis represents the
current or reported condition.
 Ho: μ = 15.
 The alternate hypothesis is that the
statement is not true: H1: μ ≠15.

10. Select a level of significance
 The level of significance is designated
α , the Greek letter alpha. It is also
sometimes called level of risk.
 There is no one level of significance that it
is applied to all tests.
 The common choices for α are .05, .01 and
.10.
LEVEL OF SIGNIFICANCE The probability of making a
Type 1 error when the null hypothesis is true as an equality.

11.  Traditionally, .05 level is selected
for consumer research projects, .01
for quality assurance, and .10 for
political polling.

12. Type I and Type II Errors
Population Condition
Ho True Hı True
Accept Ho Correct Type II
Conclusion Error
Conclusion
Reject Ho Type I Correct
Error Conclusion

13. Type I Error from Indian Epic
It may be recalled that in “Abhigyan
Shakuntalam” , king Dushyanta had married
shakuntala when he met her in her village, while
wandering in a jungle. He gave her his royal ring
as a gift which could also serve as her identity
when she would come to meet him, in future.
However, while going to meet him, she lost the
ring in the river. When she reached Dushayant’s
place and met him, he failed to recognize her
especially since she did not have the ring. Thus
Dushayant committed Type I error as he
rejected Shankutla as his wife when, in fact, she
was his true wife.

14. Type II Error from Indian Epic
In Mahabharta epic, Dronacharya – the
‘guru’ of both Pandavas and Kauravas –
was fighting from the Kaurav’s side.
However, he had taken a vow that he
would stop fighting if and when his son
Aswathama was killed in the war. It so
happened that during the war, one
elephant named Aswathama was killed.
Lord - Krishna the mentor of Pandavas –
thought of a strategy to make
Dronacharya lay down his arms.
Yudhishter on advice of lord Krishna, went
to Dronacharya and pronounced

15. Aswathama was dead-but was it a human
or an elephant? Dronacharya, on listening
the first part of Yudhishtir’s sentence,
presumed that his son was dead, and he
left for his heavenly abode without
waiting to listen to the second part of
Yudhishter’s sentence. Thus, Droncharya
could be said to have committed Type II
error i.e. accepting a statement when it
was not true.

16. Select a Test Statistic
 A value, determined from sample
information, used to determine
whether to reject the null
hypothesis.
 The test criteria that are frequently
used in hypothesis testing are Z, t,
F, Χ test.

17. Formulate the Decision Rule
 A decision rule is a statement of the
specific conditions under which the
null hypothesis is rejected and the
conditions under which it is not
rejected .

18. Step 5: Make a Decision
 Make a decision regarding the null
hypothesis on the sample
information .
 Interpret the results of the test.

19. Population Mean: known
One - tailed Test
Lower Tail Test Upper Tail Test
Ho: μ ≥ μo Ho: μ ≤ μo
H1: μ < μo H1: μ > μo
σ

20. Example:
 The Federal trade Commission (FTC)
periodically conducts statistical studies
designed to tests the claims that
manufacturers make about their products.
 For example, the label on a large can of
Hilltop Coffee States that the can contains
3 pounds of coffee.
 The FTC knows that Hilltop production
process cannot place exactly 3 pounds of
coffee, even if the mean filling weight is
for the population of all cans filled is 3
pounds per can.

21.  However, as long as the population mean
filing weight is at least 3 pounds per can,
the rights of consumers can be protected.
 Thus, the FTC interprets the label
information on a large can of coffee as a
claim by Hilltop that the population mean
is at least 3 pounds per can.
 We will show how the FTC can check the
hilltops claim by conducting the lower tail
hypothesis test?

22. Develop Null and Alternative
Hypothesis
If the population mean filling weight
is at least 3 pounds per can,
Hilltop’s claim is correct.
Ho: μ ≥ 3
Hı: μ < 3
The hypothesized value of population
mean is μo = 3

23.  Suppose a sample of 36 cans of coffee is
selected.
 Sample mean is computed as an estimate of
population mean μ.
 If < 3 pounds, the sample results will cast a
doubt on null hypothesis.
 We want to know how much less than 3 pounds
must be before we would be willing to declare
the difference significant and risk making a Type
I error by falsely accusing Hilltop of a label
violation.
x
x
x

24. The director of FTC’s program made
the following statement:
If the company is meeting its weights
specifications at µ = 3, I would like 99%
chance of not taking any action against the
company. Although I do not want to accuse
the company wrongly of under filling its
product, I am willing to risk a 1% chance of
making such an error.”
Therefore from the director’s statement we
would set a = .01
Thus we must design the hypothesis test so
that probability of making a type I error when
µ = 3 is .01

25. Test Statistic
 For the Hilltop Coffee study,
previous FTC test show that the
population standard deviation can
be assumed known with the value
of σ = .18
 In addition these tests also show
the population of filling weights can
be assumed to have a normal
distribution.

26. Sampling distribution of x
36n
03.018. ===σσx

27. TEST STATISTIC FOR HYPOTHESIS TESTS ABOUT A POPULATION
MEAN: σ KNOWN
n/
0
σ
µ−=xz

28. Suppose the sample of 36 Hilltop coffee
cans provides a sample mean of =
2.92 small enough to cause us to reject
Ho ?
= 2.92; σ = .18 and n = 36
x
x
36.18/n/
67.2392.20
σ
µ −=−=−= xz

29. Critical Value Approach
 The critical value is the value of the
test statistic that corresponds to the
area of α = .01 in the lower tail of
the standard normal distribution.
 Using standard normal distribution
table, we find that z = -2.33 provides
an area of .01 in the lower tail.

30. For Hilltop Coffee Study Critical Value
Rejection Rule for a level of significance of .
01 is
Reject Ho if Z ≤ -2.33

32. 0
z
Z=-2.33
01.=α

33.  = 2.92
 z = -2.67
 Since z = -2.67 < -2.33, we can
reject Ho and conclude that Hilltop
coffee is under filling cans.
x

34. Two – Tailed Test
Ho: μ = μo
H1: μ ≠ μo

35. Example:
 The U.S. Golf Association (USGA)
establishes rules that manufacturers of golf
equipment must meet if their products are
to be acceptable for use in USGA events.
 MaxFlight uses a high technology
manufacturing process to produce golf balls
with average distances from 295 yards.
 When the average distance passes 295
yards, MaxFlight’s golf balls may be
rejected by the USGA for exceeding the
overall distance standard concerning carry
and roll.

36.  MaxFlights’s quality control program
involves taking periodic samples of
50 golf balls to monitor the
manufacturing process.
 For each sample, a hypothesis test
is conducted to determine whether
the process has fallen out of
adjustment.

37.  We assume that the process is functioning
correctly; i.e. the golf balls produced have
a mean distance of 295 yards.
H0: μ = 295
H1: μ ≠ 295
 If the sample mean is less than is
significantly less than 295 yards or
significantly greater than 295 yards, we will
reject H0.

38.  The quality control team selected = .05 as
the level of significance for the test. Data
from previous tests conducted when the
process was known to be in adjustment
show that the population standard deviation
can be assumed known with a value of =
12. Thus, with a sample size of n = 50, the
standard error of is
x
x
50n
7.112 ===σσx

39.  Because the sample size is large, the
central limit theorem allows us to conclude
that the sampling distribution of can be
approximated by a normal distribution.
Suppose that a sample of 50 golf balls is
selected and that the sample mean is =
297.6 yards. This sample mean provides
support for the conclusion that the
population mean is larger than 295 yards.

40. Computing z - statistic
5012//
53.12956.2970
n
xz
σ
µ =−=−=

41. Critical Value Approach
 With a level of significance of = .05
 The area in each tail beyond the
critical values is
 Using the table of of areas of
standard normal distribution.
025.2/05.2/ ==α

43. 0
z
-1.96 1.96
Reject H0Reject H0

44. 96.1and96.1 025.025. =−=− zz
Reject H0 if z ≤ 1.96 or if z ≥ 1.96
Because the value of the test for the MaxFlight study is z = 1.53,
The statistical evidence will not permit us to reject the null
hypothesis at the .05 level of significance.

45. Example:
• The Jamestown Steel Company
manufactures and assembles desks
and other office equipment at several
plants in Western New York State. The
weekly production of Model A325 desk
at the Fredonia Plant follows a normal
distribution, with a mean of 200 and a
standard deviation of 16. recently
because of market expansion, new
production methods have been
introduced and new employees hired.

46.  The vice president of manufacturing
would like to investigate whether
there has been a change in the
weekly production of the model
A325 desk. To put it another way, is
the mean number of desk produced
at the Florida plant different from
200 at the .01 significance level?

47. Solution
 State the null hypothesis and
alternate hypothesis:
 This is a two-tailed test because the
alternative hypothesis does not
state a direction.
200:
200:
1
0
≠
=
µ
µ
H
H

48.  As noted, the .01 level of
significance is used.
 It is the probability of committing a
Type I error, and it is the probability
of rejecting a true null hypothesis.

49. Select the test statistic
n/σ
µ−=xz

50. Formulate the decision Rule
 The decision rule is formulated by
finding the critical values of z.
 Since it is a two tailed test, half of .
01, or .005, is placed in each tail.
 The area where H0 is not rejected,
located between the two tails, is
therefore .99.

51. Make a decision and interpret the
result
50/16n/
55.12005.203
σ
µ =−=−= xz
Because H0 does not fall in the rejection region, H0 is not
rejected
We conclude that population mean is not different from 200.