Chapter 10 – Hypothesis
Testing
• What is a hypothesis?
A statement about a population that may
or may not be true.
• What is hypothesis testing?
A statistical test to prove or disprove a
hypothesis. At the end of the test, either
the hypothesis is rejected or not rejected.
Steps in hypothesis testing
• There are 5 steps in hypothesis testing:
• Step 1 – pattern of population distribution
(normal)
• Step 2 – Formulation of hypothesis
Null Hypothesis (H0)
Alternate Hypothesis (H1)
Steps in hypothesis testing
• Step 3 - Level of significance (α )
Alpha is the probability of rejecting a null
hypothesis when it is true. Alpha is also known
as the level of risk. It is described in terms of
percent or decimals (5% or 0.05, 1% or 0.01,
and so on.)
• Step 4 - Test statistic
It is a quantity that is used to compare with
critical Z from appendix D to determine if a null
hypothesis is to be rejected or not.
Steps in hypothesis testing
• This quantity is the z-statistic
_
Z = (x – m) / ( n)
. Step 5 - Decision Rule:
It is a rule by which a null hypothesis is rejected or not rejected.
It always takes the form:
Reject H0 if Z-statistic is > Critical Z
or
< - Critical Z
Critical Z-value is obtained from appendix D.
Steps in hypothesis testing
Decision: Reject H0 or Do not reject H0
Hypothesis tests can be one-tailed or twotailed depending on how the alternate
hypothesis is written.
Let’s take an example
• Suppose:
H : μ = 100
0
H : μ > 100
1
This is a one tailed test.
• Another example: H
: μ = 100
0
: μ < 100
H
1
An example of a two tailed test
• Suppose:
: μ = 100
H
0
: μ ≠ 100
H
1
• This is a two tailed test.
• Exercises from book: Problem 7(P age 239),
Problem 8 (Page 240), Problem 9 (Page 240)
Exercises from Book
•
•
•
•
•
•
•
•
Problem 7 (pg 283):
a. ONE-TAILED b. ONE-TAILED
c. ONE-TAILED d. ONE-TAILED
e. TWO-TAILED
Problem 8 (pg 283):
a. Ho : μ = 500, H1 : μ ≠ 500
b. Ho : μ = 500, H1 : μ > 500
c. Ho : μ = 500, H1 : μ < 500
Exercises from book
• Problem 9 (pg 283)
• Ho : μ = 60; H1 : μ > 60
• Problem 10 (pg 283)
• Ho : μ = 1000; H1 : μ ≠ 1000
• Ho : μ ≤ 1000; H1 : μ > 1000
Testing for population mean
• Such testing can be made under 2
situations: (1) When population
standard deviation is known and (2)
when population standard deviation is
unknown. This chapter discusses the
first.
Refer to example problem 10-2 (page
241)
• Two hypotheses are: H : μ = 10
0
H : μ < 10
1
Testing for population mean
• Alpha ( α ) is 5% (or 0.05).
Critical Z from appendix D is –1.645
• Reject H0 if Z-statistic < - 1.645
Test statistic (Z-statistic)
_
X–μ
8.8 - 10
Z = -------- = ---------- = - 2.0
σ/√n
2.4 / 4
Testing for population mean
Decision:
Reject H0
• What does it mean? The new
manufacturing process reduces tar content
of cigarettes.
• Example problem 10-4.
Testing for population mean
: μ = 1000
H
0
H
: μ > 1000
1
α = 0.025
Critical Z = 1.96 (From Appendix D)
_
X–μ
1050 - 1000
Test Statistic Z = --------- = -------------- = 2.5
σ/n
100 / 5
Decision: Reject H
0
What does it mean? The new manufacturing process
increases life of fuses.
This is a two tailed test
Example Problem 10-7 (page 248)
H :
0
μ = 100
H : μ ≠ 100
1
Critical Z’s are – 2.575 and + 2.575
Decision Rule:
Reject H0 if Z-statistic > + 2.575 or < - 2.575
_
X - 100
Z-statistic = -------------- = 1.667 Decision: Do not reject H
√ (225 / 25)
0
IQs of the school district students are no different from the national average
Example problem 13
•
Problem 13 (page 251)
: μ ≥ 30
H
0
H
: μ < 30
1
Critical Z = - 2.326
Decision Rule: Reject H0 if Z-statistic ≤ - 2.326
27 - 30
Z-Statistic = ------------- = - 5.0
6 / √ 100
Decision: Reject H0
•
What does it mean? Vendor’s claim that mean weight of his
chickens is at least 30 ounces is rejected.
Testing for difference between 2
means
• These tests are performed when one wants to
compare 2 groups of populations.
• For example, is these a difference between
average GPAs of students of two universities?
Or, are average monthly incomes the same
between two groups of people?
• The testing procedure is similar to what has
been presented in the previous section.
Testing for difference between 2
means
• The null hypothesis is always of the form:
H
•
•
: μ = μ
0
1
2
The alternative hypothesis can be either of the 3
forms:
•
μ > μ or
1
2
μ < μ or
1
2
μ ≠ μ
1
2
Testing for difference between 2
means
Formula 10-1, p. 252
The decision rule is the same as before
Example problem 10-8 (page 252-253)
H : μ = μ
H : μ ≠ μ
0
1
2
1 1
2
Testing for difference between 2
means
0.8 – 1.0
• Z- statistic = --------------------------- = - 1.0
√ (0.36/25) + (0.64/25)
• At α = 0.05, critical Z = -1.96 and +1.96
Decision: Do not reject H
0
• Interpretation: The avg. nicotine contents of two
brands of cigarettes are equal.
Problem 4, pg 255
• Is average hourly output of male workers less
than that of female workers?
• This is a test involving difference of means
between two population groups.
• Data are given in the table
Ho: μ1 = μ2; H1: μ1 < μ2
(μ1 indicates output of males)
• Z-statistic =
(150-153)/√[(70/36) + (74/36)]
= -1.5
• At α = 0.05, critical Z is -1.645
• Decision: Do not reject Ho
• Interpretation: Average outputs of male
and female workers are the same.
Testing for population proportion
•
•
In some cases, we are interested in population proportions rather than
population means. For example, are more than 50% of CSULB
students females? Or, is a new medicine more than 60% effective?
Two hypothesis are:
H : p = a given value
0
H : p > a given value OR
1
p < a given value OR
p ≠ a given value
Example 10-11 (Page 257-258)
The test statistic
• Z-statistic = (X – n.p)/√[(n.p(1-p)]
• The decision rule is the same as before.
Example problem 10-12
(pg 259)
• A TV program attracts 50% audience. A
new anchorperson has been hired. Does
the new hire increase audience level? Use
5% level of significance.
• Ho: p = 0.5; H1: p > 0.5
• Given n = 100; critical Z = 1.645
• P = 55/100
Problem 10-12, cont.
• Z-statistic = (55-50)/√[(100)(0.5)(0.5)] = 1
• Decision: Do not reject Ho
• Interpretation:
– The audience level has remained the same
(at 50%) after hiring the new anchorperson.
Problem 6, pg 261
• An auto manufacturer claims that 20% of
customers prefer his products. In a sample
of 100 customers, 15 indicate their
preferences for his products. At 5% level
of significance, can we support his claim?
• Ho: p = 0.2; H1: p ≠ 0.2
Problem 6, cont.
• Z-statistic = (15-20)/√[(20)(0.08)] = -1.25
• Decision: Do not reject Ho.
• Interpretation
– The auto manufacturer’s claim cannot be
disputed.
Hypothesis Testing
• Strength of rejection (p-value)
– When we reject a null hypothesis, there is a
quantity that describes the strength of
rejection. This quantity is called the p-value.
– The lower the p-value, the greater the
strength of refection. In other words, the lower
the p-value, the greater the strength in the
rejection.
What is p-value?
• It is the area to right right of the calculated
value of Z-statistic when the area of
rejection is on the right side of the normal
curve. When the area of rejection is on the
left side of the normal curve, the p-value is
the area to the left of the calculated value
of the Z-statistic. If there are two area of
rejection, add the area to the right of Zstatistic and the area to the left of Zstatistic.
How do we find the P-value?
• Let’s take an example problem:
• Test if the mean waiting time is less than 3
minutes.
• α = 0.05
• μ = 3, σ = 1, n=50, X=2.75
• Ho: μ = 3; H1: μ < 3
• Decision rule: Reject Ho if z-statistic < critical Z
• Z-statistic = (2.75 - 3)/(1/√50) = -1.768
• Critical Z - -1.645 so we reject Ho.
• Interpretation: Mean waiting time is < 3 minutes.
What is the p-value?
• Find the area to the left of -1.768.
• This area is the p-value.
• P-value = 1.0 – 0.9616 = 0.0384