This document discusses hypothesis testing procedures. It defines key concepts like the null and alternative hypotheses, Type I and Type II errors, and one-tailed and two-tailed tests. It provides examples of testing hypotheses about a population mean using z-tests and t-tests, and about a population proportion using z-tests. The steps involved are outlined as establishing hypotheses, determining the test, specifying the significance level, stating the decision rule, calculating statistics, and making a conclusion. Examples demonstrate applying these procedures to test hypotheses about a soft drink filling process and a CPA's net income.

1. Hypothesis Testing
Z Test
By
Khem Raj Subedi
Associate Professor
Tikapur Multiple Campus

2. Learning Objectives
• Understand the logic of hypothesis testing,
and know how to establish null and alternate
hypotheses.
• Understand Type I and Type II errors.
• Use large samples to test hypotheses about a
single population mean and about a single
population proportion.
• Test hypotheses about a single population
mean using small samples when σ is
unknown and the population is normally
distributed.

3. Method of Indirect Proof
X
X
YEither X or Y is true but not both
X is demonstrated not to be true Y
YY is true by default

4. Hypothesis Testing
A process of testing hypotheses about
parameters by setting up null and
alternative hypotheses, gathering
sample data, computing statistics from
the samples, and using statistical
techniques to reach conclusions about
the hypotheses.

5. Steps in Testing Hypotheses
1. Establish hypotheses: state the null and
alternative hypotheses.
2. Determine the appropriate statistical test
and sampling distribution.
3. Specify the Type I error rate (α).
4. State the decision rule.
5. Gather sample data.
6. Calculate the value of the test statistic.
7. State the statistical conclusion.
8. Make a managerial decision.

6. Null and Alternative Hypotheses
• The Null and Alternative Hypotheses are
mutually exclusive. Only one of them can
be true.
• The Null and Alternative Hypotheses are
collectively exhaustive. They are stated to
include all possibilities. (An abbreviated
form of the null hypothesis is often used.)
• The Null Hypothesis is assumed to be true.
• The burden of proof falls on the Alternative
Hypothesis.

7. Null and Alternative
Hypotheses: Example
• A soft drink company is filling 12 oz.
cans with cola.
• The company hopes that the cans are
averaging 12 ounces.
H oz
H oz
o
a
:
:
µ
µ
=
≠
12
12

8. Rejection and Nonrejection Regions
µ=12 oz
Nonrejection Region
Rejection Region
Critical Value
Rejection Region
Critical Value

9. Type I and Type II Errors
• Type I Error
– Rejecting a true null hypothesis
– The probability of committing a Type I error
is called α, the level of significance.
• Type II Error
– Failing to reject a false null hypothesis
– The probability of committing a Type II
error is called β.
– Power is the probability of rejecting a false
null hypothesis, and equal to 1- β.

10. Decision Table
for Hypothesis Testing
(
( )
Null True Null False
Fail to
reject null
Correct
Decision
Type II error
β)
Reject null Type I error
α
Correct Decision
(Power)

11. • One-tailed Tests
One-tailed and Two-tailed Tests
H
H
o
a
:
:
µ
µ
≥
<
12
12
H
H
o
a
:
:
µ
µ
≤
>
12
12
H
H
o
a
:
:
µ
µ
=
≠
12
12
• Two-tailed Test

12. One-tailed Tests
H
H
o
a
:
:
µ
µ
≥
<
12
12
H
H
o
a
:
:
µ
µ
≤
>
12
12
µ=12 oz
Rejection Region
Nonrejection Region
Critical Value
µ=12 oz
Rejection Region
Nonrejection Region
Critical Value

13. Two-tailed Tests
H
H
o
a
:
:
µ
µ
=
≠
12
12
µ=12 oz
Rejection
Region
Nonrejection Region
Critical Values
Rejection
Region

14. CPA Net Income Example:
Two-tailed Test
914,74$:H
914,74$:Ho
≠
=
µ
µ
a
If reject H .
If do not reject H .
o
o
Z Z
Z Z
c
c
> =
≤ =
196
196
. ,
. ,
78,646 74,914
2.75
14,530
112
X
Z
n
µ
σ
− −
= = =
c oZ = 2.75 Z = 1.96, reject H≥
Rejection
Region
Nonrejection Region
Ζ=0
Zc = 196.
Rejection
Region
Zc = −196.
α
2
025=.α
2
025=.

15. CPA Net Income Example:
Critical Value Method (Part 1)
Upper
nc cX Z= +
= +
=
µ
σ
74 914 196
14 530
112
77 605
, .
,
,
H
H
o
a
: $74,
: $74,
µ
µ
=
≠
914
914
Lower
nc cX Z= −
= −
=
µ
σ
74 914 196
14 530
112
72 223
, .
,
,
Rejection
Region
Nonrejection Region
Ζ=0 Zc = 196.
Rejection
Region
Zc = −196.
α
2
025=.α
2
025=.
72,223 77,605

16. CPA Net Income Example:
Critical Value Method (Part 2)
If X or X reject H .
If 77,223 X do not reject H .
o
o
< >
≤ ≤
77 223 77 605
77 605
, , ,
, ,
Since X reject H .o= > =78 646 77 605, , ,cX
Rejection
Region
Nonrejection Region
Ζ=0 Zc = 196.
Rejection
Region
Zc = −196.
α
2
025=.α
2
025=.
72,223 77,605

17. Demonstration Problem (Part 1)
30.4:H
30.4:H
a
o
<
=
µ
µ
Rejection
Region
Nonrejection Region
0
α=.05
Zc = −1645.
If reject H .
If , do not reject H .
0
0
Z .
Z .
< −
≥ −
1645
1645
,
Z
X
s
n
=
−
=
−
= −
µ 4156 4 30
0574
32
142
. .
.
.
,
do not reject H .0
Z .= − ≥ −142 1645.

18. Demonstration Problem (Part 2)
H
H
o
a
: .
: .
µ
µ
=
<
4 30
4 30
Rejection
Region
Nonrejection Region
0
α=.05
Zc = −1645.
cx = 4133. 4.30
If reject H .
If , do not reject H .
0
0
X
X
<
≥
4133
4133
. ,
.
cX Z
s
n
= +
= + −
=
µ
4 30 1645
0574
32
4133
. ( . )
.
.
, do not reject H .0X = ≥4156 4133. .

19. Rejection
Region
Nonrejection Region
0
α=.05
Demonstration Problem (Part 3)
H
H
o
a
: .
: .
µ
µ
=
<
4 30
4 30
If p - value < , reject H .
If p - value , do not reject H .
o
o
α
α≥
Since p-value = .0778 > = .05,
do not reject H .o
α
Z
X
s
n
P Z
=
−
=
−
= −
< − =
µ 4156 4 30
0574
32
142
142 0778
. .
.
.
( . ) .

20. Two-tailed Test: Small Sample,
σ Unknown, α = .05 (Part 1)
Weights in Pounds of a Sample of 20 Plates
22.6 22.2 23.2 27.4 24.5
27.0 26.6 28.1 26.9 24.9
26.2 25.3 23.1 24.2 26.1
25.8 30.4 28.6 23.5 23.6
X = 2551. , S = 2.1933, and n = 20

21. Two-tailed Test: Small Sample,
σ Unknown, α = .05 (Part 2)
Critical Values
Nonrejection Region
Rejection Regions
ct = −2 093. ct = 2 093.
α
2
025=.
α
2
025=.
H
H
o
a
:
:
µ
µ
=
≠
25
25
df n= − =1 19

22. Two-tailed Test: Small Sample,
σ Unknown, α = .05 (Part 3)
t
X
S
n
=
−
=
−
=
µ 2551 250
21933
20
104
. .
.
.
Since t do not reject H .o= ≤104 2 093. . ,Critical Values
Non Rejection Region
Rejection Regions
ct = −2 093. ct = 2 093.
α
2
025=.
α
2
025=.
If t reject H .
If t do not reject H .
o
o
>
≤
2 093
2 093
. ,
. ,

23. Demonstration Problem (Part 1)
Size in Acres of 23 Farms
445 489 474 505 553 477 545
463 466 557 502 449 438 500
466 477 557 433 545 511 590
561 560
23=and46.94,=,78.498 nSX =

24. Demonstration Problem (Part 2)
471:
471:
>
=
µ
µ
a
o
H
H
df n= − =1 22
Critical Value
Nonrejection Region
Rejection Region
ct = 1717.
α =.05

25. Demonstration Problem (Part 3)
If t reject H .
If t do not reject H .
o
o
>
≤
1717
1717
. ,
. ,
84.2
23
94.46
47178.498
=
−
=
−
=
n
S
X
t
µ
.Hreject,717.184.2tSince o>=
Critical Value
Nonrejection Region
Rejection Region
ct = 1717.
α =.05

26. Z Test of Population Proportion
Z
p P
P Q
n
where
=
−
⋅

: p = sample proportion
P = population proportion
Q = 1 - P
n P
n Q
⋅ ≥
⋅ ≥
5
5
, and

27. Testing Hypotheses about a
Proportion: Manufacturer Example
(Part 1)
08.:H
08.:H
a
o
≠
=
P
P
cZ = 1645.
Critical Values
Nonrejection Region
Rejection Regions
cZ = −1645.
α
2
05=.
α
2
05=.

28. Testing Hypotheses about a
Proportion: Manufacturer Example
(Part 2)
 .
 . .
(. )(. )
.
p
Z
p P
P Q
n
= =
=
−
⋅
=
−
=
33
200
165
165 08
08 92
200
4 43
If Z reject H .
If Z do not reject H .
o
o
>
≤
1645
1645
. ,
. ,
Since Z reject H .o= >4 43 1645. . ,
cZ = 1645.
Critical Values
Nonrejection Region
Rejection Regions
cZ = −1645.
α
2
05=.
α
2
05=.

29. Demonstration Problem (Part 1)
H P
H P
o
a
: .
: .
=
>
17
17
Critical Value
Nonrejection Region
Rejection Region
cZ = 1645.
α =.05

30. Demonstration Problem (Part 2)
 .
 . .
(. )(. )
.
p
Z
p P
P Q
n
= =
=
−
⋅
=
−
=
115
550
209
209 17
17 83
550
2 44
If reject H .
If do not reject H .
o
o
Z
Z
>
≤
1645
1645
. ,
. ,
Since Z = 2.44 reject H .o> 1645. ,Critical Value
Nonrejection Region
Rejection Region
cZ = 1645.
α =.05