Download Week 9, Lecture 3, Testing hypotheses about a population mean

Business Statistics - QBM117
Testing hypotheses about a
population mean
Objectives

To test hypotheses about a population mean when sigma is
known and when sigma is unknown.
Testing a hypothesis about the population
mean,  when  is known
The operations manager is concerned with determining
whether the filling process for filling 100g boxes of
smarties is working properly. He believes the standard
deviation of the filling process is 10g.
If the manager wants to know whether the average fill
of the boxes differs from 100g, he would specify the
null and alternative hypotheses to be
H 0 :   100
H A :   100
In order to test this hypothesis the manager selected a
random sample of 25 boxes and found their average weight
to be 95g. Is their sufficient evidence at  = 0.05 to
conclude that the weight of the boxes differs from 100g?
Does the question ask us to test a hypothesis
about a mean or a proportion?
a mean
Do we know the population standard
deviation,  or do we only have the sample
standard deviation s?
we know 
x
Z 
/ n
Step 1
H 0 :   100
H A :   100
Step 2
x
Z
/ n
Step 3
  0.05
Region of non-rejection
0.95
3.00
1.96
2.50
2.00
1.50
1.00
0
0.50
0.00
-0.50
-1.00
-1.50
-2.00
-2.50
-3
-1.96
Z
Region of rejection
Region of rejection
 /2 = 0.025
 /2 = 0.025
Step 1
H 0 :   100
H A :   100
Step 2
x
z
/ n
Step 3
  0.05  / 2  0.025 z0.025  1.96
Step 4
Reject H 0 if z sample  1.96 or z sample  1.96
Step 5
  10
n  25
x
z
/ n
95  100

10 / 25
  2 .5
x  95
Region of non-rejection
0.95
3.00
1.96
2.50
2.00
1.50
1.00
0
0.50
0.00
-0.50
-1.00
-1.50
-2.00
-2.50
-3
-1.96
z
-2.5
Region of rejection
Region of rejection
 /2 = 0.025
 /2 = 0.025
Step 5
  10
n  25
x
z
/ n
95  100

10 / 25
  2 .5
x  95
Step 6
Since -2.5 < -1.96 we reject HA.
There is sufficient evidence at  = 0.05 to conclude
that the average fill is different from 100g.
Testing a hypothesis about the population
mean,  when  is unknown
Exercise 10.26 p346 (9.26 p312 abridged)
Does the question ask us to test a hypothesis
about a mean or a proportion?
a mean
Do we know the population standard
deviation,  or do we only have the
sample standard deviation s?
x
t
s/ n
Step 1
H 0 :   160
H A :   160
Step 2
x
t
s/ n
Step 3
  0.01
Region of non-rejection
0.99
3.00
2.50
2.00
1.50
1.00
0.50
0.00
α = 0.01
-0.50
-1.00
-1.50
-2.00
-2.50
-3
-2.624
Critical
value
Region of rejection
0
t
Step 1
H 0 :   160
H A :   160
Step 2
x
t
s/ n
Step 3
  0.01 t ,n1  t0.01,14  2.624
Step 4
Reject H 0 if t sample  2.624
Step 5
s  10
n 15
x
t sample 
s/ n
150  160

10 / 15
 3.87
xx 150
Region of non-rejection
0.99
3.00
2.50
2.00
1.50
1.00
0.50
0.00
-0.50
α = 0.01
-1.00
-1.50
-2.00
-2.50
-3
-3.87
-2.624
Region of
rejection
0
t
Step 5
s  10
n 15
x
t sample 
s/ n
150  160

10 / 15
 3.87
x  150
Step 6
Since -3.87 < -2.624 we reject H0.
There is sufficient evidence at  = 0.01 to conclude
that the mean is less than 160.
Testing a hypothesis about the population
mean,  when  is unknown
Exercise 10.30 p347 (9.30 p313 abridged)
Does the question ask us to test a hypothesis
about a mean or a proportion?
a mean
Do we know the population standard
deviation,  or do we only have the
sample standard deviation s?
x
t
s/ n
Step 1
H 0 :   32
H A :   32
Step 2
x
t
s/ n
Step 3
  0.05
Region of non-rejection
0.95
3.00
2.776
2.50
2.00
1.50
1.00
0
0.50
0.00
-0.50
α/2 = 0.025
-1.00
-1.50
-2.00
-2.50
-3
-2.776
t
α/2 = 0.025
Step 1
H 0 :   32
H A :   32
Step 2
x
t
s/ n
Step 3
  0.05 t / 2,n1  t0.025, 4  2.776
Step 4
Reject H 0 if t sample  2.776 or t sample  2.776
Step 5
s  6.91
n5
x
t sample 
s/ n
24.4  32

6.91 / 5
 2.46
x  24.4
Region of non-rejection
0.95
3.00
2.776
2.50
2.00
1.50
1.00
0
0.50
0.00
-0.50
-1.00
-1.50
-2.00
-2.50
-3
-2.776
-2.46
α/2 = 0.025
t
α/2 = 0.025
Step 5
s  6.91
n 5
x
t sample 
s/ n
24.4  32

6.91 / 5
 2.46
x  24.4
Step 6
Since -2.46 > -2.776 we do not reject H0.
There is insufficient evidence at  = 0.05 to
conclude that the mean is not equal to 32.
Strong and weak conclusions

Generally we will be presented with a null hypothesis,
which we will try to reject.

Before carrying out the test, we know there is a possibility
we may make a type I error.

This probability  is preset to a small number, say 0.05.

Knowing that we have a small probability of committing a
type I error, ie rejecting a null hypothesis when it is true,
makes our rejection of the null hypothesis a strong
conclusion.

Generally the same cannot be said about not rejecting the
null hypothesis.

This is because the probability of , failing to reject a null
hypothesis when it is should be rejected, is not preset to a
known small number.

Therefore, failing to reject the null hypothesis is generally
a fairly weak conclusion because we do not know the
probability that we will fail to reject a null hypothesis
when it should be rejected.
Reading for next lecture
Chapter 10 Sections 10.7 (Chapter 9 Section 9.7 abridged)
Exercises to be completed before next lecture
S&S 10.2 10.3 10.5 10.11 10.29 10.33
(9.2 9.3 9.5 9.11 9.29 9.33 abridged)