Download ISE362Chapter Eight Tests of Hypothesis

ISE 362
PROBABILISTIC
SYSTEMS II
Chapter Eight
Tests of
Hypothesis
Based on a
Single Sample
Hypothesis Testing Elements
Null Hypothesis H0:
> Prior Belief
Alternative Hypothesis Ha:
> Contradictory Belief
Test Statistic:
> Parameter used to test
Rejection Region:
> Set of values for
rejecting H0
Hypothesis Testing Errors
Type I Error:

> Rejecting H0 when it
is true.
Type II Error: 
> Not rejecting H0 when
it is false.
Example of Type I Errors
Highway engineers have found that
many factors affect the performance of
reflective highway signs. One is the
proper alignment of the car’s
headlights. It is thought that more than
50% of the cars on the road have
misaimed headlights. If this contention
can be supported statistically, then a
new tougher inspection program will be
put into operation. Let p denote the
proportion of cars in operation that
have misaimed headlights. Setup a test
of hypothesis to test this statement.
Type II Errors
The reject region for the car
test is R = 14, 15, …,20 at
 = 0.05. Suppose that the true
proportion of cars with
misaimed headlights is 0.7.
What is the probability that our
test is unable to detect this
situation?
Hypothesis Testing Protocol
Identify parameter of interest
State Null & Alternative Hypothesis
Give Test Statistic
Find Rejection Region for Level
Calculate Sample Size for
Decide if H0 is Rejected or
Accepted
()

Hypothesis Test about a
Population Mean
Normal pdf (known )
Null Hypothesis: H0: u = u0
Test Statistic:
z = x – u0
/n
Alternative Hypothesis: Reject Region
Ha: u > u0
(Upper Tailed)
z  z
Ha: u < u0
(Lower Tailed)
z  -z
Ha: u  u0 (Two-Tailed)either z  z/2
or z  -z/2
Hypothesis Testing Mean (Known )
Automotive engineers are using more
aluminum in the construction of cars in hopes
of improving gas mileage. For a particular
model the number of miles per gallon
obtained currently has a mean of 26.0 mpg
with a  of 5 mpg. It is hoped that a new
design will increase the mean mileage rating.
Assume that  is not affected by this change.
The sample mean for 49 driving tests with
this new design yielded 28.04 mpg. Use a
Hypothesis Test with  = .05 to make a
decision on the validity of this new design to
increase the mean mileage rating.
Hypothesis Testing Mean (Known )
As new engineering manager, you test
the melting point of 16 samples of
hydrogenated vegetable oil from
production, resulting in a sample mean
of 94.320F. Your company claims a
melting point of 950F to all vendors.
Using Hypothesis testing, test if your
production run meets the 950F
specification at level .01. Other
evidence indicates a Normal
distribution with  = 1.20 for the
melting point.
Determining  ( known)
Alternative
Hypothesis
Ha: u > u0
Ha: u < u0
Ha: u  u0 
Type II
Error (u)
 z+ u0- u
/n
1 -  -z+ u0- u
/n
z/2+ u0- u /n
 -z/2+ u0- u
/n
Determining n ( known)
n =  (z+ z)
u0 - u
One Sided
2
n =  (z/2+ z)
u0 - u
2
Two Sided
Example Type II Error
At test level .01, what is
the probability of a Type II
error when u is actually
940F?
What value of n is
necessary to ensure that
(94) = .10 when  = .01?
Hypothesis Test (Large Sample) Mean
Normal pdf (Unknown )
Null Hypothesis: H0: u = u0
Test Statistic:
z = x – u0
s/n
Alternative Hypothesis: Reject Region
Ha: u > u0
(Upper Tailed)
z  z
Ha: u < u0
(Lower Tailed)
z  -z
Ha: u  u0 (Two-Tailed)either z  z/2
or z  -z/2
For  & n use plausible values for  or
use Tables A.17
Hypothesis Test (Large Sample) Mean
Ozone is a component of smog that can
injure sensitive plants even at low levels. In
1979 a federal ozone standard of 0.12 ppm
was set. It is thought that the ozone level in
air currents over New England exceeds this
level. To verify this contention, air samples
are obtained from 64 monitoring stations set
up across the region. When the data are
analyzed, a sample mean of 0.135 and a
sample SD of 0.03 are obtained. Use a
Hypothesis test at a .01 level of significance
to test this theory.
Example HT (Large Sample) Mean
The VP of Sales claims that the salesmen
are only averaging 15 sales contacts per
week. Looking for ways to increase this
figure, the VP selects 49 salesmen at
random and the number of contacts is
recorded for a week. The sample data
reveals a mean of 17 contacts with a sample
variance of 9. Does the evidence contradict
the VP’s claim at the 5% level of significant?
Now the VP wants to detect a difference
equal to 1 call in the mean number of
customer contacts per week. Specifically, he
wants to test u = 15 against u = 16. With the
same test data, find  for this test.
Hypothesis Test (Small Sample) Mean
Normal pdf (Unknown )
Null Hypothesis: H0: u = u0
Test Statistic:
t = x – u0
s/n
Alternative Hypothesis: Reject Region
Ha: u > u0
(Upper Tailed)
t  t,v
Ha: u < u0
(Lower Tailed)
t  -t,v
Ha: u  u0 (Two-Tailed)either t  t/2,v
or t  -t/2,v
To find
 & Sample Size use Table A.17
Example HT Mean (Small Sample)
A new method for measuring phosphorus
levels in soil is being tested. A sample of 11
soil specimens with true phosphorus content
of 548 mg/kg is analyzed using the new
method. The resulting sample mean &
sample standard deviation for phosphorus
levels are 587 and 10, respectively. Is there
evidence that the mean phosphorus level
reported by the new method differs
significantly from the true value of
548 mg/kg?
Use  = .05 & assume measurements of this
type are Normal.
Example HT Mean (Small Sample)
The true average voltage drop from
collector to emitter of insulated gate
bipolar transistors is supposed to be at
most 2.5 volts. A sample of 10
transistors are used to test if the
H0:  = 2.5 versus Ha:  > 2.5 volts with
 = .05. If the standard deviation of the
voltage distribution is  = 0.10, how
likely is it that H0 will not be rejected
when in fact  = 2.6?
Hypothesis Test Population 2
Normal pdf (Unknown )
Null Hypothesis: H0: 2 = 02
Test Statistic:
2 = (n-1)s2
02
Alternative Hypothesis: Reject Region
2 > 02 (Upper Tailed)
2  2,v
Ha: 2 < 02 (Lower Tailed)
2  21-,v
Ha: 2  02 (Two-Tailed)either 2  2/2,v
Ha:
or
2  21-/2,v
Example HT Variance
Indoor swimming pools are noted for their
poor acoustical properties. The goal is to
design a pool in such a way that the average
time it takes a low frequency sound to die is
at most 1.3 seconds with a standard
deviation of at most 0.6 second. Computer
simulations of a preliminary design are
conducted to see whether these standards
are exceeded. The sample mean was 3.97
seconds and the sample standard deviation
was 1.89 seconds for 30 simulations. Does it
appear that the design specifications are
being met at the  = 0.01 level for ?
Example HT Variance
A new process for producing small precision
parts is being studied. The process consists
of mixing fine metal powder with a plastic
binder, injecting the mixture into a mold,
and then removing the binder with a solvent.
Sample data on parts that should have a
1” diameter and whose standard deviation
should not exceed 0.0025 inch yielded a
sample mean of 1.00084 with a sample
standard deviation of 0.00282 for 15
measured parts. Test at the  = .05 level to
see if this new process is viable for the
population .
P-Values
Smallest Level of
Significance at which H0
would be rejected when a
specified test procedure is
used on a given data set.
Calculating P-Values
One Sided Tests:
P = 1 – Φ(z) Upper-tailed
P = Φ(z) Lower-tailed
Two Sided Test:
P = 2 [ 1 – Φ(|z|) ]