Download Chapter 9

Statistics
Tests of Hypotheses for a Single Sample
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.
Hypothesis Testing

Statistical hypothesis


A statistical hypothesis is a statement about the
parameters of one or more populations.
For example,
 H 0 :   50 centimeters per second
 H1 :   50 centimeters per second
 H 0 is the null hypothesis and H1 is a twosided alternative hypothesis

Type I error


Type II error


Failing to reject the null hypothesis when it is
false is defined as a type II error
Probability of type I error


Rejecting the null hypothesis H 0 when it is true
is defined as a type I error
 = P(type I error) = P(reject H 0 when H 0 is
true)
Probability of type II error

 = P(type II error) = P(fail to reject H 0 when H 0
is false)
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.
Reject null hypothesis
Fail to reject null
hypothesis
Null hypothesis (H0) is
true
Null hypothesis (H0) is
false
Type I error
False positive
Correct outcome
True positive
Correct outcome
True negative
Type II error
False negative
From Wikipedia, http://www.wikipedia.org.

Properties




 

The size of the critical region and  can be
reduced by appropriate selection of the critical
values
Type I and type II errors are related. Decrease
one will increase the other
An increase in sample size reduces 
 increases as the true value of the parameter
approaches the value hypothesized in the null
hypothesis
= 0.05
Widely used
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.

Power
1 


The probability of correctly rejecting a false null
hypothesis
Sensitivity: the ability to detect differences
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.

Formulating one-sided hypothesis
:  = 1.5 MPa
 H1 :  > 1.5 Mpa (We want)
 Or
 H 0 :  = 1.5 MPa
 H1 :  < 1.5 Mpa (We want)
 H0
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.

Formulating one-sided hypothesis
:  = 1.5 MPa
 H1 :  > 1.5 Mpa (We want)
 Or
 H 0 :  = 1.5 MPa
 H1 :  < 1.5 Mpa (We want)
 H0

P-value

The P-value is the smallest level of significance
that would lead to rejection of the null
hypothesis H 0 with the given data
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.

General procedure for hypothesis tests




Specify the test statistic to be used (such as Z 0 )
Specify the location of the critical region (twotailed, upper-tailed, or lower-tailed)
Specify the criteria for rejection (typically, the
value of  , or the P-value at which rejection
should occur)
Practical significance

Be careful when interpreting the results from
hypothesis testing when the sample size is large,
because any small departure from the
hypothesized value  0 will probably be
detected, even when the difference is of little or
no practical significance
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.

Example 9-1 Propellant Burning Rate




Suppose that if the burning rate is less than 50
centimeters per second, we wish to show this
with a strong conclusion.
H 0 :   50 centimeters per second
H1 :   50 centimeters per second
Since the rejection of H 0 is always a strong
conclusion, this statement of the hypotheses will
produce outcome if H 0 is rejected.
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.

Exercise 9-27




A random sample of 500 registered voters in
Phoenix is asked if they favor the use of
oxygenated fuels year-round to reduce air
pollution. If more than 400 voters respond
positively, we will conclude that more than 60%
of the voters favor the use of these fuels.
(a) Find the probability of type I error if exactly
60% of the voters favor the use of these fuels.
(b) What is the type II error probability  if
75% of the voters favor this action?
Hint: use the normal approximation to the
binomial.
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.
Tests on the Mean of a Normal
Distribution,Variance Known

Hypothesis tests on the mean

Hypotheses, two-sided alternative
 H 0 :   0
 H1 :    0
x  0
 Test statistic: z0 
/ n



P-value: P  2[1  (| z0 |)]
Reject H 0 if z0  z / 2 or z0   z / 2

Hypotheses, upper-tailed alternative
 H 0 :   0
 H1 :    0
 P-value: P  1  ( z0 )
 Reject H 0 if z0  z

Hypotheses, lower-tailed alternative
 H 0 :   0
 H1 :   0


P-value: P  ( z0 )
Reject H 0 if z0   z

Type II error and choice of sample size
 Finding the probability of type II error 
 Hypotheses, two-sided alternative
 H 0 :   0
 H1 :    0
 Suppose the true value of the mean under H1 is




  0  
Test statistic:
x  0 x  ( 0   )  n
z0 



/ n
/ n
Under H1
 n 
z0  N 
,1
 




n

n
    z / 2 

   z / 2 










Type II error and choice of sample size
 Sample size formulas
 If   0



n

n
    z / 2 

   z / 2 










 n 

  z / 2 
 

 Let z  be the 100 upper percentile of the
standard normal distribution. Then   ( z  )
 z   z / 2 
 n


Note


 n 
 n 




   z / 2 



z


/
2

 
 




 n 
 n 


 1    z / 2 
 1   z / 2 











n

n
    z / 2 

  z / 2 










Sample size for a two-sided test on the mean,
variance known
n

( z / 2  z  ) 2  2

2
where      0
Sample size for a one-sided test on the mean,
variance known
n
( z  z  ) 2  2

2
where      0


Operating characteristic (OC) curves
 Curves plotting  against a parameter d for
various sample size n
|   0 | 
d



 See Appendix VII
 For a given n and d , find  .
 For a given  and d , find n
Large-sample test
 If n  40 , the sample standard deviation s can
be substituted for  in the test procedures
with little effect



Example 9-2 Propellant Burning Rate
   2 ,   0.05 , n  25 , x  51.3 ,
 Specifications require that the mean burning rate
must be 50 centimeters per second. What
conclusions should be drawn?
Example 9-3 Propellant Burning Rate Type II Error
 Suppose that the true burning rate is 49
centimeters per second. What is  for the twosided test with   0.05 ,   2 , and n  25 ?
Example 9-4 Propellant Burning Rate Type II Error
from OC Curve
 Suppose the true mean burning rate is   51
centimeters per second.
|   0 | |  | 1
d




2

Example 9-4 Propellant Burning Rate Sample Size
from OC Curve
 Design the test so that if the true mean burning
rate differs from 50 centimeters per second by
as much an 1 centimeter per second, the test
will detect this with a high probability 0.90.
1    0.90

Exercise 9-47
 Medical researchers have developed a new
artificial heart constructed primarily of titanium
and plastic. The heart will last and operate
almost indefinitely once it is implanted in the
patient’s body, but the battery pack needs to be
recharged about every four hours. A random
sample of 50 battery packs is selected and
subjected to a life test. The average life of these
batteries is 4.05 hours. Assume that battery life
is normally distributed with standard deviation
   0.2 hour.
 (a) Is there evidence to support the claim that
mean battery life exceeds 4 hours? Use   0.05 .
 (b) What is the P-value for the test in part (a)?

Exercise 9-47
 (c) Compute the power of the test if the true
mean battery life is 4.05 hours.
 (d) What sample size would be required to
detect a true mean battery life of 4.5 hours if we
wanted the power of the test to be at least 0.9?
 (e) Explain how the question in part (a) could be
answered by constructing a one-sided
confidence bound on the mean life.
Tests on the Mean of a Normal
Distribution,Variance Unknown

Hypothesis tests on the mean

Hypotheses, two-sided alternative
 H 0 :   0
 H1 :    0
x  0
 Test statistic: T0 
S/ n



P-value: P  2P(Tn1 | t0 |)
Reject H 0 if t0  t / 2,n 1 or t0  t / 2,n 1


Hypotheses, upper-tailed alternative
 H 0 :   0
 H1 :    0
 P-value: P  P(Tn 1  t0 )
 Reject H 0 if t0  t , n 1
Hypotheses, lower-tailed alternative
 H 0 :   0
 H1 :   0


P-value: P  P(Tn1  t0 )
Reject H 0 if t0  t ,n 1

Type II error and choice of sample size
 Finding the probability of type II error 
 Hypotheses, two-sided alternative
 H 0 :   0
 H1 :    0
 Suppose the true value of the mean under H1 is




  0  
Test statistic:
x  0
t0 
S/ n
n ( x  (  0   ))



(n  1) S 2 1
2
n 1
n

Under H1
 t 0 is of the noncentral t distribution with
degrees of freedom and noncentrality  n / 
parameter
.

PDF of noncentral
t
distribution
From Wikipedia, http://www.wikipedia.org.

Type II error and choice of sample size
 Finding the probability of type II error 
 Hypotheses, two-sided alternative
  P{t / 2,n1  T0  t / 2,n1 |   0}

 P{t / 2,n1  T0 '  t / 2,n1}
 where T0 ' denotes the noncentral t random
variable


Operating characteristic (OC) curves
 Curves plotting  against a parameter d for
various sample size n
|   0 | 
d



 See Appendix VII
 Note that d depends on the unknown
parameter  2

Example 9-6 Golf Club Design
n  15
 It is of interest to determine if there is evidence
(with   0.05) to support a claim that the mean
coefficient of restitution exceeds 0.82.
 Data: 0.8411, …
 x  0.83725 and s  0.02456
 Example 9-7 Golf Club Design Sample Size
 If the mean coefficient of restitution exceeds 0.82
by as much as 0.02, is the sample size n  15
adequately to ensure that H 0 :   0.82 will be
rejected with probability at least 0.8?

Exercise 9-59
 A 1992 article in the Journal of the American
Medical Association (“A Critical Appraisal of 98.6
Degrees F, the Upper Limit of the Normal Body
Temperature, and Other Legacies of Carl Reinhold
August Wunderlich”) reported body temperature,
gender, and heart rate for a number of subjects. The
body temperatures for 25 female subjects follow:
97.8, …
 (a) Test the hypothesis H 0 :   98.6 versus H1 :   98.6
using   0.05 . Find the P-value.
 (b) Check the assumption that female body
temperature is normally distributed.
 (c) Compute the power of the test if the true mean
female body temperature is as low as 98.0.

Exercise 9-59
 (d) What sample size would be required to detect a
true mean female body temperature as low as 98.2
if we wanted the power of the test to be at least
0.9?
 (e) Explain how the question in part (a) could be
answered by constructing a two-sided confidence
interval on the mean female body temperature.

Exercise 9-59
 Normality plot
Tests on the Variance and Standard
Deviation of a Normal Distribution

Hypothesis tests on the variance

Hypotheses, two-sided alternative
 H 0 :  2   02
 H1 :  2   02
2
(
n

1
)
S
 Test statistic: X 02 
 02



P-value: P  P( X n21  2 / 2,n1 )  P( X n21  12 / 2,n1 )
Reject H 0 if  02  2 / 2,n1 or 02  12 / 2,n1

Hypotheses, upper-tailed alternative
 H 0 :  2   02
 H 1 :  2   02



P-value: P  P( X n21  2,n1 )
Reject H 0 if 02  2,n1
Hypotheses, lower-tailed alternative
 H 0 :  2   02
 H 1 :  2   02


P-value: P  P( X n21  12 ,n1 )
Reject H 0 if 02  12 ,n1

Type II error and choice of sample size
 Finding the probability of type II error 
 Hypotheses, two-sided alternative
 H 0 :  2   02
 H1 :  2   02


Suppose the true value of the variance under H1
is  2
(n  1) s 2
2
2
  P{1 / 2,n 1 


 / 2 , n 1 | H 1}
2
0
 02 2
(n  1) s 2  02 2
 P{ 2 1 / 2,n 1 
 2  / 2,n 1 | H1}
2




Type II error and choice of sample size
 Finding the probability of type II error 
 Hypotheses, upper-tailed alternative
 H 0 :  2   02
 H 1 :  2   02


Suppose the true value of the variance under H1
is  2
(n  1) s 2
2
  P{


 , n 1 | H 1}
2
0
 P{
(n  1) s 2
2
 02 2
 2  ,n 1 | H1}


Type II error and choice of sample size
 Finding the probability of type II error 
 Hypotheses, lower-tailed alternative
 H 0 :  2   02
 H 1 :  2   02


Suppose the true value of the variance under H1
is  2
(n  1) s 2
2
  P{1 ,n 1 
| H 1}
2
0
 02 2
(n  1) s 2
 P{ 2 1 ,n 1 
| H 1}
2



Type II error and choice of sample size
 Finding the probability of type II error 
 Hypotheses, two-sided alternative
 Operating characteristic (OC) curves
 Curves plotting  against a parameter  for
various sample size n


0

See Appendix VII


Example 9-8 Automated Filling
 n  20 , s 2  0.0153 ,   0.05.
 Is there evidence in the sample data to suggest that
the manufacture has a problem with underfilled or
overfilled bottles? (  2  0.01 )
Example 9-8 Automated Filling Sample Size
  0  0.10 ,   0.125
 Find 

Exercise 9-83
 Recall the sugar content of the syrup in canned
peaches from Exercise 8-46. Suppose that the
variance is thought to be  2  18 (milligrams)2. Recall
that a random sample of n  10 cans yields a sample
standard deviation of s  4.8 milligrams.
2
 (a) Test the hypothesis H 0 :   18 versus H1 :  2  18
using   0.05
. Find the P-value for this test.
 (b) Suppose that the actual standard deviation is
twice as large as the hypothesized value. What is the
probability that this difference will be detected by
the test described in part (a)?
2
 (c) Suppose that the true variance is   40 . How
large a sample would be required to detect this
difference with probability at least 0.90?
Tests on a Population Proportion

Large-sample tests on a proportion

Hypotheses, two-sided alternative
 H 0 : p  p0
 H1 : p  p0
X  np0
z

 Test statistic: 0
np0 (1  p0 )



P-value: P  2[1  (| z0 |)]
Reject H 0 if z0  z / 2 or z0   z / 2

Hypotheses, upper-tailed alternative
 H 0 : p  p0
 H1 : p  p0
 P-value: P  1  ( z0 )
 Reject H 0 if z0  z

Hypotheses, lower-tailed alternative
 H 0 : p  p0
 H1 : p  p0


P-value: P  ( z0 )
Reject H 0 if z0   z

Type II error and choice of sample size
 Finding the probability of type II error 
 Hypotheses, two-sided alternative
 H 0 : p  p0
 H1 : p  p0
 Suppose the true value of the proportion under H1
is p

  P{ p0  z / 2 p0 (1  p0 ) / n  p  p0  z / 2 p0 (1  p0 ) / n | H1}
 p0  z / 2 p0 (1  p0 ) / n  p 
 p0  z / 2 p0 (1  p0 ) / n  p 
  

 




p
(
1

p
)
/
n
p
(
1

p
)
/
n





Type II error and choice of sample size
 Finding the probability of type II error 
 Hypotheses, upper-tailed alternative
 H 0 : p  p0
 H1 : p  p0
 Suppose the true value of the proportion under H1
is p

  P{ p  p0  z p0 (1  p0 ) / n | H1}
 p0  z p0 (1  p0 ) / n  p 

 


p
(
1

p
)
/
n



Type II error and choice of sample size
 Finding the probability of type II error 
 Hypotheses, lower-tailed alternative
 H 0 : p  p0
 H1 : p  p0
 Suppose the true value of the proportion under H1
is p

  P{ p0  z p0 (1  p0 ) / n  p | H1}
 p0  z p0 (1  p0 ) / n  p 

 1  


p
(
1

p
)
/
n



Type II error and choice of sample size
 Two-sided alternative
 Let z  be the 100 upper percentile of the
standard normal distribution. Then   ( z  )
 p0  z / 2 p0 (1  p0 ) / n  p 
 p0  z / 2 p0 (1  p0 ) / n  p 
  

  




p
(
1

p
)
/
n
p
(
1

p
)
/
n




 p0  z / 2 p0 (1  p0 ) / n  p 

 


p(1  p) / n


 z 
p0  z / 2 p0 (1  p0 ) / n  p
p(1  p) / n
 z / 2 p0 (1  p0 )  z 
n
p  p0

p(1  p) 


2

Type II error and choice of sample size
 Upper-tailed alternative
 Let z  be the 100 upper percentile of the
standard normal distribution. Then   ( z  )
 p0  z p0 (1  p0 ) / n  p 

  


p
(
1

p
)
/
n


 z 
p0  z p0 (1  p0 ) / n  p
p(1  p) / n
 z p0 (1  p0 )  z 
n
p  p0

p(1  p) 


2

Type II error and choice of sample size
 Lower-tailed alternative
 Let z  be the 100 upper percentile of the
standard normal distribution. Then   ( z  )
 p0  z p0 (1  p0 ) / n  p 

  1  


p
(
1

p
)
/
n


  p0  z p0 (1  p0 ) / n  p 

 


p
(
1

p
)
/
n


 z 
 p0  z p0 (1  p0 ) / n  p
p(1  p) / n
 z p0 (1  p0 )  z 
n
p  p0

p(1  p) 


2


Example 9-10 Automobile Engine Controller
 p  0.05 ,   0.05 , n  200
 The semiconductor manufacturer takes a random
sample of 200 devices and finds that four of them
are defective. Can the manufacturer demonstrate
process capability for the customer? ( p  0.05 )
Example 9-11 Automobile Engine Controller Type II
Error
 Suppose that its process fallout is really p  0.03
.
What is the  -error for a test of process capability
that uses n  200 and   0.05 ?

Exercise 9-95
 In a random sample of 85 automobile engine
crankshaft bearings, 10 have a surface finish
roughness that exceeds the specifications. Does this
data present strong evidence that the proportion of
crankshaft bearings exhibiting excess surface
roughness exceeds 0.10?
 (a) State and test the appropriate hypotheses using
  0.05 .

 (b) If it is really the situation that p  0.15, how likely
is it that the test procedure in part (a) will not
reject the null hypotheses?
 (c) If p  0.15 , how large would the sample size
have to be for us to have a probability of correctly
rejecting the null hypothesis of 0.9?
,
,
Testing for Goodness of Fit








Test the hypothesis that a particular distribution will
be satisfactory as a population model
Based on the chi-square distribution
n observations, p is the number of parameters of
the hypothesized distribution estimated by sample
statistics
Oi : the observed frequency in the i th class
interval
Ei : the expected frequency in the i th class
interval
2
k
(
O

E
)
2
i
i
Test statistic: X 0  
Ei
i 1
2
2
P-value: P  P(  k  p1  0 )
2
2



Reject the hypothesis if 0
 ,k  p 1


Example 9-12 Printed Circuit Board Defects, Poisson
Distribution
 Number of defects: 0, observed frequency: 32
 Number of defects: 1, observed frequency: 15
 Number of defects: 2, observed frequency: 9
 Number of defects: 3, observed frequency: 4
Example 9-13 Power Supply Distribution, Continuous
Distribution
 x  5.04 , s  0.08 , n  100
 A manufacturer engineer is testing a power supply
used in a notebook computer and, using   0.05 ,
wishes to determine whether output voltage is
adequately described by a normal distribution.

Exercise 9-101
 The number of cars passing eastbound through the
intersection of Mill and University Avenues has been
tabulated by a group of civil engineering students.
They have obtained the data in the adjacent table:
 (a) Does the assumption of a Poisson distribution
seem appropriate as a probability model for this
process? Use   0.05.
 (b) Calculate the P-value for this test.
 Data: (40, 14), (41, 24), …
Contingency Table Tests
Test the hypothesis that two methods of
classification are statistically independent
 Based on the chi-square distribution
 n observations, r  c contingency table
 Oij : the observed frequency for level i of the first
classification and level j for the second
classification
c
1 r
1

Eij  nuˆi vˆ j
uˆi   Oij , vˆ j   Oij ,
n i 1
n j 1
r
c (O  E ) 2
ij
ij
 Test statistic: X 02  
Eij
i 1 j 1



2
2
P-value: P  P( ( r 1)(c1)  0 )
2
2
Reject the hypothesis if 0   ,( r 1)(c1)


Example 9-13 Health Insurance Plan Preference
 A company has to choose among three health
insurance plans. Management wishes to know
whether the preference for plans is independent of
job classification and wants to use   0.05 .
 n  500 , data: …
Exercise 9-107
 A study is being made of the failure of an electronic
component. There are four types of failures possible
and two mounting positions for the device

A
B
C
D
1
20
48
20
7
2
4
17
6
12
Would you conclude that the type of failure is
independent of the mounting position? Use   0.01.
Find the P-value for this test.
Nonparametric Procedures

The sign test
Test hypotheses about the median ~ of a
continuous distribution
~ 0 )
 r  : the observed number of plus signs ( X i  
0
 Hypotheses, two-sided alternative
~ ~
 H 0 :   0
~  ~
 H :

1

0

P-value: P  2 P( R  r






P

2
P
(
R

r
or
Reject H 0 if P  
1
when p  ) if r   n / 2
2
1
when p  ) if r   n / 2
2

Hypotheses, upper-tailed alternative
~  ~
 H0 : 
0
~  ~
 H :
1



0
1
P-value: P  P( R   r  when p  )
2
Reject H 0 if P  
Hypotheses, lower-tailed alternative
~ ~
 H 0 :   0
~  ~
 H :
1
0
1


P

P
(
R

r
when
p

)
 P-value:
2
 Reject H 0 if P  




Appendix Table VIII ( r )
Hypotheses, two-sided alternative
~  ~
 H0 : 
0
~  ~
 H1 : 
0



 Reject H 0 if min( r , r )  r
Hypotheses, upper-tailed alternative
~ ~
 H 0 :   0
~ ~
 H1 :    0
Reject H 0 if r   r
Hypotheses, lower-tailed alternative
~  ~
H
:

0
0

~ ~
 H :  


1
0


r

r
H
 Reject

0 if


Ties in the sign test
~ should be set
 Values of X i exactly equal to 
0
aside and the sign test applied to the remaining
data
Normal approximation for sign test statistic

R
 0.5n

Z0 
0.5 n

 Reject H if | z | z
for H : ~  ~
0


0
 /2
or if z0  z
or if z0   z
1
0
for H1 : ~  ~0
for H1 : ~  ~0

Type II error for the sign test
 Finding the probability of type II error 
~ , say, ~   , must
 Not only a particular value of 
be used but also the form of the underlying
distribution will affect the calculations

Wilcoxon signed-rank test

w
 Appendix Table IX (  )
 Rank the absolute differences | X i  0 | in ascending
order, and then give the ranks the signs of their
corresponding differences

 w : the sum of the positive ranks

 w : the absolute value of the sum of negative ranks
 Hypotheses, two-sided alternative
 H 0 :   0
 H1 :    0

Reject H 0 if min( w , w )  w

Wilcoxon signed-rank test

w
 Appendix Table IX (  )
 Hypotheses, upper-tailed alternative
 H 0 :   0
 H1 :    0


 Reject H 0 if w  w
 Hypotheses, lower-tailed alternative
 H 0 :   0
 H1 :   0


 Reject H 0 if w  w


Ties in the Wilcoxon signed-rank test
 If several observations have the same absolute
magnitude, they are assigned the average of the
ranks that they would receive if they differed
slightly from one another
Normal approximation for Wiocoxon signen-rank
test statistic





Z0 
W   n(n  1) / 4
n(n  1)( 2n  1) / 24
~ ~
Reject H 0 if | z0 | z / 2 for H1 :   0
or if z0  z for H1 : ~  ~0
or if z0   z for H1 : ~  ~0

Example 9-15 Propellant Shear Strength Sign Test
n  20
 We would like to test the hypothesis that the
median shear strength is 13790 kN/m2, using   0.05
 Example 9-16 Propellant Shear Strength Wilcoxon
Signed-Rank Test
n  20
 We would like to test the hypothesis that the
median shear strength is 13790 kN/m2, using   0.05

Exercise 9-117
 A primer paint can be used on aluminum panels. The
drying time of the primer is an important
consideration in the manufacturing process. Twenty
panels are selected and the drying times are as
follows: 1.6, …
 Is there evidence that the mean drying time of the
primer exceeds 1.5 hr?