Transcript distribution study

Chapter 9
Test of Hypotheses for a
Single Sample
Learning Objectives
• Structure hypothesis tests
• Test hypotheses on the mean of a normal
distribution using either a Z-test or a t-test
procedure
• Test hypotheses on the variance or standard
deviation of a normal distribution
• Test hypotheses on a population proportion
• Use the P-value approach for making
decisions in hypotheses tests
Learning Objectives
• Compute power, type II error probability,
and make sample size selection decisions
for tests on means, variances, and
proportions
• Explain and use the relationship between
confidence intervals and hypothesis tests
• Use the chi-square goodness of fit test to
check distributional assumptions
• Use contingency table tests
Statistical Hypotheses
• Construct a confidence interval estimate of a
parameter from sample data
• Many problems require that we decide
whether to accept or reject a statement about
some parameter
– Called a hypothesis
• Decision-making procedure about the
hypothesis is called hypothesis testing
Statistical Hypotheses
• Useful aspects of statistical inference
• Relationship between hypothesis testing and
confidence intervals
• Focus is on testing hypotheses concerning the
parameters of one or more populations
Sources of Null Hypothesis
• Three ways to specify the null hypothesis
– May result from past experience or knowledge
– May be determined from some theory or model
regarding the process under study
– May be determined from external
considerations
Sample, Population and Statistical
Inference
• Statements about the population, not statements
about the sample
• If this information is consistent with the hypothesis
– Conclude that the hypothesis is true
• If this information is inconsistent with the
hypothesis
– Conclude that the hypothesis is false
• Truth or falsity of a particular hypothesis can never
be known with certainty
• Unless we can examine the entire population
Structure of Hypothesis-testing
• Identical in all the applications
• The null hypothesis
– Hypothesis we wish to test
– Rejection of the null hypothesis always leads to
accepting the alternative hypothesis
• The alternate hypothesis
–
–
–
–
Takes on several values
Involves taking a random sample
Computing a test statistic from the sample data
Make a decision about the null hypothesis
Structure of Hypothesis-testing
• Suppose a manufacturer is interested in the
output voltage of a power supply
– Null hypothesis :H0: =50 v
– Alt. hypothesis: H1: 50 v
– A sample of n=10 specimens is tested
• If the sample mean is close to 50 v
– Such evidence supports the null hypothesis H0
• If the sample mean is different from 50 v
– Such evidence is in support of the alternative
hypothesis
Critical Regions
• The sample mean can take on many different
values
Suppose 48.5 x  51.5
Constitutes the critical region for the test
Acceptance region
The boundaries between the critical regions and the
acceptance region are called the critical values
– Critical values are 48.5 and 51.5
–
–
–
–
Reject H0
50 Volts
48.5
Fail to
reject H0
=50 Volts
50
Reject H0
50 Volts
51.5
Two Types of Error
• Type I Error
– Occurs when a true null hypothesis is rejected
– Value of  represents the probability of committing this
type of error
•  = P (Ho is rejected / Ho is true)
•  is called the significance level of the test
• Type II Error
– Occurs when a false null hypothesis is not rejected
– Value of  represents the probability of committing a type
II error
– = P(fail to reject Ho / Ho is false)
Two Types of Error
• In testing any statistical hypothesis, four different
situations determine whether the final decision is
correct or in error
Do not reject H0
H0 is True
Correct
Decision
H0 is False
Type II or 
error
Reject H0
Type  or 
error
Correct
Decision
• Probabilities can be associated with the type I
and type II errors
Probability of Making a
Type I Error
• Probability of making a type I error is
denoted
–  = P (type I error) =P (reject H0 when H0 is
true)
• Sometimes the type I error probability is
called the significance level or the -error
• Previous example, a type I error will occur
when either sample mean > 51.5 or <48.5
or when the true voltage is =50 v
Probability of Type I Error
• Probability of type I error can be shown by the
tails of normal distribution
•  = P(X<48.5 when =50) +P(X>51.5 when =50)
• Using the corresponding z-values, =0.0574
• 5.74% of all random samples would lead to rejection of
the hypothesis when the true mean is really 50 v
Probability of Type II Error
• Examine the probability of a Type II error
•  = P(type II error) = P(fail to reject H0 when H0 is
false)
• Have a specific alternative hypothesis
• Alterative hypothesis: H1: =52 v
• A type II error will be committed if the sample mean
falls between 48.5 and 51.5 when =52 v
• Probability that 48.5x-bar 51.5 when H0 is false
(because =52)
• Using the z-values, the =0.2643
• Corresponds to testing H0: =50 against H1: 50
with n =10, when the true value of the mean=52
Power of a Statistical Test
• Power = 1-
• Power=1-0.2643=0.7357
• Power can be interpreted as the probability
of correctly rejecting a false null hypothesis
• If it is too low, the analyst can increase
either  or the sample size n
Example-1
•
•
A textile fiber manufacturer is investigating
a new drapery yarn, which the company
claims has a mean thread elongation of 12
kg with a standard deviation of 0.5 kg.
The company wishes to test the hypothesis
H0: =12 against H1<12, using a random
sample of four specimens.
1. What is the type I error probability if the critical
region is defined as sample mean <11.5 kg?
2. Find  for the case where the true mean
elongation is 11.25 kg.
Solution-Part 1
•
 = P(reject H0 when H0 is true)
= P(X  11.5 when  = 12)
X   11.5  12 

 P


  / n 0.5 / 4 
= P(Z  2)= 1  P(Z  2)
= 1  0.97725 = 0.02275
•
The probability of rejecting the null
hypothesis when it is true is 0.02275
Solution-Part 2
•
 = P(accept H0 when  = 11.25)
. when   1125
. 
= P X  115
X   11.5  11.25 

 P


0.5 / 4 
 / n
•
=P(Z > 1.0) = 1  P(Z  1.0)
=1  0.84134 = 0.15866
The probability of accepting the null
hypothesis when it is false is 0.15866.
Tests on The Mean of a Normal
Distribution, Variance Known
• Consider hypothesis testing about the mean  of a single,
normal population where the variance of the population 2 is
known
• Random sample X1, X2, … , Xn has been taken from the
population
• x is an unbiased point estimator of  with variance 2/n
• Test the hypotheses
• H0: =0
• H1: 0
• Test procedure for H0:=0 uses the test statistic
Cont.
• If  is the significance level, the probability
that the test statistic Z0 falls between -Z/2
and Z/2 will be 1-
• Regions associated with -Z/2 and Z/2
• Reject H0 if the observed value of the test
statistic z0 is either > Z/2 or <-Z/2
Location of Critical Region for
One-sided Tests
The distribution of Z0
when H0: =0 is true,
for the one-sided
alternative H1: >0
The distribution of Z0
when H0: =0 is true,
for the one-sided
alternative H1: <0
Relationship between the Signs
in Ho And H1
Two-sided
Test
Left-sided
Test
Right-sided
Test
Sign in the H0
=
= or 
= or 
Sign in the H1

<
>
Rejection
Region
On both
sides
On the left
side
On the right
side
General Procedure for
Hypothesis Tests
•
Following steps will be used
1. From the problem, identify the parameter
of interest
2. State the null hypothesis, H0
3. Specify an appropriate alternative
hypothesis, H1
4. Choose a significance level 
5. Determine an appropriate test statistic
General Procedure for
Hypothesis Tests
6. State the rejection region for the statistic
7. Compute any necessary sample quantities,
substitute these into the equation for the test
statistic, and compute that value
8. Decide whether or not H0 should be rejected
Example-2
• A manufacturer produces crankshafts for an
automobile engine
• The wear of the crankshaft after 100,000 miles
(0.0001 inch) is of interest because it is likely to
have an impact on warranty claims
• A random sample of n=15 shafts is tested and
sample mean is 2.78
• It is known that =0.9 and that wear is normally
distributed
– Test H0: =3 versus H1: 3 using =0.05
Solution
•
Using the general procedure for hypothesis
testing
1.
2.
3.
4.
5.
The parameter of interest is the true mean crankshaft wear, 
H0 :  = 3
H1 :   3
 = 0.05
x
Test statistic is z0 
/ n
6. Reject H0 if z0 < z /2 where z0.005 = 1.96 or z0 > z/2 where
z0.005 = 1.96
7.
x  2.78 and  = 0.9
2.78  3
z 
 0.95
0.9 / 15
0
8. Since –0.95 > -1.96, do not reject the null hypothesis and
conclude there is not sufficient evidence to support the claim the
mean crankshaft wear is not equal to 3 at  = 0.05
P-Values in Hypothesis Tests
• P-value is the smallest level of significance that
would lead to rejection of the null hypothesis H0
• Conveys much information about the weight of
evidence against H0
• Adopted widely in practice
• P-value is
Example-3
• What is the P-value in Example-2?
• Solution
P-value = 2[1  (|-0.95|)]
=2[1-0.8289]
=0.34
Probability of Type II Error
• Analyst directly selects the type I error probability
• Probability of type II error depends on the choice of
sample size
• Considering a two-sided test, the probability of the
type II error is the probability that Z0 falls between Z/2 and Z/2 given that H1 is true
• Expressed mathematically, this probability is
Example-4
• What is the power of the test in Example-2 if
=3.25?
• Solution
3  3.25 
3  3.25 


   z 
    z 

0.9 / 15 
0.9 / 15 


0.025
0.025
= (1.96 + 1.075)  (1.96 + 1.075)
= (0.884)  (3.035)
= 0.81057—0.001223
=0.8098
• Power=1-0.8098
=0.1902
Choice of Sample Size
• Suppose that the null hypothesis is false and that the true
value of the mean is = +0, where >0
• Formulas that determine the appropriate sample size to
obtain a particular value of  for a given  and  are
• Two sided-test alternative hypotheses
• For either of the one-sided alternative hypotheses
• Where =-0
Example-5
• Considering the problem in Example-2, what
sample size would be required to detect a true
mean of 3.75 if we wanted the power to be at
least 0.9?
• Solution
n
z
/2
 z   2
2

2

z 0.025  z 0.10 2  2
(3.75  3) 2
(1.96  1.29) 2 (0.9) 2

 20.218,
2
(0.75)
Using Operating
Characteristic Curves
• Convenient to use the operating
characteristic curves in Appendix Charts
VIa and VIb to find 
• Curves plot  against a parameter d for
various sample sizes n
• d is defined as
Using Operating
Characteristic Curves
• Curves are provided for both =0.05 and =0.01
• When d=0.5, n =25, and =0.05, Then =0.3
• 30% chance that the true will not be detected by the test
with n =25
Large-Sample Test
• In most practical situations 2 will be unknown
• May not be certain that the population is well
modeled by a normal distribution
• If n is large (say n>40)
• Sample standard deviation s can be substituted
for 
• Valid regardless of the form of the distribution of
the population
• Relies on the central limit theorem just as we did
for the large sample confidence interval
Tests On The Mean Of A Normal
Distribution, Variance Unknown
• Situation is analogous to Ch.8, where we
constructed a C.I.
• S2 replaces 2
• Use the test statistic
• Use the critical values -tα/2,n-1 and tα/2,n-1 as the
boundaries of the critical region
• Reject H0: μ=μ0 if t0> tα/2,n-1 or if t0<-tα/2,n-1
Location of the Critical Region
• H1: μ > μ0
• H1: μ  μ0
• H1: μ < μ0
Choice of Sample Size
• Situation is analogous to the case, where
variance was known
• Use the sample variance s2 to estimate 2
• Charts VIe, VIf, VIg, and VIh plot β for the ttest against a parameter d for various sample
sizes n
Hypothesis Tests on the Standard
Deviation of a Normal Population
• Wish to test the hypotheses on the population
variance
• Suppose
– H0: 2=20
– H1: 2#20
• Use the test statistic
• X2 follows the chi-square distribution with n-1
degrees of freedom
• Reject H0: 2=20 if X20> X2α/2,n-1 or if X20 <X21-α/2,n-1
Location of the Critical Region
• Reference distribution for the test of H0: 2=20
with critical region values
• H1: 2  20
• H1: 2 <20
• H1: 2 >20
Tests on a Population Proportion
• Test hypotheses on a population proportion
• Recall Pˆ =X/n is a point estimator of the
proportion of the population p
• Sampling distribution of Pˆ is approximately
normal with mean p and variance p(1-p)/n
• Now consider testing the hypotheses
– H0: p=p0
– H1: p#p0
• Use
• Reject H0: p=p0 if z0>zα/2 or z0<-zα/2
Type II Error and Choice of
Sample Size
• Obtain closed-form equations for the approximate
β-error for the tests
• For the two-sided alternative, sample size
equation
• For the one-sided alternative, sample size
equation
Example-6
• An article in Fortune (September 21, 1992) claimed
that nearly one-half of all engineers continue
academic studies beyond the B.S. degree,
ultimately receiving either an M.S. or a Ph.D.
degree
• Data from an article in Engineering Horizons
(Spring 1990) indicated that 117 of 484 new
engineering graduates were planning graduate
study
– Are the data from Engineering Horizons consistent with
the claim reported by Fortune? Use =0.05 in reaching
your conclusions
– Find the P-value for this test
Solution
•
Part 1
1.
2.
3.
4.
5.
True proportion of engineers, p
H0 : p = 0.50
H1 : p  0.50
 = 0.05
p  p0
z

Test statistic 0
p 1 p
0

0

n
6.
7.
Reject H0 if z0 <  z/2 where z/2 = z0.025 = 1.96 or z0 > z/2
where z/2 = z0.025 = 1.96
x = 117 n = 484
z0 
p  p0
p0 1  p0 
n
8.
•

p 
117
 0.242
484
0.242  0.50
0.5(1  0.5)
484
 11352
.
Since 11.352 < 1.96, reject the null hypothesis
Part 2
1.
P-value = 2(1  (11.352)) = 2(1  1) = 0
Testing for Goodness of Fit
• Not know the underlying distribution of the
population
• Wish to test the hypothesis that a particular
distribution will be satisfactory as a population
model
– For example, test the hypothesis that the population is
normal
• Used a very useful graphical technique called
probability plotting
• A formal goodness-of-fit test procedure based
on the chi-square distribution
Test Procedure
• Requires a random sample of size n from the population
whose probability distribution is unknown
• Arranged the n observations in a frequency histogram
– k bins or class intervals
• Let Oi be the observed frequency in the ith class interval
• Compute the expected frequency, Ei, in the ith class interval
• Test statistic
• If the population follows the hypothesized distribution, X2 has
a chi-square distribution with k-p-1 d.o.f.
– p represents the number of parameters of the hypothesized
distribution
• Reject the hypothesis that the distribution if the calculated
value of the test statistic
Magnitude of Expected
Frequencies
• Application of this test procedure concerns
the magnitude of the expected frequencies
• There is no general agreement regarding
the minimum value of expected frequencies
• But values of 3, 4, and 5 are widely used as
minimal
• Use 3 as minimal
Example-7
• Consider the following frequency table of
observations on the random variable X
Values
0
1
2
3
4
Observed Frequencies
24
30
31
11
4
• Based on these 100 observations, is a
Poisson distribution with a mean of 1.2 an
appropriate model? Perform a goodnessof-fit procedure with α=0.05
Solution
•
•
•
•
•
Mean=1.2
Degrees of freedom=k-p-1=4-0-1=3
Compute pi
P1=P(X=0)=[e-01.2 (0.75)0]/0! =0.3012
Similarly, p2, p3, p4, and p5 are 0.3614, 0.2169,
0.0867, 0.0260, respectively
• Expected frequency Ei=npi
• Expected frequency for class 1 =0.3012*100=30.12
• Similarly, the E2,E3, E4, and E5 are 36.14, 21.69,
8.67, and 2.60, respectively
Solution-Cont.
• Summarize the expected and observed frequencies are
ValueObserved frequency
0
1
2
3
4
24
30
31
11
4
Expected frequency
30.12
36.14
21.69
8.67
2.69
• Expected frequency in the last interval is < 3, we combine
the last two rows
ValueObserved frequency
0
24
1
30
2
31
3-4
15
Expected frequency
30.12
36.14
21.69
11.67
Solution-Cont.
1.
2.
3.
4.
5.
6.
7.
Variable of interest
H0: Form of the distribution is Poisson
H1: Form of the distribution is not Poisson
 = 0.05
k O  E 2
Test statistic is 20  i1 i Ei i
Reject H0 if 2o  20.05,3  7.81
Computations
2
2
24

3012
.
15

1167
.




2 

0
3012
.
1167
.
 7.23
8. Since 7.23 < 7.81 do not reject H0
Contingency Table Tests
• Classify n elements of a sample to two different criteria
• Know whether the two methods of classification are
statistically independent
• Example
– Consider the population of graduating engineers
– May wish to determine whether starting salary is independent of
academic disciplines
• First method of classification has r levels
• Second method has c levels
• Let Oij be the observed frequency for level i of the first
classification and level j on the second classification
• Construct an r x c table
• Called an r x c contingency table
Testing the Hypothesis
• Interested in testing the hypothesis that the methods of
classification are independent
• By rejecting the hypothesis
– There is some interaction between the two criteria of classification
• The statistic
• Has an chi-square distribution with (r-1)(c-1) d.o.f.
• Reject the hypothesis of independence if
Example
•
•
•
•
Patients in a hospital are classified as surgical or medical
A record is kept of the number of times patients require
nursing service during the night and whether or not these
patients are on Medicare. The data are presented here:
• Test the hypothesis (using =0.01) that calls by surgical
medical patients are independent of whether the patients
are receiving Medicare. Find the P-value for this test
• Solution