Download Chapter 5. Inferences about Population Central Values

Chapter 5.
Inferences about Population
Central Values
Estimation versus Hypothesis Testing
• Estimation
– Information from the sample is used to estimate the
parameters of population.
• Hypothesis testing
– Information from the sample is used to determine whether the
parameters of population is greater, equal or less than a
certain value.
Estimation of µ (1)
• Point estimation
– To compute a single value (statistic) from the sample
to estimate a population parameter.
– Point estimator
• The formula for calculating specific point estimates from
sample data.
• For most situations, we will use sample mean as a point
estimate of µ. From the Central Limit Theorem for the
sample mean, we know that for large n (crudely, n ≥ 30), y
will be approximately normally distributed, with a mean µ
and a standard error σ y .
Estimation of µ (1)
• Level of confidence
– When the value of y is reported, we should also provide an
indication of how accurately y estimate µ. We will accomplish
this by considering an interval of possible values for µ in place
of using just a single value y.
– According to the Central Limit Theorem, the probability of y
falling in the interval µ ± 1.96σ y is 0.95, so state that y ± 1.96σ y
is an interval estimate of µ with level of confidence 0.95. This
interval is called the 95% confidence interval of y .
0.2
0.1
y − 1.96σ y
y + 1.96σ y
Observed y
0.0
Probability
0.3
0.4
Confidence Interval
µ − 1 .96 σ
y
µ
µ + 1.96σ y
Confidence Coefficient
•
The evaluation of the goodness of an interval estimation procedure
is examining the fraction of times in repeated sampling that
interval estimates would encompass the parameter to be estimated.
This fraction is called confidence coefficient.
•
The interval y ± 1.96σ y is constructed by setting up the confidence
coefficient 0.95. Thus, 95% of the time in repeated sampling
intervals will contain the mean µ.
Example 5.1
•
•
A study was conducted to determine the percent calories from fat
in the diet of a population of women. The Food frequency
questionnaire (FFQ) which uses a carefully designed series of
questions to determine the dietary intakes of participants in the
study. A sample of 168 women who represented a random sample
from a population of female nurses completed a single FFQ. From
the information gather from the questionnaire, the percentage of
calories from fat (PCF) was computed. The procedure for
constructing the confidence interval for µ requires that the sample
size be reasonably large or that the population distribution to have
a normal distribution. Please construct the normal probability
plot to see whether the population distribution is really a normal
distribution.
Answer
Confidence Interval for µ
•
A general formula for a confidence interval for µ with a confidence
coefficient of (1-α), where α is between 0 and 1, is:
100 ⋅ (1 − α )% C.I. = y ± zα 2σ y
where σ y = σ
•
•
n
The quantity zα 2 is a value of z having a tail area of α/2 to its
right.
In practice, it is difficult to find situations which the population
standard deviation is known. Usually both the mean and the
standard deviation must be estimated from the sample. Because σ
is estimated by the sample standard deviation s, the actual
standard error of the mean, σ n , is naturally estimated
by s n .
Choosing the Sample Size for Estimating µ (1)
•
There are two considerations in determining the appropriate
sample size for estimating µ using a confidence interval.
– First, the tolerable error establishes the desired width of the interval.
– The second consideration is the level of confidence.
•
To obtain a confidence interval having a narrow width and a high
level of confidence may require a large value for the sample size
and hence be unreasonable in terms of cost and/or time.
•
In most situation, the confidence level is set at 95% or 90%, partly
because of tradition and partly because these levels represent a
reasonable level of certainty.
Choosing the Sample Size for Estimating µ (2)
•
Suppose we want to estimate µ using a 100(1-α)% confidence
interval having tolerable error W. Our interval will be of the
form y ± E , where E=W/2. Note that W is the width of the
confidence interval. To determine the sample size n, we solve the
equation:
σ
E = zα / 2σ y = zα / 2
n
•
This formula for n is:
( zα / 2 ) 2 σ 2
n=
E2
– Employ information from a prior experiment to calculate a sample
variance s2. This value is used to approximate σ2.
– Use information on the range of the observations to obtain an
estimate of σ.
Example 5.2
• A federal agency has decided to investigate the
advertised weight printed on cartons of a certain brand
of cereal. The company in question periodically
samples cartons of cereal coming off the production
line to check their weight. A summary of 1500 of the
weights made available to the agency indicates a mean
weight of 11.80 ounces per carton and a standard
deviation of 0.75 ounce. Use this information to
determine the number of cereal cartons the federal
agency must examine to estimate the average weight of
cartons being produced now, using 99% confidence
interval of width 0.50.
• Answer
Statistical Testing or Hypothesis Testing (1)
• In hypothesis testing, there is a preconceived idea about
the value of the population parameter.
• Research hypothesis versus null hypothesis
– The hypothesis being proposed by the person conducting the
study is called the research hypothesis or alternative hypothesis.
– The negation of the research hypothesis is called null hypothesis.
– The goal of the study is to decide whether the data tend to
support the research hypothesis.
Statistical Testing or Hypothesis Testing (2)
• A statistical test is based on the concept of
proof by contradiction and is composed of the
five parts list below:
–
–
–
–
–
Research hypothesis or alternative hypothesis denoted by Ha.
Null hypothesis, denoted by H0.
Test statistics, denoted by T.S.
Rejection region, denoted by R.R.
Check assumptions and draw conclusions
Statistical Testing for µ
•
The statement that µ equals to a specific value will always be included
in H0. The particular value specified for µ is called its null value and
is denoted µ0.
•
The statement about µ that the researcher is attempting to support or
detect with the data from the study is the research hypothesis, Ha.
•
The negation of Ha is the null hypothesis, H0.
•
The null hypothesis is presumed correct unless there is overwhelming
evidence in the data that the research hypothesis is supported.
Test Statistic
• The decision to state whether or not the data support
the research hypothesis is based on a quantity
computed from the sample data called the test statistic.
• For example, a test statistic for µ is
of y .
z=
y−µ
σ
n
y or some function
Rejection Region (1)
•
The rejection region contains the values of y that support the
research hypothesis and contradict the null hypothesis, hence the
region of values for y that reject the null hypothesis.
•
The rejection region will be the values of
null distribution.
y in the upper tail of the
0.0010
0.0005
Contradictory values of y
0.0000
Probability
0.0015
Rejection Region (2)
Acceptance region
µ = 520
Rejection region
Type I and II Errors (1)
• Type I error
– To reject the null hypothesis when it is true.
– The probability of a Type I error is denoted by the symbol α.
• Type II error
– To accept the null hypothesis when it is false.
– The probability of a Type II error is denoted by the symbol β.
Type I and II Errors (2)
Null Hypothesis
Decision
True
Reject H0 Type I error
Accept H0
False
Correct
α
1-β
Correct
Type II error
1-α
β
Level of Significance
• The decision as to which values go into the rejection
region and which ones go into the non-rejection region
is made on the basis of the desired level of significance,
designated by α.
• The weight of evidence for rejecting the null hypothesis
is called the p-value.
– The probability of obtaining a value of the test statistic that is
as likely or more likely to reject H0 as the actual observed
value of the test statistic. This probability is computed
assuming that the null hypothesis is true.
One and Two Tailed Tests
• When the rejection region is located in only one tail of
the distribution of y , the statistical test is called an
one-tailed test. (H0: µ < or ≥ a certain value)
• When the rejection region is located in both tails of the
distribution of y , the statistical test is called a twotailed test. (H0: µ = or ≠a certain value)
Example 5.3
•
A corporation maintains a large fleet of company cars for its
salespeople. To check the average number of miles driven per
month per car, a random sample of n = 40 cars is examined. The
mean and standard deviation for the sample are 2752 miles and
350 miles, respectively. Records for previous years indicate that
the average number of miles driven per car per month was 2600.
Use the sample data to test the research hypothesis that the
current mean µ differs from 2600. Set α = 0.05 and assume that σ
can be estimated by s.
•
Answer
Example 5.4
•
A city manager audits the parking tickets issued by city parking
officers to determine the number of tickets that were contested by
the car owner and found to be improperly issued. In past years,
the number of improperly issued tickets per officer had a normal
distribution with mean µ=380 and σ=35.2. Because there has
recently been a change in the city’s parking regulations, the city
manager suspects that the mean number of improperly issued
tickets has increased. An audit of 50 randomly selected officers is
conducted to test whether there has been an increase in
improperly tickets. Use the sample data given here and α=0.01 to
test the research hypothesis that the mean number of improperly
issued tickets is greater than 380. The audit generates the
following data: n = 50 and y =390.
•
Answer
Computation of β
• β
– The probability of a Type II error
– The probability of incorrectly accepting the null hypothesis
• Operational characteristic curve (OC curve)
– The value of β for a number of values of µ in the alternative
hypothesis Ha and plot β versus µ in a graph called the OC
curve.
• Power
– Tests of hypotheses are evaluated by computing the probability
that the test rejects false null hypotheses.
– Power = 1 - β
– Power curve : the plot of power versus the value of µ
The Probability β (1)
Acceptance
region
Distributi on of y
Distributi on of y
under µ 0
under µ a
0.000
0.005
0.010
0.015
0.020
0.025
• β when µ0 = 410
350
375
380
420425
400
2 .0 σ
y
450
The Probability β (2)
0.025
• β when µ0 = 395
Distributi on of y
under µ 0
Distribution of y
under µ a
0.000
0.005
0.010
0.015
0.020
Acceptance
region
350
375
380
395
400
2.0σ y
425
450
The Probability β (3)
Acceptance
region
Distributi on of y
under µ 0
Distribution of y
under µ a
0.000
0.005
0.010
0.015
0.020
0.025
• β when µ0 = 435
350
380
400
2.0σ y
435
450
500
The Probability β (4)
•
The probability of a Type II error decreases (and hence power
increases) the further µ lies away from the hypothesized means
under H0.
•
For one-tailed test (H0: µ ≤ µ0 or H0: µ ≥ µ0 ), the value of β at µa
is the probability that z is less than zα − | µ0 − µ a | σ y .
µ0 − µ a
β ( µ a ) = p[ z < zα −
]
σy
β ( µ a ) : the probability of a Type II error if the actual value of
the mean is µ a
µ 0 : the null value of µ
µ a : the actual value of the mean in H a
The Probability β (5)
•
For two-tailed test (H0: µ = µ0 ), the value of β at µa can be
calculated as the following:
µ0 − µa
β ( µ a ) ≈ p[ z < zα 2 −
]
σy
β ( µ a ) : the probability of a Type II error if the actual value of
the mean is µ a
µ 0 : the null value of µ
µ a : the actual value of the mean in H a
Example 5.5
•
•
•
Prospective salespeople for an encyclopedia company are now
being offered a sales training program. Previous data indicate
that the average number of sales per month for those who do not
participate in the program is 33. To determine whether the
training program is effective, a random sample of 35 new
employees us given the sales training and then sent out into the
field. One month later, the mean and standard deviation for the
number of sets of encyclopedias sold are 35 and 8.4, respectively.
Do the data present sufficient evidence to indicate that the training
program enhances sales? (use α = 0.05)
Suppose that the encyclopedia company thinks that the cost of
financing the sales program will be offset by increased sales if
those in the program average 38 sales per month. Compute β for
µa = 38 and, based on the value β(38), indicate whether you would
accept the null hypothesis.
Answer
0.000 0.125 0.250 0.375 0.500 0.625 0.750 0.875 1.000
Probability of Type II Error
OC Curve (1)
n = 10
n = 18
n = 25
84.00
84.25
84.50
84.75
85.00
Mean
85.25
85.50
85.75
86.00
0.9
1.0
OC Curve (2)
0.7
0.6
α = 0.01
0.1
0.2
0.3
0.4
0.5
α = 0.05
0.0
Probability of Type II Error
0.8
α = 0.001
84.00
84.25
84.50
84.75
85.00
Mean
85.25
85.50
85.75
86.00
0.9
1.0
OC Curve (3)
0.8
α = 0.05
0.6
0.1
0.2
0.3
0.4
0.5
α = 0.001
0.0
Power
0.7
α = 0.01
84.00
84.25
84.50
84.75
85.00
Mean
85.25
85.50
85.75
86.00
Choosing the Sample Size for µ (1)
• The information required for determining the sample
size for a statistical test about µ is:
– the value of α
– the value of µ in the alternative, that is, µ1
– the value of β(µ1 ), that is, β.
• For any value of µ larger than µ1 , the sample size
necessary to meet the requirements with a predetermined αand β is:
– One-sided test
n=
( zα + z β ) 2 σ 2
∆2
∆ = µ1 − µ 0
Choosing the Sample Size for µ (2)
– For two-sided test
n=
( zα / 2 + z β ) 2 σ 2
∆2
∆ = µ − µ0
• If is σ2 unknown, substitute an estimated value to get an
approximate sample size.
Example 5.6
•
•
A cereal manufacturer produces cereal in boxes having a labeled
weight of 12 ounces. The boxes are filled by machines that are set
to have a mean fill per box of 16.37 ounces. Because the actual
weight of a box filled by these machines has a normal distribution
with a standard deviation approximately 0.225 ounces, the
percentage of boxes having weight less than 16 ounces is 5% using
this setting. The manufacturer is concerned that one of its
machines is underfilling the boxes and wants to sample boxes from
the machine’s output to determine whether the mean weight µ is
less than 16.37 with α = 0.05. If the true mean weight is 16.27 or
less, the manufacturer needs the probability of falling to detect this
underfilling of the boxes with a probability of at most 0.01, or risk
incurring a civil penalty from state regulators. Thus, we need to
determine the sample size n such that our test of H0 versus Ha has
α = 0.05 and β(µ) less than 0.01 whenever µ is less than 16.27
ounces.
Answer
Example 5.7
• Refer to example 5.4
• Determine the p-value for the statistical test and reach
a decision concerning the research hypothesis using
α=0.01.
• If the present value of α is 0.05 instead of 0.01, is your
decision concerning Ha change?
• Answer
Inferences about µ for a Normal
Population, σ Unknown (1)
• Gosset’s finding
– When the σ is replaced by s for small sample sizes, the test
statistic
z=
y − µ0
σ n
falsely rejecting the null hypothesis H0: µ = µ0 at a slightly
higher rate than that specified by α.
– He derived the distribution and percentage points of the test
statistic
y − µ0
s n
for n < 30.
Inferences about µ for a Normal
Population, σ Unknown (2)
• Student’s t
y − µ0
t=
s n
– This quantity from the equation above is called t statistic and
its distribution is called the Student’s t distribution or simply,
Student’s t.
Inferences about µ for a Normal
Population, σ Unknown (3)
0.4
• Comparison of two t distribution and a standard
normal distribution
0.2
0.1
0.0
Probability
0.3
Normal distribution
t distribution, df = 5
t distribution, df = 10
-6
-4
-2
0
y
2
4
6
Properties of Student’s t Distribution
•
•
•
•
•
•
There are many different t distributions. We specify a particular
one by a parameter called the degrees of freedom (df).
The t distribution is symmetrical about 0 and hence has mean
equal to 0, the same as the z distribution.
The t distribution has variance df/(df-2), and hence is more
variable than the z distribution, which has variance equal to 1.
As the df increases, the t distribution approaches the z distribution.
(Note that as df increases, the variance df/(df-2) approaches 1.)
Thus, with
y − µ0
t=
s n
We conclude that t has a t distribution with df = n-1, and, as n
increases, the distribution of t approaches the distribution of z.
Special Concerns to Student’s t
Distribution (1)
•
Degrees of freedom are pieces of information for estimating σ using
s. The standard deviation s for a sample of n measurements is
based on the deviations yi − y . Because∑ ( yi − y ) = 0 always, if
n-1 of the deviations are known, the last (nth) is fixed
mathematically to make the sum equal to 0. Thus, in a sample of n
measurements there are n – 1 pieces of information (degrees of
freedom).
•
A second method of explaining degrees of freedom is to recall that σ
measures the dispersion of the population values about µ, so prior to
estimating σ we must first estimate µ. Hence, the number of pieces
of information (degrees of freedom) in the data can be used to
estimate σ is n-1, the number of original data values minus the
number of parameters estimated prior to estimating σ.
Special Concerns to Student’s t
Distribution (2)
•
The t test (rather than z test ) should be used any time σ is
unknown and the distribution of y values is bell-shaped.
•
The probability of Type II error is dependent on three quantities,
df = n-1, α and the difference between µa and µ0 in σ units.
•
The confidence interval for µwith σ unknown is identical to the
corresponding confidence interval for µ when σ is known, with z
replaced by t and σ replaced by s.
s
n
Note : df = n - 1 and the confidence coefficient is (1 - α ).
C.I . = y ± tα 2
Example 5.8
•
A massive multistate outbreak of food-borne illness was attributed
to Salmonella enteritidis. Epidemiologists determined that the
source of the illness was ice cream. They sampled nine production
runs from the company that had produced the ice cream to
determine the level of Salmonella enteritidis in the ice cream.
These levels (MPN/g) are as follows: 0.593 0.142 0.329 0.691
0.231 0.793 0.519 0.392 0.418
Use these data to determine whether the average level of
Salmonella enteritidis in the ice cream is greater than 0.3 MPN/g a
level that is considered to be very dangerous. Set α = 0.01.
•
Answer
Influence of Nonnormality on t test (1)
•
A simulation study of the effect of skewness and heavy-tailedness
on the level and power of the t test yielded the results.
Population
Distribution
0
n = 10
n = 15
n = 20
Shift d
Shift d
Shift d
0.2
0.6
0.8
0
0.2
0.6
0.8
0
0.2
0.6
0.8
Normal
0.050 0.145 0.534 0.754 0.050 0.182 0.714 0.903 0.050 0.217 0.827 0.964
Heavy
Tailed
0.035 0.104 0.371 0.510 0.049 0.115 0.456 0.648 0.045 0.163 0.554 0.736
Light
Skewness
0.025 0.079 0.437 0.672 0.037 0.129 0.614 0.864 0.041 0.159 0.762 0.935
Heavy
Skewness
0.007 0.055 0.277 0.463 0.006 0.078 0.515 0.733 0.011 0.104 0.658 0.873
Influence of Nonnormality on t test (2)
•
Because t procedures have reduced power when sampling from
skewed populations with small sample sizes, robust methods have
been developed that are not affected by the skewness or extreme
heavy-tailedness of the population distribution.
•
Two robust methods, the sign test and Wilcoxon signed test, can be
used when the t test is not appropriate.
Inferences about the Median
Homework
•
•
•
•
•
5.52 (p.239)
5.88 (p.255)
5.95 (p.257)
5.101(p.259)
5.112 (p.260)
Answer to Example 5.1
50
s = 6.73
6.73
= 36.92 ± 1.02
168
30
40
95% C.I. = 36.92 ± 1.96 ⋅
20
Percentage Of Calories from Fat (PCF)
y = 36.92
-2.00
-1.50
-1.00
-0.50
0.00
0.50
1.00
Quantiles of Standard Normal
1.50
2.00
Answer to Example 5.2
W = 2 E = 0.5, E = 0.25
n=
zα 2 ⋅ σ 2
E2
2.582 0.752
=
= 59.91 ≈ 60
2
0.25
Answer to Example 5.3
H 0 : µ = 2600
H a : µ ≠ 2600
∴ p − value < 0.05
(In fact, p − value = 0.00596)
0.0010
0.0008
0.0006
0.0004
40
α = 0.025
α = 0.025
0.0002
n
z > 1.96
f (y)
0.0000
α = 0.05
y − µ 2752 − 2600
z=
=
= 2.75
350
σ
1500
2000
2500
3000
µ=2600
3500
4000
Answer to Example 5.4
H 0 : µ ≤ 380
H a : µ > 380
α = 0.01
y−µ
390 − 380
10
=
=
= 2.01
T .S . : z =
35.2
σ
4.979
50
n
Critical point : zα =0.01 = 2.33
z = 2.01 < 2.33, accept H 0 .
Conclusion : The number of improperly issued parking
tickets has not increased.
Answer to Example 5.4
0.004
0.006
0.008
0.010
f (y)
0.000
0.002
α = 0.01
300
400
µ = 380
500
2.33σ y
Answer to Example 5.5 (1)
H 0 : µ ≤ 33
H a : µ > 33
α = 0.01
y − µ0
35 − 33
2
T .S . : z =
=
=
= 1.41
8.4
σ
1.42
n
35
Critical point : zα =0.05 = 1.645
z = 1.41 < 1.645, accept H 0 .
Conclusion : Persons in the sales program have the same
or a smaller mean number of sales per month
as those not in the program.
Answer to Example 5.5 (2)
⎡
µ0 − µa
β (38) = P ⎢ z ≤ z0.05 −
σy
⎢⎣
⎡
⎤
⎤
33 − 38 ⎥
⎢
⎥ = P ⎢ z ≤ 1.645 −
⎥
8.4
⎥⎦
⎢⎣
35 ⎥⎦
= P[z ≤ −1.88]
= 0.0301
PWR(38) = 1 − 0.0301 = 0.9699
Conclusion :
Becasue β is relatively small, we accept the null hypothesis that
the training program has not increased the average sales per month
above the point at which increased sales would offset the cost of the
training program.
Answer to Example 5.6 (1)
H 0 : µ ≥ 16.37
H a : µ < 16.37
n=
( zα + z β ) 2 σ 2
∆2
(1.645 + 2.33) 2 ⋅ 0.2252
= 79.99 ≈ 80
=
(16.37 − 16.27) 2
Suppose that after obtaining the sample, we compute y = 16.35 ounces.
The computed value of the test statistic is :
z=
y − 16.37
σy
=
16.35 − 16.37
= −0.795 (< z0.05 = 1.645)
0.225 / 80
Conclusion :
The manufacutrer is somewhat secure in concluding that the mean fill from the examined
machine is at least 16.37 ounces.
Answer to Example 5.7
(1) H 0 : µ ≤ 380
H a : µ > 380
α = 0.01
y−µ
10
390 − 380
=
= 2.01
σ
35.2
4.979
50
n
p - value = P ( y ≥ 390) = P( z ≥ 2.01) = 0.0222 > α = 0.01
Accept H 0 .
T .S . : z =
=
Conclusion : The number of improperly issued parking
tickets has not increased.
(2)
p - value = P( y ≥ 390) = P ( z ≥ 2.01) = 0.0222 < α = 0.05
Reject H 0 .
Conclusion : The number of improperly issued parking
tickets has increased.
Answer to Example 5.8 (1)
0.2
0.3
0.4
0.5
0.6
0.7
0.8
To determine whether the data appear to have been randomly
sampled from a normal distribution.
Salmonella level
•
-1.5
-1.0
-0.5
0.0
0.5
Quantiles of Standard Normal
1.0
1.5
Answer to Example 5.8 (2)
H 0 : µ ≤ 0.3
H a : µ > 0.3
α = 0.01
Rejection region : t0.01,df =9−1=8 = 2.896
y = 0.456, s = 0.2128
y − µ00
0.456 − 0.3
= 2.21 < t0.01,8 = 2.896
s/ n
0.2128 / 9
p − value = P(t > 2.21) = 0.029
t=
=
95%C.I . = y ± tα
s
0.2128
= 0.45 ± 1.86 ⋅
= 0.45 ± 0.132
n
9
Accept H 0 .
Conclusion : The average level of Salmonella enteritidis in the ice
cream is not greater than 0.3 MPN/g.
Answer to Example 5.8 (3)
•
To calculate the probability of Type II error for the test.
µ0 − µa
]
β ( µ a ) = p[t < tα −
σ/ n
β ( µ a ) : the probability of a Type II error if the actual value of
the mean is µ a
µ 0 : the null value of µ
µ a : the actual value of the mean in H a
0.8
0.7
0.4
0.5
0.6
df = 8
df = 17
0.1
0.2
0.3
df = 26
0.0
Probability of Type II error
0.9
1.0
Answer to Example 5.8 (4)
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0
Difference standardized by σ