Stat 491: Biostatistics Chapter 7: Testing Hypothesis–One Sample Inference Solomon W. Harrar

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A Statistical Test for µ
Test for the Binomial Distribution
Stat 491: Biostatistics
Chapter 7: Testing Hypothesis–One Sample Inference
Solomon W. Harrar
The University of Montana
Fall 2012
Elements of a Test of Hypothesis
There is a preconceived idea (set of values) about µ.
A statistical test is an approach used to see whether there is a
compelling evidence in a random sample in favor of these
preconceived values for µ.
Five components of a statistical test are
a. The research hypothesis Ha known as the Alternative
Hypothesis. Another commonly used notation for the
alternative hypothesis is H1 .
b. The negation of the research hypothesis (or the status quo) H0
known as the Null Hypothesis.
c. The test statistic (T.S.): a quantity computed from the
sample used to state whether or not the data supports Ha .
d. Rejection Region (R.R): very unlikely values of the test
statistic if H0 is true.
e. Check assumptions and draw conclusions.
Elements of a Test of Hypothesis Cont’d ...
Example: Suppose average cholesterol level in children is 175
mg/dL. A group of men who have died from heart disease
within the past year are identified, and cholesterol levels of
their offsprings are measured. Let µ be the mean cholesterol
level (mg/dL) of all children whose fathers have died from
heart disease within the past year. The main question here is
”Is there a familial aggregation of cardiovascular risk
factors?”. State the null and alternative hypothesis.
The two hypotheses are:
H0 : µ = 175
and Ha : µ > 175.
We have seen how to address this problem using the CI
method. Another way is in terms of hypothesis testing.
¯ to base our inference on. That is,
It is reasonable to use X
¯.
our test statistic will be based on X
Two Types of Errors
The four possible scenarios in the hypothesis testing are given
in the following table.
Test Decision
Reject H0
Accept H0
State of Nature
H0 is True
H0 is False
Type I error
Correct
Correct
Type II error
Denote by α the Probability of Type I Error. That is,
α = P(Rejecting H0 |H0 is true).
Denote by β the Probability of Type II Error. That is,
β = P(Accepting H0 |H0 is false).
Example: Interpret α and β in the context of the familial
aggregation of cardiovascular risk factor.
Forming the Rejection Region (R.R.)
Ideally, we would like to construct our decision rule such that
both α and β are small.
This is impossible for a fixed sample size because the two are
inversely related.
We specify a tolerable probability for type I error α = 0.05,
say, and then locate the rejection region R. R.
For the familial aggregation of cardiovascular-risk-factor
example, if the distribution of cholesterol of children who lost
their father within the past year due to heart disease is
¯ is normal.
normally distributed, the sampling distribution of X
More precisely,
¯ ∼ N(µ, σ 2 /n).
X
Forming the Rejection Region (R.R.) Cont’d...
Let us now consider the test statistic
t=
¯ − µ0
X
√
s/ n
H0
∼
tn−1
where µ0 is the hypothesized value and for this example
µ0 = 175. This test statistic is known as the t-test statistic.
It would make sense to reject H0 for large values of t.
We want now to define our rejection region so that the
probability of type I error is 0.05.
We know that
P(t > tn−1,0.95 |H0 is True) = 0.05.
We use the rejection region, t > tn−1,0.95 which guarantee the
probability of falsely rejecting H0 does not exceed 0.05.
Graphical Display of the Rejection Region
density
tn−−1
0
t1−−α
x
Summary of a Statistical Test for µ
Assume the population is normally distributed
Hypotheses:
Case 1. H0 : µ ≤ µ0
Case 2. H0 : µ ≥ µ0
Case 3. H0 : µ = µ0
T.S.: t =
vs
vs
vs
Ha : µ > µ0 (One-Sided Alternative)
Ha : µ < µ0 (One-Sided Alternative)
Ha : µ =
6 µ0 (Two-Sided Alternative)
¯ −µ0
X
√
s/ n
R.R.:
Case 1. Reject H0 if t ≥ tn−1,1−α
Case 2. Reject H0 if t ≤ −tn−1,1−α
Case 3. Reject H0 if t ≤ −tn−1,1−α/2 or t ≥ tn−1,1−α/2 . In short,
|t| ≥ tn−1,1−α/2 .
Example: Cardiovascular Disease
Suppose we want to compare fasting serum-cholesterol levels
among recent Asian immigrants to the united states with typical
levels found in the general US population. The mean cholesterol
levels in women ages 21-40 in the US is 190 mg/dL. It is unknown
whether cholesterol levels among recent Asian immigrants are
higher or lower than those in the general US population. Let us
assume that levels among recent female Asian immigrants are
normally distributed. Blood tests are performed on 100 female
Asian immigrants ages 21-40, and the mean and standard deviation
of these 100 people are 181.52 mg/dL and 40 mg/dL, respectively.
What can we conclude on the basis of this sample evidence?
Example: Salmonella Enteritidis
A massive multi-state 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 in MPN/g (most
probable number per gram) 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.
R-code
t.test(x=c(0.593, 0.142, 0.329, 0.691, 0.231,
0.793, 0.519, 0.392, 0.418), mu=0.3,
alternative="greater")
Example: Salmonella Enteritidis Cont’d ...
Box Plot of Salmonella enteritidis Data
0.8
1.0
Normal Q−Q Plot
●
0.6
●
●
0.4
Sample Quantiles
●
●
●
●
0.2
●
0.0
●
−1.5
−1.0
−0.5
0.0
0.5
1.0
1.5
0.2
0.3
0.4
Theoretical Quantiles
0.5
0.6
0.7
0.8
Level of Significance (p-value)
Reporting the result of the hypothesis test by specifying the
rejection region (as we have been doing so far) is called
critical-value method
An alternative is the p-value method which allows you to
state the weight of evidence in the sample against H0 .
A p-value (level of significance) is the probability of obtaining
a value of the test statistic that is as or more extreme than the
actual value obtained, given that the null hypothesis is true.
In other words, it is the probability of observing a sample
outcome as or more contradictory to H0 than the observed
sample result, given the null hypothesis is true.
The smaller the p-value is the heavier the evidence in the
sample against H0 .
Level of Significance (p-value) Cont’d ...
For a specified probability of type I error α,
Reject H0 if p-value ≤ α
Fail to Reject H0 if p-value > α
Summary of how to calculate p-value
Case 1
H0 : µ ≤ µ0
Ha : µ > µ 0
P(t ≥ computed t)
Case 2
H0 : µ ≥ µ0
Ha : µ < µ 0
P(t ≤ computed t)
case 3
H0 : µ = µ0
Ha : µ 6= µ0
2P(t ≥ |computed t|)
In R, to compute p-value you use the command
pt(computed t, n-1) #for lower-sided test
1-pt(computed t,n-1)#for upped-sided test
2*(1-pt(abs(computed t),n-1)) #for two-sided test
Graphical Display of Two-Sided p-value
density
tn−−1
−t
0
t
x
Level of Significance (p-value) Cont’d ...
The entries of the last row in the above table are the p-values.
Example: For the Salmonella-Enteritidis example, compute
the p-value.
For the Asian-immigrants example, compute the p-value.
Remarks: Compared to the critical-value method, the p-value
method is more precise in that we are reporting the weight of
the evidence against H0 . The decision of whether the evidence
is enough is up to the reader.
The Z -test
Assume the standard deviation of the population σ is known.
Consider the statistic
Z=
¯ − µ0 )
(X
√
σ/ n
H0
∼
N(0, 1).
This statistic is less variable than the t-statistic.
With the Z -statistic, our decision rule for testing H0 : µ ≤ µ0
vs Ha : µ > µ0 would be to reject H0 if
Z ≥ z1−α .
Summary of the Z -test
Applicable when the population is normally distributed and the
standard deviation is known OR n is large.
Hypotheses:
Case 1. H0 : µ ≤ µ0
Case 2. H0 : µ ≥ µ0
Case 3. H0 : µ = µ0
T.S.: Z =
R.R.:
vs
vs
vs
Ha : µ > µ0
Ha : µ < µ0
Ha : µ =
6 µ0
¯ −µ0
X
√
σ/ n
Case 1. Reject H0 if Z ≥ z1−α .
Case 2. Reject H0 if z ≤ z1−α .
Case 3. Reject H0 if |z| ≥ z1−α/2 .
p-values
Case 1
H0 : µ ≤ µ0
Ha : µ > µ 0
P(Z ≥ computed z)
Case 2
H0 : µ ≥ µ0
Ha : µ < µ 0
P(Z ≤ computed z)
case 3
H0 : µ = µ0
Ha : µ 6= µ0
2P(Z ≥ |computed z|)
Standard Deviation and Normality Known?
Z -test will have advantage over t-test when the standard
deviation is known.
Previous census or large scale studies may be used to
determine σ.
For large n, the assumption of normality and known variance
are less important.
Use t-test when n is small and normality assumption is
tenable.
For smaller sample size, if the normality appear violated (from
Q-Q plot, Box plot, Histogram and Empirical Rule),
Non-parametric methods should be used.
Example: Congestive Heart Failure
A study was conducted of 25 adult male patients following a new
treatment for congestive heart failure. One of the variables
measured on the patients was increase in exercise capacity (in
minutes) over a 4-week treatment period. The previous treatment
regime had produced an average increase of 2 minutes. The
researchers wanted to evaluate whether the new treatment has on
average increased the exercise capacity. The data yielded x¯ = 2.17.
From a previous large study it is known that σ = 1.05. Using
α = 0.05, what conclusion can we reach about the research
hypothesis?
Power Analysis
Recall
β = P(Accept H0 |H0 is false)
That is, β is the probability of falsely accepting H0 .
The power of a test is defined by
PWR = 1 − β
which is the probability of rejecting H0 when it is false.
Tests of hypothesis are evaluated by calculating the power.
More importantly, calculation of power is used to plan a study,
usually before any data have been obtained, except possibly
from a pilot study.
Power Analysis: Notations
Notations
Let µ0 denote the null value of the population mean.
Let µa denote the actual value of the mean in Ha .
Let β(µa ) denote the power when the actual value of the
mean is µa .
Let PWR(µa ) denote the power when the actual value of the
mean is µa .
Under the Null
Under the Alternative
0.00
0.05
0.10
0.15
Distribution of X
µ0
µ0 + Z1−ασ
n
µa
X
Power Analysis: Upper-Sided Alternative
For a one-tailed (upper-sided) test
PWR(µa ) = P(Z ≤ −z1−α +
µ a − µ0
√ )
σ/ n
and
β(µa ) = 1 − PWR(µa ).
In R this area can be computed by using the pnorm command.
Example: For the congestive-heart-failure example, compute
the β(µa ) and plot the power PWR(µa ) when
µa = 1.6, 1.8, 2, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 3
Example: Plotting the Power Upper Sided Alternative
0.0
0.2
0.4
Power
0.6
0.8
1.0
The Power Curve
1.6
1.8
2.0
2.2
2.4
2.6
µa
2.8
3.0
The Power Analysis: Lower-Sided and Two-Sided
Alternatives
For one-tailed (lower-sided) test
PWR(µa ) = P(Z ≤ −z1−α +
µ 0 − µa
√ )
σ/ n
and
β(µa )Z = 1 − PWR(µa ).
For a two-tailed test
µ0 − µa
√ )
σ/ n
µa − µ0
√ )
+ P(Z ≤ −z1−α/2 +
σ/ n
β(µa ) = 1 − PWR(µa )
PWR(µa ) = P(Z ≤ −z1−α/2 +
and
The Power Analysis: Examples
Example: Using 5% level of significance and sample of size
10, compute the power of the test for the
cholesterol-aggregation data, with an alternative mean of 190
mg/dL, and a standard deviation (σ) of 50 mg/dL.
Example: A new drug in the class of calcium-channel blockers
is to be tested for the treatment of patients with unstable
angina, a severe form of angina. The effect this drug will have
on heart rate is unknown. Suppose 20 patients are to be
studied and the change in heart rate after 48 hours is known
to have a standard deviation of 10 beats per minute. What
power would such a study have of detecting a significant
difference in heart rate over 48 hours if it is hypothesized that
the true mean change in the heart rate from base line to 48
hours could be either a mean increase or decrease of 5 beats
per minute?
Factors Affecting the Power
The actual value of the mean in the alternative. For fixed α
and n, the power increases as the µa moves away from the
values in the null hypothesis.
Sample size. For a fixed α and µa , the power increases as
sample size increases.
Standard Deviation. For fixed n, α and µa , power is higher
when standard deviation is low.
Probability of type I error (α). For fixed n and µa , the power
decreases as α increases.
Sample Size Determination
Suppose we are interested in H0 : µ ≤ µ0 vs Ha : µ > µ0 .
Specify the value of α.
Then the sample size needed to guarantee a probability of
type II error at most β when the actual mean differs from µ0
by at least ∆ is
(z1−α + z1−β )2
.
∆2
Effect Size = ∆/σ is may be obtained by
n = σ2
1
2
3
using µa and σ from previous studies or
assessing the smallest clinically important difference expressed
in standard deviation units or
conducting a pilot study.
Sample size obtained in this way is only a ballpark because of
the inaccuracy in estimating µa and σ. They are useful,
though, in checking a proposed sample size.
Sample Size Determination Cont’d...
The same formula also holds to determine the sample size
needed to guarantee a probability of type II error at most β
when the actual mean differs from µ0 by at least ∆ in the
case H0 : µ ≥ µ0 vs Ha : µ < µ0 .
For the two sided hypothesis, the approximate sample size
needed to guarantee a probability of type II error at most β
when the actual mean differs from µ0 by at least ∆ is:
n=σ
+ z1−β )2
.
∆2
2 (z1−α/2
Example
Consider a study of the effect of a calcium-channel-blocking agent
on heart rate for patients with unstable angina. Suppose we want
at least 80% power for detecting a significant difference if the
effect of the drug is to change mean heart rate by 5 beats per
minute over 48 hours in either direction and σ = 10 beats per
minute. How many patients should be enrolled in such a study?
What if the direction of the effect of the drug on heart rate was
well known?
Relationship b/n Hypothesis Testing and CIs
Suppose we are testing H0 : µ = µ0 versus Ha : µ 6= µ0 .
H0 is rejected with a two-sided level α test if and only if the
two-sided 100(1 − α)% confidence interval does not contain
µ0 .
H0 is accepted with a two-sided level α test if and only if the
two-sided 100(1 − α)% confidence interval does contain µ0 .
To see these, rejecting H0 happens if and only if t < −t1−α/2
or t > t1−α/2 . This happens if and only if µ0 lies outside
s
x¯ ± t1−α/2 √ .
n
This relationship is the rationale for using CIs to decide on the
reasonableness of specific values for the parameter µ.
Similar relationship exists between a one-sided test and
one-sided confidence interval.
p-value versus Confidence Interval
p-value tells us precisely how statistically significant the
results are.
However, often results that are statistically significant
(because of a large sample) are not clinically important.
A confidence interval would give additional information
because it would tell you the range of values within which µ is
likely to fall.
On the other hand, confidence interval does not precisely tell
us how significant the results are.
Hence, it is good practice to compute both a p-value and
confidence interval.
Example: For the Cholesterol of Asian-Immigrants, 95% CI for
µ is (173.58, 189.46) and p-value is 0.037. These two types of
information are complementary.
Sample-Size Estimation Based on CI Width
In some situations, it is known that the treatment has a
significant effect.
Interest focuses instead on estimating the effect with a given
degree of precision.
Suppose it is well known that propranolol lowers heart rate
over 48 hours when given to patients with angina at standard
dosage level. A new study is proposed using a higher dose of
propranolol than the standard one. Investigators are interested
in estimating the drop in heart rate with high precision. How
can this be done?
The precision of a confidence interval is judged by its width.
Sample-Size Estimation Based on CI Width Cont’d...
The length of a 100(1 − α)% two-sided confidence interval is
√
L = 2tn−1,1−α/2 s/ n.
2
This implies, n = 4tn−1,1−α/2
s 2 /L2 .
We usually approximate tn−1,1−α/2 by z1−α/2 and obtain
n = 4z1−α/2 s 2 /L2 .
s 2 can be obtained from a previous study or a pilot study.
Example: For the propranolol example, find the minimum
sample size needed to estimate the mean change in heart rate
(µ) if we require that the two-sided 95% CI for µ be no wider
than 5 beat per minute and the sample standard deviation for
the change in the heart rate equals 10 beats per minute.
Test for Proportion
Suppose p denotes the prevalence of a certain disease in a
population.
We are interested in testing
H0 : p = p 0
vs
Ha : p 6= p0
or a one-sided alternative.
The number of people who have the disease X in a random
sample of n individuals has a binomial distribution.
Example: We are interested in the effect of having a family history
of breast cancer on the incidence of breast cancer. Suppose that
400 of the 10,000 women of ages 50-54 samples whose mothers had
breast cancer had breast cancer themselves at some time in their
lives. It is known that prevalence rate of breast cancer for US
women in this age group is 2%. Does this result indicate a link
between family history and incidence of breast cancer?
Approximate Test: Two Sided
Let pˆ = X /n be the proportion of individuals in the sample
who have the disease.
If np0 (1 − p0 ) > 5 then, using the normal approximation to
·
the binomial, pˆ ∼ N(p0 , p0 (1 − p0 )/n) when H0 is true.
Define the test-statistic Z as follows:
pˆ − p0
Z=p
p0 (1 − p0 )/n
·
∼
N(0, 1)
if H0 is true.
We reject H0 if z < −z1−α/2 or z > z1−α/2 .
p-value= 2 × P(Z > |computed z|)
Approximate Test: One-Sided Alternative
Let pˆ = X /n be the proportion of individuals in the sample
who have the disease.
If np0 (1 − p0 ) > 5 then pˆ ∼ N(p0 , p0 (1 − p0 )/n).
Define the test-statistic Z as follows:
pˆ − p0
Z=p
p0 (1 − p0 )/n
·
∼
N(0, 1)
if H0 is true.
We reject H0 if z < −z1−α for lower-sided alternative and
z > z1−α for upper-sided alternative.
p-value= P(Z < computed z) for lower-sided and
p-value= P(Z > computed z) for upper-sided alternative.
Assess the statistical significance of the breast cancer data.
Exact Test
Lower-sided alternative
p-value = P(≤ x successes in n trials |H0 is true).
Upper-sided alternative
p-value = P(≥ x successes in n trials |H0 is true).
Two-sided alternative
(
2 × P(≤ x successes in n trials |H0 is true)
p-value =
2 × P(≥ x successes in n trials |H0 is true)
if pˆ ≤ p0
.
if pˆ > p0
p − value = 1 if the above formula gives a number greater
than 1.
The exact p-value can be computed in R using the binom.test
command.
Example: Occupational Health
The safety of people who work at or live close to nuclear power plants
has been the subject of widely publicized debate in recent years. One
possible health hazard from radiation exposure is an excess of cancer
deaths among those exposed. One problem with studying this question is
that because the number of deaths attributed to either cancer in general
or specific type of cancer is small, reaching statistically significant
conclusion is difficult except after long periods of follow-up. An
alternative approach is to perform a proportional mortality study,
whereby the proportion of deaths attributed to a specific cause in an
exposed group is compared with the corresponding proportion in a large
population. Suppose, for example, 13 deaths have occurred among 55- to
60-year-old male workers in a nuclear-power plant and that in 5 of them
the cause of death was cancer. Assume, based on vital-statistics reports,
that approximately 20% of all deaths can be attributed to some form of
cancer. Is this result significant?
Power Analysis
For two-tailed alternative, the power of the one-sample
binomial test the power is given by
s
√ !
p0 (1 − p0 )
|p0 − pa | n
PWR(pa ) = P Z ≤
(−z1−α/2 + p
)
pa (1 − pa )
p0 (1 − p0 )
provided np0 (1 − p0 ) > 5 and npa (1 − pa ) > 5.
For one-tailed alternative we replace α/2 by α.
Example: Suppose we wish to test the hypothesis that women with
a sister history of breast cancer are at higher risk of developing
breast cancer themselves. Suppose the prevalence of breast cancer
among the general population of 50- to 54- US women is 2%. We
propose to interview 500 women 50 to 54 years of age with sister
history of the disease. What is the power of such a study if the
prevalence of breast cancer among women 50- to 54- years of age
with sister history of breast cancer was 5%?
Sample-Size Estimation
The sample size needed to conduct a two-sided test with
significance level α and power 1 − β versus the specified
alternative hypothesis of p = pa is
p0 (1 − p0 ) z1−α/2 + z1−β
n=
q
pa (1−pa )
p0 (1−p0 )
(pa − p0 )2
2
.
For one-tailed test, we replace α/2 with α.
Example: In sister’s-history breast cancer example, how many
women should we interview to achieve 90% power with
α = 0.05.

Stat 491: Biostatistics Chapter 7: Testing Hypothesis–One Sample Inference Solomon W. Harrar

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