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Pharmaceutical Statistics
Lecture 12
Hypothesis Testing: Introduction
Hypothesis testing
• A hypothesis may be defined as a statement about one or more
population.
• The hypothesis is frequently concerned with the parameters of
populations about which the statement is made. (we use samples, but we
do care about the parent pop).
• Hypothesis testing: Determination whether or not such statements are
compatible with the available data.
• Hypothesis testing includes different sequential logical steps:
Null & Alternative
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Hypothesis
Null hypothesis or the hypothesis to be tested (H0)
a claim that there is NO difference between the population mean and the
hypothesized value.
The complement of the conclusion that the researcher is seeking to reach
becomes the statement of the null hypothesis.
In the testing process the null hypothesis either is rejected or is not rejected
(we do not say: accepted!!!)
Alternative hypothesis HA
a statement of what we will believe is true if our sample data cause us to
reject the null hypothesis.
Usually the alternative hypothesis is the research hypothesis (the
conclusion that the researcher is seeking to reach).
H0
HA
Gabapentin has no pharmacological effect
Gabapentin has a pharmacological effect
The mean for population A is 20 (H0: μ = 20)
The mean for population A is not 20 (HA: μ ≠ 20)
The mean for population A is less than or equal to 20 (H0:
μ ≤ 20)
The mean for population A is larger than 20 (HA: μ > 20)
The mean for population A is larger than or equal to 20
((H0: μ ≥ 20)
The mean for population A is less than 20 (HA: μ < 20)
Null & Alternative Hypothesis
•
•
•
An indication of equality (=, ≥, ≤) must appear in the null hypothesis.
Rules of thumb to conclude what statements goes in null hypothesis and what
statements goes in the alternative hypothesis:
– The null hypothesis should contain a statement of equality
– The null hypothesis is the hypothesis to be tested
– What you hope or expect to be able to conclude as a result of the test usually
should be placed in the alternative hypothesis.
– The null and alternative hypothesis are complementary. The two together
exhaust all possibilities regarding the value that the hypothesized parameter
can assume.
Accepting or rejecting a hypothesis is not a proof of the hypothesis !!!!!!! (Null
hypothesis can be true or false, we only can reject it or not to reject it!!)
Example: How do things flow?!!
Researchers postulate that drug A201bc may reduce blood glucose level.
• Null hypothesis (H0): Drug A201bc does not reduce blood glucose level
• Alternative hypothesis (HA): Drug A201bc does reduce blood glucose level (note
this is the researchers wishes)
We need two groups of patients (one receives drug A201bc ; the second receives placebo)
We measure blood glucose level for the two groups (treated & control)
If the blood glucose level data show significant difference
we will reject H0
Concept of Hypothesis Testing
What is the average weight of the second-yr pharmacy students in Jordan?
I think it is 60 kg…
So you hypothesize that the mean = 60, right?
Yes, and we can test this hypothesis…we have learned this in a previous statistic course
Really!! How do you test this hypothesis?
First, we need to gather student sample (n= 30). Then we need to measure the weight of each and
compute the mean of their weights and the standard error of themean
It sounds easy. let us go to UJ/Pharmacy school and dothis…
After measuring student sample weights
O.k. we got a mean of 58 and SE of 0.8 Kg
mmm..What we can conclude now??..58 is not 60, do we reject your hypothesis?
mmm, 58 kg is different than 60 kg..but the important point here is: How big or how significant this
difference is?… in other words: is the difference is due a real difference between my hypothesized
mean value and the true mean or it is due to sampling errorsonly?
How can we determine if this difference is big/significant?
Since 58 kg is the sample mean, we can use the standard normal distribution of the sample mean and
the z-score to find how far is 58 kg from the hypothesized mean value (60 kg)

x  0
58  60
z

 2.5
0.8
/ n
As you see below, the Z-score of the mean 58 kg is extremely to the left. (in the left gray area
below!!)…this means that the occurrence of mean=58 kg is rare and unlikely if the hypothesized
population mean is 60 kg!!!..which indicates that the population from which the sample was taken may
have different mean value of 60 kg!!! May be the true pop mean is 59, then the z-score of the sample
mean then will be(-1.25) a value that fall in the white area (likely to occur and not rare anymore)
This is intersting..but the gray area is a part of the sample mean distributionand values in this area are
possible, what about the sample we picked is just odd and the true is 60kg??
You are right…always there is a possibility of error here. The true mean can be 60 kg and our sample
just was extreme. This error is called α (rejecting a true null hypothesis). The good thing here that we
choose α to be small for more certainty (5%, 1%).
I start to understand now…you mean that we have no idea about the population mean, so we hypothesize
a value to be (60 kg) and then we draw a sample from the population. If this sample was withdrawn from a
population that has a mean of 60 (as we think), the mean of this sample should fall in the white area [58.461.6 kg] with a probability of (P=95%). Since our sample mean felt in the gray area, we conclude that the
population mean may not be the hypothesized 60 kg. The error in our conclusion is only 5%, right?
Exactly..the white area is the area where we do not reject our hypothesis..where the gray area where
we do. Remember it is possible always to reject a true null hypothesis (probability is α, type I error) and
to fail to reject false hypothesis (probability is β, type II error). It all about probability, certainty,
confidence, and significance..
That was a good brain exercise.. What else we can use hypothesis testing for? (topics for coming
lectures)
Well, we can use hypothesis testing to test hypothesized values for single population mean, the
difference between two population means, single pop proportion, and difference between two pop
proportions
Thanks for this info
You are welcome
Note: we used z-stat since we
assumed that the parent pop is
normal and sample size is 30 (large)
© Alkilany 2012
Test Statistic& Decision rules
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All possible values of the test statistic are points on the horizontal axis of the statistic
distribution graph. These points are divided into a rejection region and a nonrejection
region.
The values of the test statistic forming the rejection region are those values that are
less likely to occur if the null hypothesis is true, while the values forming the
nonrejection (acceptance) region are more likely to occur if the null hypothesis is
true.
The decision rule tells us to reject the null hypothesis if the value of the test statistic
that we compute from our sample is one of the values in the rejection region and to
not reject the null hypothesis if the computed value of the test statistic is one of the
values in the nonrejection region.
The decision of which values go into the rejection region and which ones go into the
Critical
nonrejection region is based on the desired level of significance α.
values
The level of significance α is the probability of rejecting a true null hypothesis.
Since rejection a true null hypothesis is an error, we should make the level of
significance small.
The most frequently encountered α values are 0.01, 0.05 and 0.1.
Note: using z- or t-statistic depends on
the assumptions (normality, known
variance, sample size, equality of
variance ..etc
Rare, unlikely
values
Example on the Concept of Hypothesis Testing
•
•
We postulate that average Lead concentration in human blood is 18.5 ppb. We want
to investigate our postulation!!! H0: μ[Pb in blood]=18.5ppb
We will take a random blood sample and measure the lead levels and then analyze
the collected data:
Hypothesized Mean (μH) 18.5
Sample Mean X
23.25
n
28
D.F
27
s
9.49
SE (standard error)
1.794
Confidence level (1-α)
95%
Significance (α)
5%
α/2=0.025
t=-2.052
X=14.82
±2.05
X 2
Lower Critical value of
14.82
t 2 3.2 5
t=+2.052
X=22.16
x  0 23.25 18.5


 2.65
s/ n
1.794
The sample mean (X =23.35, t=+2.65) is
less likely to occur (in the rejection
rejoin). Thus we reject the null
hypothesis and conclude that the pop
mean for Lead levels in human blood
may not equal 18.5 ppb
X
aslutaetosfinc
α/2=0.025

Critical t-value
N otLeo:wweerCuristiecdaltv-
H  0
e22w.1e6a
ssumed that
the parent pop is normal with unknown
variance +small sample size
Summary…..!!
1. State null and the alternative hypotheses
2. Choose a significance level  (usually .05)
3. Determine the critical region
4. Computed the test statistic (based on assumptions!!). This
is the calculated value that we compare with the critical
value
5. Reject the null hypothesiswhen the test statistic felt in the
rejection region (otherwise do not reject null hypo).
6. State the appropriate conclusions
x
t
s n
Types of Hypothesis Tests
Two-tailed test (not equal to)
T ≠ H
Left-tailed test (greater than)
Too small
H>T
T ≠ H
Right-tailed test (Less than)
Too large
H<T
P-Value
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It is a number that tells us how unusual our sample results are, given that the null
hypothesis is true.
It is the probability of obtaining a value of the test statistics as extreme and even
more extreme than the computed one.
If P-value≤α, we do reject the null hypothesis. If P-value>α we do not reject it.
Researcher prefer to report P-value as an indication of the significance of their test
statistics
P-value<0.05
In (Two-tailed test), P-value=
probability of observing a Z≥+ZT or probability of observing Z≤-ZT
probability of observing a Z≤-ZT
In (left-tailed test), P-value=
In (right-tailed test), P-value= probability of observing a Z≥+ZT
These probabilities are computed using software packages or from Tables
Types of errors (α&β)
Type I error (α): Rejecting a true null hypothesis (Ho). The true population is the
Ho, but the sampling error resulted in an odd statistic test value in the blue area
(rejection area for Ho)
Type II error (β): Failure to reject (accepting) a false null hypothesis. The true
population is the HA, but the sampling error resulted in an odd static test value in
the brown area (in the acceptance area for Ho)
Pharmaceutical Statistics
Lecture 13
Hypothesis Testing:
A Single Population Mean
Case I: A Single Normal Population Mean with
known variance
Researchers are interested in the mean age of a certain population. They are
wounding if the mean age is 30 yrs. Assuming that the population is normally
distributed with variance equal to 20. A simple random sample of 10 individuals
drawn from the population of interest. From this sample, a mean of 27 is
calculated. Construct the proper hypothesis, test your hypothesis, and then state
the proper conclusion?
• Hypotheses:
The null hypothesis is H0: μage = 30
The alternative hypothesis is HA: μage ≠ 30
To reduce the risk of rejecting a true null hypothesis we choose a small
value of α=0.05 (confidence level= 95%)
• Test statistic:
– Since the population is normally distributed and since the population
variance is known, we use the z-statistic.

x 0
z 
/ n
Case I: A Single Normal Population Mean with
known variance
• Distribution of test statistic:
– The test statistic that we have chosen (z-dist of the sample
mean) is normally distributed with a mean of 0 and a
variance of 1 (standard normal distribution).
• Decision rule:
– The decision rule tells us to reject H0 if the computed
values falls in the rejection region and to fail to reject H0 if
it falls in the nonrejection region.
– So either sufficiently small values or sufficiently large
values will cause rejection of null hypothesis (two-tailed
test). But how large or small??
Case I: A Single Normal Population Mean with known
variance
• Decision rule: (Contd)
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Since our rejection region consists of two regions, part of α will have to be associated with the large
values and part with the small valuest.
So we divide α equally and let α/2=0.025 associated with large values and α/2=0.025 associated with
small values.
The critical values of z becomes then the value to the right of which lies 0.025 of the area under the
SND and the one to the left of which lies 0.025 of the area under the SND.
Z=1.96 and -1.96 (we determined the critical values here)
Reject H0 if the computed value of the Z-score is either ≥1.96 or ≤-1.96

3
x 0 27  30

 2.12
z


• Calculation of the test statistic: 27
 /n
20 /10 1.4142
•
Statistical decision:
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We are able to reject the null hypothesis since -2.12 is in the rejection region.
We can say that the computed values is significant at the 0.05 level.
Case I: A Single Normal Population Mean with known
• Conclusion:
variance
•
– We can coclude that μ is not equal to 30 (confidence level 95%;
significance=5%)
p value:
– The p value indicates the probability of observing z ≥2.12 or z≤-2.12 (twotailed test, both areas in green).
– p value = P(z ≥2.12 ) + P(z≤-2.12)= 0.017+0.017=0.034 [AUC obtained from ztables]
– The p value of the hypothesis testing is the probability of obtaining, when H0 is
true, a value of the test statistic as extreme or more extreme (in the direction
of supporting HA) than the one actually computed.
– General rule: if the p value is less than or equal to α, we reject the null
hypothesis, if it is greater than α, we do not reject the null hypothesis
Note: Testing H0 by Means of A Confidence
Interval
• Treating the previous example using the confidence interval approach:
X  z(1 /2) *

n
27  1.96 20/10
27 1.96(1.4142)
27  2.7718
(24.2282_29.7718)
• Since the interval does not include 30, we can say that 30 is not a
candidate for the mean we are estimating and therefore μ is not equal to
30 and H0 is rejected.
Testing H0 by Means of A Confidence Interval
• When testing a null hypothesis by means of a two sided confidence
interval, we reject H0 at the α level of significance if the hypothesized
parameter is not contained within the 100(1-α) percent confidence
interval.
• If the hypothesized parameter is contained within the interval, H0 can not
be rejected at the α level of significance.
One-Sided Hypothesis Test
• In this case either significantly “small” values only or significantly “large”
values only will cause rejection of the null hypothesis.
• In the previous example, the researchers are asking: can we conclude that
μ30 (left-tailed test).
The null hypothesis is H0: μ ≥30
The alternative hypothesis is HA: μ  30
Left-tailed test (Greater than)
Too small
T < H

x 0
z 
/ n
One-Sided Hypothesis Test
• The critical z value is one to the left of which lies 0.05 of the area under
the SND. (-1.645)
• Our decision rule tells us to reject H0 if the computed value of the test
statistic is less than or equal to -1.645.

3
x  0 27  30

 2.12
Test statistic: z27   / n
20 /10 1.4142
•
• We reject the null hypothesis since -2.12-1.645
• We conclude that the population mean may be smaller than 30.
• p value = P(z≤-2.12)= 0.0170 (just one tail)
Case II: A Single Normal Population Mean with unknown
variance
The t Distribution (remember this slide from previouslect)
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•
•
In the previous example, the population standard deviation σ is known, we have
learned how to test the hypothesis on a population mean based on sample mean.
However, usually the population standard deviation is unknown as well as the
population mean (what do we need to do in this case?).
We may use the sample standard deviation to replace σ. Here the z-statistics will
convert to t-statistics :
Note:
-When we have small
samples, it becomes
necessary for us to use the tdistribution in hypothesis
testing.
- When the sample size is
large (>30), our faith in s as
an approximation of σ is
usually substantial, and we
may be justified in using
standard normal distribution
theory to test the
hypothesized value for the
population mean.
Standard Normal Distribution

x 
z 
 / n
z-distribution if σ is known
Student’s t distribution

t 
x 
s / n
t-distribution if σ is unknown
[pop variance is unknown+small n
Case II: A Single Normal Population Mean with unknown
variance
• The body mass index (BMI) of a group of 14 healthy adult
males has a mean of 30.5 and a standard deviation of 10.6392,
can we conclude that the mean BMI of the population is equal
to 35?
t-distribution since σ is unknown
[pop variance is unknown+small n]
The null hypothesis is H0: μ = 35
The alternative hypothesis is HA: μ ≠ 35

Test statistic
x 0
t 
s/ n
α=0.05
Case II: A Single Normal Population Mean with unknown variance
• Decision rule: this is a two-tailed test and so we put α/2 (0.025) in each
tail. The t values to the right and left of which 0.025 of the area are 2.1604
and -2.1604.

• Calculation of the test statistic: t  x 0 
s/ n
30.5  35
10.6392/ 14

 4.5
 1.58
2.8434
• We do not reject H0 since -1.58 falls in the nonrejection region.
• We can conclude that the mean of the population from which the sample
came may be 35.
• P not available directly
Case III: A Single Non-normal Population Mean with unknown
variance and large sample
We may, if our sample is large (greater than or equal to 30), take advantage of
the central limit theorem and use z-stat as the test statistic.
• The mean maximum oxygen uptake (VO2max) for a sample of 242 women
was 33.3 with a standard deviation of 12.14. we wish to know if, on the
basis of the data, Can we conclude that the mean score for a population of
such women is greater than 30?
The null hypothesis is H0: μ ≤ 30
The alternative hypothesis is HA: μ > 30
Test statistic
α=0.05

x  0
z 
/ n
We use sample’s S since we do not
know σ. This approximation is valid due
to the large sample size. Also, we have
learned that when n is large t-distribution
is close to z-distribution, and here we
can use the z-distribution as a valid
approximation
Case III: A Single Non-normal Population Mean with
unknown variance and large sample
•
We are unwilling to assume that scores are normally distributed in such population,
however, because of the central limit theorem, the test statistic is at worst
approximately normally distributed with μ=0.
•
By letting α=0.05, the critical value of the test statistic is 1.645. Reject H0 if
computed z≥1.645
33.3  30
3.3
z 

 4.23
12.14 / 242 0.7804
•
•
•
Reject H0 since 4.23>1.645
Conclude that the mean Vo2max score for the sampled population is greater than 30
P value for this test is 0.00005 since 4.23 is greater than 3.9
Conventions for interpreting P values
P Value
Interpretation
P > .05
Result is not significant; usually indicated by no asterisk
P < .05
Result is significant; usually indicated by *
P < .01
Result is highly significant; usually indicated by **
P < .001
Result is very highly significant; usually indicated by ***