Hypothesis Testing
(Inference)
By
Abebe Megerso
(BSc in PH., MPH in Epid., Asst. Prof.)
1
Session Objectives
• Define hypothesis & describe types of
hypothesis,
• Describe steps in hypothesis testing,
• Discus rules for stating statistical hypotheses,
• Explain hypothesis testing process,
• Describe types of errors in hypothesis tests,
• Test hypothesis on single & double population,
2
Definition of Hypothesis
• Is a claim (assumption) about a population
parameter.
• Is a statement about one or more population.
• Is frequently concerned with the parameters of
the population about which the statement is
made.
3
Examples of Hypotheses:
Population Mean:
• The average length of stay of patients admitted
to the hospital is five days.
• The mean birth weight of babies delivered by
mothers with low socioeconomic status (SES) is
lower than those from higher SES.
• Etc.
4
Examples …
Population Proportion:
• The proportion of adult smokers in Adama
is 40% (p = 0.40).
• The prevalence of HIV among non-married
adults is higher than that in married adults.
• Etc.
5
Types of Hypothesis
1. The Null Hypothesis, HO:
 Is a statement claiming that there is no
difference between the hypothesized value &
the population value.
(The effect of interest is zero  no difference)
 States the assumption (hypothesis) to be
tested.
6
Types of Hypothesis…
 H0 is a statement of agreement (or no difference),
 H0 is always about a population parameter, not
about a sample statistic,
 Always contains “=” , “ ≤” or “≥ ” sign.
7
Types of Hypothesis…
2. The Alternative Hypothesis, HA
• Is a statement of what we will believe is true if
our sample data causes us to reject Ho.
• Is generally the hypothesis that is believed (or
that needs to be supported) by the researcher.
8
Types of Hypothesis…
• Is a statement that disagrees (opposes) with
H o.
(The effect of interest is not zero),
• Never contains “=” , “ ≤” or “≥ ” sign.
• May or may not be accepted.
9
Hypothesis Testing
• The majority of statistical analyses involve
comparison, (e.g. between treatments or
procedures or between groups of subjects).
• Hypotheses are formulated, experiments are
performed, & results are evaluated for their
consistency or non-consistency with a
hypothesis.
10
Hypothesis Testing…
• Hypothesis Testing (HT) provides an objective
framework for making decisions using
probabilistic methods.
• The purpose of HT is to aid the clinicians,
researchers or administrators in reaching a
decision concerning a population by examining
a sample from that population.
11
Hypothesis Testing…
• Begin with the assumption that the Ho is
true:
– Similar to the notion of innocent until
proven guilty.
• Ho may or may not be rejected.
12
Steps in Hypothesis Testing
1. Formulate the appropriate statistical hypotheses
clearly.
•
Specify HO & HA
H0:  = 0
H0:  ≤ 0
H1:   0
H1:  > 0
two-tailed
one-tailed
H0:  ≥ 0
H1:  < 0
one - tailed
2. State the assumptions necessary for computing
probabilities.
•
•
A distribution is approximately normal.
Variance is known or unknown.
13
Steps …
3. Select a sample & collect data.
• Categorical, continuous;
4. Decide on the appropriate test statistic for
the hypothesis.
E.g., One population:
Or
14
Steps …
5. Specify the desired level of significance for
the statistical test (=0.05, 0.01, etc.).
6. Determine the critical value.
–
-1.96
A value the test statistic must attain to be
declared significant (=0.05).
1.96
1.645
-1.645
15
Steps …
7. Obtain sample evidence & compute the test
statistic.
8. Reach a decision & draw the conclusion.
•
If Ho is rejected, we conclude that HA is true (or
accepted).
•
If Ho is not rejected, we conclude that Ho may be
true.
16
Rules for Stating Statistical
Hypotheses
1. One population:
• Indication of equality (either =, ≤ or ≥) must appear in
Ho.
Ho: μ = μo,
HA: μ ≠ μo
• Can we conclude that a certain population mean is
– not 30?
Ho: μ = 30 & HA: μ ≠ 30
OR
– greater than 50?
Ho: μ ≤ 50
&
HA: μ > 50
17
Rules …
Population Proportions:
Ho: P = Po, HA: P ≠ Po
E.g. Can we conclude that the proportion of
patients with leukemia who survive more than
six years is not 60%?
Ho: P = 0.6 & HA: P ≠ 0.6
18
Rules …
2. Two populations:
Mean Difference:
 Ho: μ1 = μ2
&
HA: μ1 ≠ μ2
Proportion Difference:
 Ho: P1 = P2
&
HA: P1 ≠ P2
19
In summary:
1. What you hope to conclude as a researcher
should be placed in the HA.
2. The Ho should have a statement of equality,
either =, ≤ or ≥.
3. The Ho is the hypothesis that is tested.
4. The Ho & HA are complementary.
20
Hypothesis Testing Process
• Now think about how the hypothesis test should
be carried out.
• We draw a random sample of size n from the
underlying population & calculate its sample
mean.
• We compare the sample mean to the postulated
mean μ0.
• Is the difference between sample mean & μ0 too
large to be attributed to chance alone?
21
Process …
22
Decision Rule:
• Results used for decision are computed from
the data of the sample.
• The decision to reject or not to reject the Ho is
based on the magnitude of the test statistic.
23
Decision Rule …
• An example of a test statistic is the
quantity obtained from:
• When the variance of the population is
unknown & sample is small, we use.
24
Rejection & Non-Rejection Regions
• The values of the test statistic assume the points on
the horizontal axis of the normal distribution & are
divided into two groups:
 Rejection region, &
Non-rejection region.
• The values of the test statistic forming the rejection region are
less likely to occur if the Ho is true.
• The values making the acceptance (non-rejection) region are
more likely to occur if the Ho is true.
25
Example: Two-sided test at α 5%
= 0.025
-1.96
Rejection region
= 0.025
0.95
1.96
Non-rejection region
Rejection region
26
Statistical Decision
• Reject Ho if the value of the test statistic that
we compute from our sample is one of the
values in the rejection region.
• Don’t reject Ho if the computed value of the
test statistic is one of the values in the nonrejection region.
27
Level of Significance, α
• Is the probability of rejecting a true Ho.
• Defines unlikely values of sample statistic if Ho is true.
– Defines rejection region of the sampling distribution.
• The decision is made on the basis of the level of
significance, designated by α.
• More frequently used values of α are 0.01, 0.05 &
0.10.
• α is selected by the researcher at the beginning.
28
One tail & two tail tests
• In a one tail test, the rejection region is at
one end of the distribution or the other.
• In a two tail test, the rejection region is split
between the two tails.
• Which one to use depends on the way the Ho
is written.
29
Level of Significance
& the Rejection Region
Example:
• The average survival year after cancer
diagnosis is less than 3 years.
• See the next slide:
30
Level of Significance …
31
Another way to state conclusion
• Reject Ho if P-value < α
• Accept Ho if P-value ≥ α (fail to reject)
P-value is the probability of obtaining a test statistic
as extreme as or more extreme than the actual test
statistic obtained if the Ho is true.
The larger the test statistic, the smaller is the P-value.
OR, the smaller the P-value the stronger the evidence
against the Ho.
32
Types of Errors in Hypothesis
Tests
•
Whenever we reject or fail to reject the Ho,
we commit errors.
•
Two types of errors are committed:
 Type I Error,
 Type II Error,
33
Type I Error
• The error committed when a true Ho is rejected.
• Considered as a serious type of error.
• The probability of a type I error is the
probability of rejecting the Ho when it is true.
• The probability of type I error is α.
• Called level of significance of the test.
• Set by researcher in advance.
34
Type II Error
• The error committed when a false Ho is
not rejected (fail to reject false Ho).
• The probability of type II Error is .
• Usually unknown but larger than α.
35
Power
• The probability of rejecting the Ho when it is
false.
• Power = 1 – β = 1- probability of type II error
• We would like to maintain low probability of a
type I error (α) & low probability of a type II
error (β)  [high power = 1 - β].
36
Action
(Conclusio
n)
Reality
Ho True
Ho False
Do not
reject Ho
Correct action
Type II error (β)
(Prob.= 1-α)
(Prob. = β= 1-Power)
Reject Ho
Type I error (α)
Correct action
(Prob. = α = Sign. level)
(Prob. = Power = 1-β)
37
Type I & II Error Relationship
38
Factors Affecting Type II Error
39
Factors affecting the Power of the
Test
The power depends on the following:
1. As n↑, power ↑
2. As |µ1-µo|↑, power ↑
3. As ↑, power ↓
4. As α↓, power ↓
40
Hypothesis Testing approach
Hypothesis Test for One Samples
• Test for single mean,
• Test for single proportion,
Hypothesis Test for Two Samples
• Test for the difference between two population
means,
• Test for the difference between two population
proportions,
41
1.
Hypothesis Testing of a Single Mean
(Normally Distributed)
42
1.1 Known Variance
43
Example: Two-Tailed Test
1. A simple random sample of 10 people from a certain
population has a mean age of 27. Can we conclude that
the mean age of the population is not 30? The variance
is known to be 20; & let level of significance be = .05.
A.
Hypothesis
 Ho: µ = 30
 HA: µ ≠ 30
B.
Assumptions
• Simple random sample,
• Normally distributed population,
44
Example …
C. Data:
n = 10, sample mean = 27, 2 = 20, α = 0.05
D. Test statistic:
 As the population variance is known, we use Z as
the test statistic.
45
Example …
E. Decision Rule:
•
Reject Ho if the Z value falls in the rejection
region.
•
Don’t reject Ho if the Z value falls in the nonrejection region.
•
Because of the structure of Ho it is a two tail test.
•
Therefore, reject Ho if Z ≤ -1.96 or Z ≥ 1.96.
46
Example …
F. Calculation of test statistic:
G. Statistical decision:
 We reject the Ho because Z = -2.12 is in the rejection
region; the value is significant at 5% = α.
H. Conclusion:
 We conclude that µ is not 30; P-value = 0.0340 < α.
 A Z value of -2.12 corresponds to an area of 0.0170.
 Since there are two parts to the rejection region in a
two tail test, the P-value is twice this which is .0340.
47
Hypothesis test using confidence interval
• A problem like the above example can also be solved
using a confidence interval.
• A confidence interval will show that the calculated
value of Z does not fall within the boundaries of the
interval; however, it will not give a probability.
• Confidence interval:
48
Example: One -Tailed Test
• A simple random sample of 10 people from a certain
population has a mean age of 27.
• Can we conclude that the mean age of the population is
less than 30? The variance is known to be 20; let α =
0.05.
• Data:
n = 10, sample mean = 27, 2 = 20, α = 0.05
• Hypotheses:
Ho: µ ≥ 30, HA: µ < 30
49
Example …
• Test statistic:
• Rejection Region:
Lower tail test
• With α = 0.05 & the inequality, we have the entire rejection
region at the left.
• The critical value is be Z = -1.645; we reject Ho if Z < 1.645.
50
Example …
• Statistical Decision:
 We reject the Ho because -2.12 & < -1.645.
• Conclusion:
 We conclude that µ < 30.
 p = .0170; this time because it is only a
one tail test & not a two tail test.
51
Example …
• Suppose that the Ho & Ha take the form
Ho: µ = µo, Ha: µ > µo
• In this case, Ho would be rejected for large
values of test statistic (critical values >0).
• The P-value would correspond to the area in the
upper tail of the SND, to the right of the value of
the test statistic.
Upper tail test
52
1.2 Unknown Variance
• In most practical applications the standard
deviation of the underlying population is not
known.
• In this case,  can be estimated by the sample
standard deviation s.
• If the underlying population is normally
distributed, then the test statistic is:
53
Example: Two-Tailed Test
• A simple random sample of 14 people from a certain
population gives a sample mean body mass index (BMI)
of 30.5 & s of 10.64.
• Can we conclude that the BMI is not 35 at α = 5%?
• Ho: µ = 35, Ha: µ ≠35
• Test statistic
• If the assumptions are correct & Ho is true, the test
statistic follows student's t distribution with 13 degrees
of freedom.
54
• Decision rule:
Example …
 We have a two tailed test; with α = 0.05 & it means that
each tail is 0.025.
 The critical t values with 13df are -2.1604 & 2.1604.
 We reject Ho if the t ≤ -2.1604 or t ≥ 2.1604.
• Do not reject Ho because -1.58 is not in the rejection
region.
• Based on the data of the sample, it is possible that µ =
35. P-value = 0.1375
55
2. Hypothesis Testing about the Difference
Between Two Population Means
• When studying one-sample tests for a
continuous random variable, the unknown mean
μ of a single population was compared to some
known value μo.
• We are usually interested in comparing the
means of two different populations when the
values of both means are unknown.
56
Two Population Means, Independent
Samples
57
Two Sample Means,
Independent Samples
Two Population Means …
58
2.1 Known Variances
(Independent Samples)
• When two independent samples are drawn from
a normally distributed population with known
variance, the test statistic for testing the Ho of
equal population means is:
59
Example:
• Researchers wish to know difference in mean serum uric
acid (SUA) levels between normal individuals & those
with Down’s syndrome.
• The means SUA levels on 12 individuals with Down’s
syndrome & 15 normal individuals are 4.5 & 3.4 mg/100
ml, respectively, with variances (2=1, 2=1.5,
respectively).
• Is there a difference between the means of both groups at
α = 5%?
• Hypotheses:
 Ho: µ1- µ2 = 0 or Ho: µ1 = µ2
 HA: µ1 - µ2 ≠ 0 or HA: µ1 ≠ µ2
60
Example …
• With α = 0.05, the critical values of Z are -1.96 &
+1.96. We reject Ho if Z < -1.96 or Z > +1.96.
• Reject Ho because 2.57 > 1.96.
• From these data, it can be concluded that the
population means are not equal.
• A 95% CI would give the same conclusion; & P-value
= 0.01.
61
2.2 Unknown Variances
i. Equal variances (Independent samples)
• With equal population variances, we can obtain a
pooled value from the sample variances.
• The test statistic for µ1 - µ2 is:
• Where tα/2
has (n1 + n2 – 2) df., &
62
Example:
• We wish to know if we may conclude, at the 95%
confidence level that smokers, in general, have
greater damaged lung cells than do non-smokers.
• Calculation of Pooled Variance:
63
Example …
• Hypotheses:
Ho: µ1 ≤ µ2 = 0, HA: µ1 > µ2
• With α = 0.05 & df = 23, the critical value of t is
1.7139; we reject Ho if t > 1.7139.
• Test statistic:
• Reject Ho because 2.6563 > 1.7139; & on the basis of
the data, we conclude that µ1 > µ2.
64
ii. Unequal variances (Independent
samples)
• We are still interested in testing
H0 : μ1 = μ2 vs HA: μ1 ≠ μ2
• The test statistic used is:
• To compute a test statistic, we simply
substitute s 2 for  2 & s 2 for  2.
1
1
2
2
65
Unequal variances …
• Where the degree of freedom (d’) is given by:
66
Unequal variances …
• If t > td’’,1-α/2 or t < -td’’,1-α/2 then
reject Ho.
• If -td’’,1-α/2 ≤ t ≤ td’’,1-α/2, then accept
Ho
67
Example:
• Suppose we want to compare the characteristics of
tuberculosis meningitis for patients infected with HIV
& those not infected with HIV.
• In particular, we are interested in comparing age at
diagnosis.
• A random sample of n1 = 37 HIV infected patients has
mean age at diagnosis x1 = 27.9 years & s1 = 5.6 years.
• A sample of n2 = 19 uninfected patients has mean age
at diagnosis x2 = 38.8 years & s2 = 21.7 years.
68
Example …
• The test statistic is:
69
Example …
• Note that:
• And
70
Example …
• For a t distribution with 19 df, the area to the left
of −2.15 is between 0.01 & 0.025.
• Therefore, 0.02 < p < 0.05
• For a test conducted at α= 0.05, H0 is rejected.
• We conclude that among patients diagnosed with
tuberculosis meningitis, those who are infected
with HIV tend to be younger than those who are
not.
71
Hypothesis Testing for
Paired Samples
• Two samples are paired when each data point of
the first sample is matched & is related to a
unique data point of the second sample.
• Tests means of two related populations:
 Paired or matched samples,
 Repeated measures (before/after),
• Longitudinal or follow-up study,
72
Paired Samples …
• Assumptions:
– Both populations are normally distributed,
– Or, if not normal, use large samples,
73
The Paired t Test
n = number of pairs in the paired sample
Sd = Sample standard deviation
74
Paired t Test …
75
Paired t Test …
76
Example:
• The following data show the SBP levels (mm Hg) in 10 women
while not using (baseline) & while using (follow-up) oral
contraceptives (OC).
• Can we conclude that there is a difference between mean
baseline & follow-up SBP at α 5%? di = baseline – follow-up,
i
1
2
3
4
5
6
7
8
9
10
SBP (baseline)
115
112
107
119
115
138
126
105
104
115
SBP (follow-up)
128
115
106
128
122
145
132
109
102
117
di
13
3
-1
9
7
7
6
4
-2
2
77
Example:
= (13 + 3 + …. + 2)/10 = 4.80
S2d = [(13-4.8)2 + … + (2-4.8)2]/9 = 20.844
Sd = √20.844 = 4.566
t = 4.80/(4.566/√10) = 4.80/1.44 = 3.32
• From the Table, t9,α/2 = 2.262
• Since t = 3.32 > t9,α/2 (2.262) Ho is rejected
• P-value is between 0.001 & 0.01
• Since 3.32 falls in the rejection region, there is a
significant difference between the population means of
SBP while not using & using OC.
78
Hypothesis Tests for Proportions
• Involves categorical values,
• Two possible outcomes:
– “Success” (possesses a certain characteristic)
– “Failure” (does not possesses that
characteristic)
• Fraction or proportion of population in the
“success” category is denoted by p.
79
Proportions …
80
3. Hypothesis Testing about a Single
Population Proportion
(Normal Approximation to Binomial Distribution)
81
Single Population Proportion…
82
Example
• We are interested in the probability of developing
asthma over a given one-year period for children 0 to 4
years of age whose mothers smoke in the home.
• In the general population of 0 to 4-year-olds, the annual
incidence of asthma is 1.4%.
• If 10 cases of asthma are observed over a single year in
a sample of 500 children whose mothers smoke, can we
conclude that this is different from the underlying
probability of p0 = 0.014? α = 5%
H0 : p = 0.014
HA: p ≠ 0.014
83
Example …
• The test statistic is given by:
84
Example …
• The critical value of Zα/2 at α=5% is ±1.96.
• Don’t reject Ho since Z =1.14 is in the nonrejection region between ±1.96.
• P-value = 0.2548
• We do not have sufficient evidence to conclude
that the probability of developing asthma for
children whose mothers smoke in the home is
different from the probability in the general
population.
85
4. Hypothesis Tests about the Difference
Between Two Population Proportions
86
Two Population Proportions…
Where X1 = the observed number of events in the first sample
& X2 = the observed number of events in the second sample
87
Two Population Proportions…
88
Example
• A study was conducted to investigate the possible cause
of gastroenteritis outbreak following a lunch served in a
high school cafeteria.
• Among the 225 students who ate the sandwiches, 109
became ill; while, among the 38 students who did not
eat the sandwiches, 4 became ill.
• Is there a significant difference between the two groups
at α =5%.
• We wish to test:
 Ho: p1 = p2 against the alternative
 HA: p1 ≠ p2
89
Example …
90
Example …
• Assume that the sample sizes are large enough,
& the normal approximation to the binomial
distribution is valid.
• If the Ho is true, then p1 = p2 = p
91
The area under the standard normal curve to the right of
4.36 is less than 0.0001; &, p < 0.0002.
We reject H0 at the 0.05 level.
We conclude that the proportion of students who
became ill differs in the two groups; those who ate
the prepared sandwiches were more likely to
develop gastroenteritis.
92
Thank You !
93