1. Inferential Statistics
Session-II
2009
Dr. Arshad Sabir A.P

2. Issues in epidemiological Research
a. Studies undertaken to assess population characteristics like
age, vaccination status, prevalence of malnutrition, KAP of
Contraception etc (sample based, variations occur normally)
Issues: To what extent study findings are a true estimate of
reference population ?
b. Compare groups to study associations (Cases & Controls,
Exposed & Unexposed, Efficacy of a drug etc)
Issues: Are the differences observed hold true for the
differences in total population? ( differences observed may
be due sampling, variations occur normally) .
VRIATIONS:
Normal / Biological,………..Real………………Experimental
ISSUE: we want to be as much precise as possible.

3. Central limit theorem (CLT)
• Suppose, we want to know weight of adult
population of Rawalpindi city.
• Take multiple, Random, large ( >30) samples ( say
1000) are taken. Calculate mean wt. in each case.
• We will have 1000 mean wt.(X1-n)
• If all the sample means are presented by frequency
distribution curve. It will follow a normal distribution
pattern. Known as “Sampling distribution of means”.
• 68% samples means will fall X ±1SD , 95% means will
be X ±2SD. And 99.7% mean will be X ±3SD.
• Summary values of such a dist. i.e. mean, SD are very
close to population values.
• Its mean is almost equal to Pop. mean ( X = µ )
• Its SD is known as “Standard Error” (SE)

4. CLT
• Formula for SE is ;
SE = SD / √ n
• SE is a unit of measure of variability that can
happen due to sampling (sampling variation).
• SE error is based upon “Normal distribution” so
follows rules of “normal distribution curve”.
• In actuality we take only one sample and use its
SD as Standard Error.
• So, we can be 95% confident that pop. mean will
be within range of dist. mean ±2SE and its
chances of falling beyond this range are only ≤5%.
• SE is measure of “Chance variation” or normal
variation from sample to population or b/w two
samples or groups.

5. Confidence Limits and Confidence Interval
• When assessing Pop. mean on the basis of one sample. It
has its mean X and SD (SE). Its mean is not equal to µ.
• According to CLT, SE is a tool to measure variations that
can happen due to sampling.
• Sample mean (X) is not exactly equal to pop. mean but with
help of sample SD .i.e. SE we can construct a range of
values around sample mean within which pop. mean
would fall with certain degree of confidence.
• These limits worked out on both sides of sample mean on
the basis of CLT are called “confidence limits” {CLs}. And
the range between these limits is known as “confidence
interval” {CI}.
Formula for Pop. mean µci = X ± 2SE = X ± 2 ( SD)
(95% CI) √n

6. Estimation of population parameter from a sample
statistic
• As per CLT we are sure that 95% of sample
means will be within confidence limits of µ ±
2SE .
• 95% confidence interval means that there is
95% probability that Pop. mean (µ) lies 2SE
below or above the sample mean and 5%
probability that it lies outside this interval
(P = 0.05). We can say that we are 95%
confident in making this statement .
• CI is related to size of sample (n). Larger the
sample, smaller the CI for a given level of
significance.

7. Estimation of pop. Parameters (say mean) form sample
statistics
Apparently if there is large SE will have wide range of estimate (CI )
or vice versa. We desire a precise estimate.
SE basically depends (depends upon 02 factors)
Variability: How is dispersion of attribute in the actual Pop.
( reflected by σ ). If SD is large ,
estimate will far away or wide ( it inherent property , can
not be changed) and if it is small, estimate will be close to
true value.
Sample size: A small sample (n) with no or small variability
is good to estimate µ but larger samples are needed to
accommodate higher variability in data.
This relationship of SD to Sample size (n) is expressed as
Standard Error ; SE = SD
√n

8. Exercise
• 16Kg is mean Ht. of 3y old children obtained from
a sample of 11 from a village.( SD=2kg)
• How is this estimate? (sampling variation)
• To what extent this mean is representative of
actual pop. mean ?
• SE = 2/ √ 11 = 0.6 Kg
• 95% CI = 16 ± 2 x 0.6
14.8kg---------17.2kg
Role of sample size:
If n= 20, SE= 2/ √ 20 = 0.45kg
95%CI = 16 ± 2 x 0.45 = 15.1-----------16.9kg

9. Standard error of proportion (SEP)
• Similarly , Normal distribution of samples
proportions around the proportions of pop. may
be expressed arithmetically in term of SE of
proportion with confidence limits. [ Central limit
theorem ]
• SEP is also measure of variation due to sampling
• 95% of sample proportions will lie within limits of
population proportion as P ± 2 SEP {95% CLs}.
• Samples with larger or smaller than this range will
be rare or only 5%. And such values will taken as
statistically significant at 5% level of significance.
Formula: SEP = √ p x q / n
12/10/12 Dr. Arshad Sabir 9

10. 95% CI for a proportion (percentage)
( categorical variable) Exercise.2
• In sample of 120 T.B pts. drawn from country,
23.3%(28) had compliance with treatment.
• Is this finding holds true for whole population ?
Standard Error for Proportion/Percentage (SEP)
if p = one of the percentage (23.3%)
100-p = other percentage = 100-23.3 = 76.7% (q)
SEP = √ p x q / n = √ 23.3-76.7/ 120 =3.8
95% CI for SEP = p ± 2 x SEP = 23.3 ± (2 x 3.8)
95% CI for SEP = 15.5%----31.1%

11. Standard error of difference b/w two
proportions [SE(p1 –p2)] (02 samples)
Essentials:
1. Samples are large
2. Samples are selected at random
observed difference = p1- p2
Z =
Standard error of diff. SE (p1- p2)
if observed difference is more than 2 SE, it is statistically
significant or real difference, at 5% level of significance
other wise is “normal” difference
12/10/12 Dr. Arshad Sabir 11

12. Calculation of SE of difference b/w two proportions
[SE(p1 –p2)]
SE(p1 –p2) = sum of the square root of the sum of
the squares of SEs of the two proportions.
SE(p1 –p2) = ( p1 x q1) + ( p2 x q2)
n1 n2
Observed Difference (p1 –p2)
Z = ------------------------------------- = (LOS ≥ 2)
SE of the difference (SE(p1 –p2))
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13. SE of difference b/w two proportions: Exercise
Morality in Pyomeningitis with B. Penicillin 30% and was 20%
with Ceftrioxone in sample of 100 in both cases.
SE(p1 –p2) = (30 x 70) + (20 x 80)
100 + 100
SE(p1 –p2) = 37 = 6.08
Z = Obs. diff = 30 – 20 = 10/ 6.08 = 1.64 ( critical LOS is 2)
SE of diff. 6.08
Z = less than 2 (95% confidence limits)
Hence difference is insignificant at 95%
confidence limits or at 5% level of significance.
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14. Uses of SEP
1. To find confidence limits for population
proportions (P) when only sample proportion (p)
is known.
2. To determine if a sample was drawn from a
known population or not when the population
proportion is known……… Z = p-P/SEP
( should by within 2SEP at 5%LOS).
3. To find out standard error of the difference b/w
the two proportions ( significant or not sig.)
4. To find the size of the sample. n = 4pq / L2
(margin of error, say 5% of proportion p [0.05])
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15. Decision making in Health
1. Standard error for Mean
2. Standard error of difference b/w two
means
3. Students t-test
4. Standard Error for Proportion.
5. Standard Error of the difference b/w
two proportions
6. Chi – square test

16. Testing a statistical Hypothesis.
“Hypothesis” is a statement which is to be tested
under the assumption of to be true .
In statistical testing 02 Hypothesis are formulated:
1. Null Hypothesis ( Ho )-there is no difference
between characteristics of a two samples or
both are from same population.
{No difference Hypothesis}
2. Alternate Hypothesis (HA). Sample value is
“significantly” different from pop. OR from
other sample value.
{Hypo. of significant difference}

17. Hypothesis testing……
Ho is against the claim of the researcher.
Researcher desires to reject Ho and in
doing so he may commit error-
Type-I error or alpha -error………………
Rejecting Ho when it was actually true
( No significant differences exist )
OR
Type-II error or Beta-error .
Accepting Ho when it was not true ……
(Significant difference do exist)

19. Hypothesis testing
Decision True situation
based of Difference Difference not
study results Exist Exist
Difference Correct decision Type-I error
exist: {α error}
H0 Rejected
Difference Type –II error Correct decision
don't exist: {β-error}
H0 Accepted

20. Tests of significance
• Whether a study result can be considered as result
which indeed exist in study population from where
sample was drawn?
• Whether the differences observed are due to
chance variation(normal) or are true due to play
some external factor (significantly different).
• These tests are mathematical procedures by which
likelihood (probability) of an observed study
results (differences)occurring by chance is found.
• POWRE OF THE TEST: is its ability to detect
differences between groups if such differences
actually exist.

21. Tests of significance
When 02 or more groups are compared,
possibility could be;
– There is no difference [reject null Hypothesis]
– There is some difference:
• Slight difference (normal or by chance difference)
• Large (sig.)difference not explainable by chance or
that may be due to play of some external factor.
Extent of an observed diff. of being “normal”
and not normal beyond that (significant) is
decided on the basis of certain cut off values
obtained by applying some statistical test or
procedure.
Selection of tests depends upon type of data.

22. Level of significance
Study results are sample based
We can never 100% sure about study result ( many
sources of variation )
By convention we accept results if we have 95%
confidence upon results (diff. exist) or if chances of
having results by chance (actually no diff.) are less
than 5%.
We allow 5% level of accepting results that might have
occurred by chance . This is called level of
significance (LOS) or level of alpha.

23. Level of significance (α) and P-value
Probability of committing α-Error or getting the results
by chance or wrongly rejecting Ho is fixed before the
start of the experiment. (LOS). A max. level is fixed.
It is usually fixed at .01 (1%) or .05 (5%) LOS
But the p-value is obtained after completing the
experiment. It is derived (from a table)after applying
some suitable statistical test to the study results. It is
not fixed. It may assume any value more, or less or
equal to the LOS (5%).
Obtained p-value is compared with LOS. If is (.03,.02 0r .
01)equal or less than 0.05, we will reject Ho and
accept HA and if comes more than fixed LOS like 0.06.
0.1, 0.5 etc, we will accept Ho.

24. Important tests of significance.
Data information n Tests
(Qualitative) Frequencies as Small ( less Fisher exact
categorical percentages or than 40) test
-Nominal Proportions Chi-square
Large (more
etc than 40)
test
(Quantitative) -Means, 02 groups Students t-test
Numeric Multiple gps F-test
Interval, If linear Person’s
ratio scale relationship is Correlation-
data suspected Co- efficient.
ANVO

25. Chi-square test (x2)
ESSENTIALS:
Used to find out whether the observed differences b/w
proportions of events in 2 or more groups may
considered statistically sig.
• It was developed by Karl Pearson
• Non-parametric test. Not based on any / normal
distribution of the variable under study.
• Used qualitative, discrete data in frequencies or
proportions ( not in percentages)
• Involves calculation of a quantity called Chi-square (x2)
• This test is based on measuring diff b/w observed
frequencies and expected frequencies.
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26. Steps of applying Chi-square test (x2)
An assumption of f no difference is made which is then
proved or disproved with the x2 test. (Null hypothesis)
• Steps:
– Fix a level of sig. (.05) for tab. P-value.
– Enter study data in the table, observed Frequency (O)
– Calculate expected frequency for each cell (E)
– Formula for x 2 value of each cell = (O-E) 2/E
E f = (RT x CT / GT)
– Add up results of all cells X 2cal = ∑ (O-E) 2/E
– Df = (C-1) x ( R-1) ( it is 1 in 2X2 table)
– Compare X 2cal value with value X 2tab as pre decided LOS in
the table for given DF , if it is equal or larger than it ,. that
means p-value for this data is smaller than LOS p-value, we
reject H0 and accept HA otherwise we accept H0
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27. Is the use of ANS is associated with shorter distance ?
Distance from ANS Used ANS Not Used ANS Total
Less than 10 Km (O) 51(E= 44.4) (O)29 (E= 35.6) 80
10Km or more (O)35 (E = 41.6) (O)40 (E= 33.4) 75
86 69 155
E or Expected values are calculated on the basis of supposition
(H0 )of no difference in utilization of ANS in the two groups of
women
X 2cal = ( 51-44.4)2 +(29- 53.6)2+(35-41.6)2+(40-33.4)2 = 4.55
44.4 35.6 41.6 33.4
X 2cal at 2DF = 4.55 while X 2tab at 0.05 LOS at 2DF is 3.84
was as the cal value is larger than tab value that means
P-value in this case is less than 0.05 hence the diff
observed is sig. and H0 is rejected.

28. Chi-square (x2) as a test of “Goodness of fit
• Ratio of male to female birth is universally expected 1:1
(50% to 50%).
• Observed ratio in a village was M=52 & F=48
• Is the difference is normal or significant?
• Male Female
• Obs-freq. 52 48
• Expect-freq. 50 50 ( 50% 50%)
(52-50)2 + ( 48-50)2
X2 ______________ ____________
= 8/50 = 0.16
50 50
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29. Chi-square (x2) as a test of “Goodness of fit”.
• Degree of freedom = (No. of classes--1) K—1 =
2-1 = 1 OR DF = (R-1) x (C-1)
• At 5% LOS expected value of X2 = 3.841 (table
value) while calculated value of chi-square is (
X2cal = 0.16) much lower than it.
Hence the observed difference in births is normal
or by chance and not significant.
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30. Student’s t-test
• Numerical data (mean values),
• Normal Variable, Compare 02 groups
• Random sampling
Steps:
1. Calculate t-value (from data)
2. Chose a level of Significance (LOS) usually .05 which actually
means probability of having difference by chance (P-value)
3. Determine DF (= sum of two sample sizes minus 2)
4. Locate t-value corresponding to LOS at the given degree or
freedom. If cal-t value is equal to larger than table value of t
means P-value in this case is significant or less than chosen
LOS (indicated at the top of column), Hence H0 is rejected

31. How to calculate t-value
1. Calculate means of the two groups
(x1 and x2 )
2. Calculate difference b/w means of the two groups.
(x1 – x2 ).
3. Calculate standard Deviation of each study group
(SD1 & SD2 )
4. Calculate the Standard Error for the both groups
(SE1 & SE2)
SE = SD / √ n.
5. Formula for t- value is
x1 – x2
t= -----------------
√ SD12 /n1 + SD22 /n2

32. Exercise
Delivery outcome n Mean Ht. SD
Normal B wt 60 156cm 3.1
LBW 52 152cm 2.8
H0 : there is no difference in mean hts. of the two gp?
Diff. may be by chance but acceptable LOS is 0.05 (p < .05)
x1 – x 2 2 2
t = ----------------- = -------------------------- = ------ = 3.6
√ SD12 /n1 + SD22 /n2 √ 3.1 2 /60 + 2.8 2 /52 0.56
Calculated value of t = 3.6, Tab value of t at DF 110 at .05 LOS is
1.98 . Hence Cal- t value is larger than tab value of t hence the
difference is sig. and H is rejected.

33. OSPE Questions
Example: 1, In a sample(n=1000) obesity in man was
found 20% and30% in women. Is the difference is
has reflected actual diff in the total pop. or has
occurred by chance.
Calculate SE of the diff. b/w two proportions at
5%LOS ?
Example.2, Average B.P of bank cashier (170) as
compare to that of PRO staff (150). Is the
difference is normal or is real due to play some
external factor (stress).
Calculate SE of the difference b/w two means at
5%LOS