Introduction
 Compare two sets of measurements
 Sample size required for comparing two measurements
 When the measurements are parametric, normally distributed
 When the measurements are nonparametric, when normal distribution cannot be assumed
 Compare two regressions
 Comparing two regression coefficients
 Covariance analysis. Comparing two means adjusted for common covariance
 Compare two proportions
 Sample size required for comparing two proportions
 Fisher's Exact Probability
 Chi Square for 2x2 table
 Three models of comparing proportions
 Risks and odds
 Risk Difference and Number Needed to Treat
 Risk Ratio
 Odds Ratio
 Exercises
The programs for calculations are
Comparing 2 parametric means : Sample Size
 Calculating sample size for comparing two means
 Requires the following parameters
 The expected Standard Deviation of the measurement
 The difference we wish to detect
 The probability of Type I Error we will use to determine statistical significance. Default p=0.05
 The power we wish to use. Default = 0.8
 One or two tail model
 Results : sample size per group
 Approximate sample size
 Large Effect : clinically obvious (Difference / SD = 1) : 14(one tail) or 17 (two tail) cases per group
 Moderate Effect : not obvious but clinically meaningful (Difference / SD = 0.5) : 51 (one tail) or 64 (two tail) cases per group
 Small Effect : too small to matter in most clinical situations, but important for
scientific reasons (Difference / SD = 0.2) : 310 (one tail) or 394 (two tail) cases per group
 Example in the program : comparing birth weight between boys and girls
 Data Input
 We expect the Standard Deviation of birth weight to be 400g
 We will conclude that a difference of 150 gram will be clinically meaningful
 Result : 89 per group (1 tail), 113 per group (2 tail)
 Interpretation
 We will need 113 boys and 113 girls to find out whether such a difference exists (two tail)
 If we are interested in whether boys are heavier than girls, but do not care whether girls
are heavier than boys, then we will need 89 boys and 89 girls in the study (1 tail)
Compare Two Parametric Means
 if necessary, Calculate n, mean, and Standard Deviation from raw data in StatPgm_7_Utilities.php
 Data entry : n, mean, and SD of two groups
 Results :
 Difference between two means d = mean 1  mean 2
 Standard Error of the difference : SE
 Degrees of freedom df = n1 + n2  2
 t Test (old standard test)
 t = difference / SE
 Probabilities (1 and 2 tail) of t, df is probability of Type I Error (p, α)
 Difference is statistically significant if p<0.05
 95% confidence interval of the difference (more recent standard)
 Required t (1 and 2 tail) value for p=0.05 and df of result calculated in StatPgm_1_Prob_z_t.php
 95% confidence interval = d ± t_{2 tail} * SE for the two tail model
 95% confidence interval = d  t_{1 tail} * SE to + ∞ if whether mean2<mean1 is of interest
 95% confidence interval = ∞ to d + t_{1 tail} * SE if whether mean1<mean2 is of interest
 Assumptions of comparing two means
 measurements are parametric (continuous and normally distributed)
 The Standard Deviations in the two groups are similar
 Example in the program : comparing Haemoglobin levels between vegetarians and nonvegetarians
Grp  n  mean  SD 
Grp 1 vegetarian  10  11.2  1.04 
Grp 2 nonvegetarian  30  13.8  3.03 
 Data input : n, mean, and SD of the two groups, as above
 Result Output
 difference d = mean1  mean2 = 2.6. SE of difference = 0.98
 df = 10 + 30 2 = 38
 Statistical significance (t test)
 t = d / SE = 2.6 / 0.984 = 2.6421
 Probability (t=2.64, df=38) = 0.012 (1 tail), 0.006 (2 tail)
 Statistical significance (95% confidence interval of d)
 Student's t )p=0.05, df=38) = 1.6859 (1 tail), 2.0244 (2 tail)
 95% confidence interval (1 tail) = ∞ to mean+t*SE = ∞ to 2.6 + 1.6859*0.984
= ∞ to 0.94
 For the 2 tail model, mean±t*(SE = 2.6±2.0244*0.984 = 4.5921 to 0.6079
 The 95% confidence interval for both models did not overlap the null value (0), so the difference is statistically significant
Comparing Two Regressions : Covariance Analysis
 Comparing two regression lines is the simplest model of covariance analysis
 Dependent variable, y, in two groups are compared after adjusting for a common influence (independent variable, x, covariate).
 Example from StatPgm 3c. Compare Two Regressions
 Example compares 22 newborns, 10 boys and 12 girls, arranged in a 3 column table.
 Column 1 is group designation, in this example boy or girl
 Column 2 is the covariate (x in regression), in this example gestation in weeks
 Column 3 is the dependent variable (y in regression), in this example birth weight in g.
 Program computes as follows
 The two regression lines are computed, y_{1} = a_{1} + b_{1}x and y_{2} =
a_{2} + b_{2}x. In this example
 Group 1 (Boys) y_{1}=3772 + 185.3x_{1}, or Birth weight (g) = 3772 + 185.3 Gestation (week)
 Group 2 (Girls) y_{2}=3999 + 186.9x_{2}, or Birth weight (g) = 3999 + 186.9 Gestation (week)
 The regression coefficients from the two groups are compared to see if they are statistically significant.
 185.3g/week from group 1 (Boys)
 186.9g/week from group 2 (girls)
 The difference was 1.6g / week, statistically not significant
 If the regression coefficients are significantly different
 Boys and girls grow at different rate near term
 Gestation cannot be used as a covariate
 A significant interaction exists between grouping and covariate.
 If the regression coefficients are not significantly different
 Boys and girls grow at same rate near term
 Gestation can be used as a covariate
 No significant interaction exists between grouping and covariate.
 With no significant interaction, a combined regression coefficient is calculated, both sexes grow 186.2g/week
 The birth weights are recalculated (actual birth weight  expected birth weight from regression)
 Recalculated birth weights from the two groups are then compared
Comparing Two Sets of Nonparametric Measurements : Robust Signed Ranked Test (MannWhitney U)
 Nonparametric measurements are measurements that are not normally distributed: Examples are
 Likert Scale : 0=Strongly disagree, 1=disagree, 2=neutral, 3=agree, 4=strongly agree
 Pain scale : 0=no pain, 1=some pain, 2=severe pain
 Cancer staging : 0=no invasion, 1=local invasion, 2=beyond organ 3=lymph node, 4=distant metastasis
 Labour : 0=no labour, 1=before dull dilatation, 2=before baby delivered, 3=before placenta delivered
 Sample size : Not calculable. Common practice
 Use approximate sample size for parametric comparison, but adds 10%
 Calculations in program : Comparing Likert Response from doctors (Dr) and midwives (MS) to the statement "midwives should provide care to normal pregnancy and labour independent of doctor's supervision"
 1=strongly disagree, 2=disagree, 3=neutral, 4=agree, 5=strongly agree
 Data entry : 2 column.
 Col 1 = 1 or 2 for group designation
 Col 2 = measurements
 Each row data from a case
 Result : U = 1.7704 p<0.05
 Significant difference between doctors and midwives
 U<0, indicating group 1 (doctors) provide a lower score (less agreeable to the statement)
Compare two proportions
 Definitions
 Number of cases with positive outcome : NP
 Number of cases with negative outcome : NN
 Total = NP + NN
 Risk : clinical term for proportion. Risk = NP / (NP+NN)
 Odd = NP / NN
 Odd = Risk / (1  Risk)
 Risk = Odd / (1 + Odd)
 95% confidence interval of proportions
 Proportion is the summary of individual positive and negative cases in the data
 It is assumed to be population based, and normally distributed
 In calculations of confidence interval, the population z is used instead of t
 For 1 tail model, 95% confidence interval = proportion+1.65SE or proportion  1.65SE
 For 2 tail model, 95% confidence interval = proportion±1.96SE
 The same applies for odd, risk difference, Log(risk ratio), and Log(odds ratio)
Compare two proportions : Sample Size
 Calculations in StatPgm_3b_2Proportions.php
 Estimation of sample size is in the planning stage of research
 Information required
 Probability of type I error use to determine statistical significance (p, α).
the module default value is p=0.05
 Power of the model. the module default value is power=0.8
 The two proportions (risks) to be compared
 Output is sample size per group for one and two tail models
 Example in program : comparing postpartum haemorrhage in controlled trial. Expected rate=6% (0.06) in control group, and haved to 3% (0.03) in treastment group
 Information required
 α : default value p=0.05
 power : default value 0.8
 The two proportions are 0.06 (6%) and 0.03 (3%)
 Results are 590 per group (1 tail), 749 per group (2 tail)
Compare two proportions : Earlier Statistical Tests
Fisher's Exact probability
 This was devised by Fisher nearly 200 years ago
 It is more mathematical than statistical.
 Using a mix of combination and permutation, it estimates the probability that the counts in a 2x2 table is randomly distributed
 A low probability (p<0.05) indicates that the 4 counts are nonrandom
 The test is highly precise and valid, but requires multiple repeated calculation, especially if the sample size is large
 Before computers it is not possible to handle sample size greater than 1215
 Before modern computers (>16 bits processor) it was not possible to handle sample size >20
 With modern high speed computer with 64 bit processor or more, sample size up to 500 can be handled
 In StatPgm_3b_2Proportions.php, we calculate Fisher's Exact probability
 In group 1, 10 positives and 50 negatives, proportion = 10 / (10 + 50) = 0.167 (16.7%)
 In group 2, 18 positives and 222 negatives, proportion = 18 / (18 + 222) = 0.075 (7.5%)
 Fisher's Exact Probability = 0.044, <0.05, so statistically significant
Chi Square Test for 2x2 contingency tables
 This is an adaptation of Fisher's Exact Probability to overcome long computation with large numbers
 The logic however is similar to Fisher's Test
 In StatPgm_3b_2Proportions.php,
the Chi Square Test results are
 Chi Square = 3.74 p=0.053, not statistically significant
 This result shows that the Chi Square Test is less powerful than Fisher's Test
Compare two proportions : Risk Difference and Numbers Needed to Treat
 Risk difference usually used in controlled trials
 Data input : Numbers of cases with positive and negative outcomes in the two groups
 Result output
 Risks in group 1 (r1) and group 2 (r2)
 Risk difference rd = r1r2
 Standard Error of the difference
 95% confidence interval of the difference
 rd±1.96SE for the two tail model
 0 to rd + 1.65SE for one tail model testing whether r1<r2
 rd + 1.65SE to 1 for one tail model testing whether r1>r2
 Numbers needed to treat NNT = 1/rd, rounded to nearest whole number
 Example : Risk of Postpartum Haemorrhage in 2 treatment groups in a controlled trial.
 Group 1 (treatment) : Total = 94, PPH = 2, Risk r1 = 2/94 = 0.021 (2.1%)
 Group 2 (control) : Total 104, PPH = 15, Risk r2 = 15/104 = 0.144 (14.4%)
 Risk Difference rd = 0.021  0.144 = 0.123, Standard Error SE = 0.038.
 95% confidence interval of rd (2 tail) = 0.123±1.96*0.038 = 0.197 to 0.049.
 The Number Needed to Treat NNT = 1/0.123 = 8.13, rounded up to 9.
 An extra 9 patient to be given oxytocic is needed to reduce one case of haemorrhage
Compare two proportions : Risk Ratio
 Risk ratios are more suitable for epidemiological studies
 Risk Ratio is clinically easier to understand, but
 Ratios are lognormally distributed
 all the statistics are done using log(risk ratio) and its Standard Error
 when all calculations are concluded, the 95% confidence interval is translated back
to nonlog values.
 this means that the 95% confidence interval of risk ratio is asymmetric around the central
value, being longer at the higher value end
 Data input : Numbers of cases with positive and negative outcomes in the two groups
 Result output
 Risks in group 1 (r1) and group 2 (r2)
 Risk ratio rr = r1 / r2
 Log(Risk ratio) Lrr = log(rr)
 Standard Error of Lrr
 95% confidence interval of Lrr (null value = 0)
 95% confidence interval of rr (null value = 1)
 Example : Compare CS rate in public and private patients
 Group 1, public patient, had 8 CS out of 98 cases. r1=8/98=0.082 (8.2%)
 Group 2,private patient, had 40 CS out of 240 cases. r2=40/240=0.167(16.7%)
 Risk ratio RR=0.082/0.167=0.49
 95% Confidence Interval of Risk Ratio (2 tail) = 0.238 to 1.0081
 Significant difference between groups? no, 95% CI overlaps null (1), 23.8% to 100.8% of private
 95% Confidence Interval of Risk Ratio (1 tail) = <0.2673 or >0.8976
 Public patient less likely to have CS? yes, 95%CI does not overlap null (1), 0% to 26.7% that of private patients
 Public patient more likely to have CS? no, 95%CI overlaps null (1), 90% to ∞% that of private patients
Compare two proportion : Odds Ratio
 Odd is the ratio of number of positives and negatives in a group
 It is more flexible to use, and often used in multivariate statistics
 It can examine data retrospectively, grouping data by outcomes, and examine differences in causes
 Odds Ratio is clinically easier to understand, but
 Ratios are lognormally distributed
 all the statistics are done using log(odds ratio) and its Standard Error
 when all calculations are concluded, the 95% confidence interval is translated back
to nonlog values.
 this means that the 95% confidence interval of odds ratio is asymmetric around the central
value, being longer at the higher value end
 Data input : Numbers of cases with positive and negative outcomes in the two groups
 Result output
 Odds in group 1 (o1) and group 2 (o2)
 Odds ratio or = o1 / o2
 Log(odds ratio) Lor = log(or)
 Standard Error of Lor
 95% confidence interval of Lor (null value = 0)
 95% confidence interval of or (null value = 1)
 Example : Examine two groups of children, with or without learning problems,
and compare their intrauterine exposure to cigarette smoke environment
 Problem children (Group 1), Total=60, exposed=10, not exposed=50, Odd o1=10/50=0.2
 Normal children (Group 2), Total=240, exposed=18, not exposed=222, Odd o2=18/222=0.0811
 odd ratio OR=0.2/0.0811=2.4667
 95% Confidence Interval of Odds Ratio (2 tail) = 1.0738 to 5.6663
 Are there different to smoking exposure between the two groups of children? yes, the 95%CI does not overlap null (1). children with learning difficulties are 1.07 to 5.67 times more likely to be exposed
 95% Confidence Interval of Odds Ratio (1 tail) = >1.2274 or <4.9572
 Are children with learning difficulties more exposed to smoking? yes, the 95%CI does not overlap null (1). They are 1.23 or more times
as likely to be exposed to smoking than normal children
 Are children with learning difficulties less exposed to smoking? no, the 95%CI overlaps null (1). They are 4.96 or less times as likely to be exposed to smoking than normal children
Why 3 ways of comparing two proportions
 Historically, the 3 methods were developed by different researcher
 Risk Difference was developed for clinical trials, where the risk is the outcome. An example is
to compare the cure rates (risks of getting better) of two antibiotics for a severe infection
 Risk Ratio was developed by epidemiologists, to examine which causal factors are related to which
outcome measurements. An example is whether obesity increases the risk of high blood pressure
 The Odds Ratio was adapted from gambling to solve the problems of retrospective matched pair controlled
studies, as this procedure does not follow the cause to effect sequence, so groups can be classified according
to outcome, and the causes are compared. An example is whether babies with limb defects have greater odd of
being exposed to Thalidomide in utero.
 The advantages and disadvantages are
 Risk Difference
 The most powerful test, more likely to detect a difference if it is there.
 More likely to produce erroneous conclusions if
 Sample sizes in the groups are not approximately the same
 If the proportions involved are too close to 0 or 1
 Risk Ratio
 A more flexible test, suited for the widely variable environment of epidemiological research
 It copes well with Large discrepancies in sample size (e.g. diabetic and nondiabetic pregnancies)
 It copes well with very small proportions (e.g. perinatal death, 25%)
 Odds Ratio
 Has all the advantages of Risk Ratio, plus
 It can classify according to outcome and examine causes (e.g. thalidomide and limb defect)
 Its simplicity allows it to adapt to analyse multiple causes (e.g. Logistic Regression)
 Changing practices
 All 3 tests have been used for clinical trials, Odds Ratio is used when there are large differences in group sizes
 Epidemiological studies use Risk Ratio and Odds Ratio. Risk Difference is generally not used
 Only Odds Ratio is used for retrospective matched pair controlled trials
Exercises
Calculations for these exercises require 4 programs
Q 1. We wish to estimate the difference in birth weight between boys and girls. We will use the common p=0.05 to determine statistical
significance, and a power of 0.8. What is the sample size for each sex we will need for the following combinations
 if expected Standard Deviation of birth weight is 350,400,450,500g
 if expected difference of 50, 100, 150, or 200g
A 1. Click to show contents
The number of babies for each sex required are in the following table, rows for Standard Deviation and columns for expected difference
 One Tail  Two Tail 
 diff=50g  diff=50g100g  diff=50g150g  diff=50g200g  diff=50g50g  diff=50g100g  diff=50g150g  diff=50g200g 
SD=350g  607  153  68  39  771  194  87  50 
SD=400g  793  199  89  51  1006  253  113  64 
SD=450g  1003  252  112  64  1273  319  143  81 
SD=500g  1238  310  139  78  1571  394  176  100 
Q 2. We wish to conduct a controlled trial to see if elective episiotomy in the primipara can reduce the
prevalence of third degree tears during normal vaginal delivery.
 We feel that if elective episiotomy can reduce third degree tear by half, it would be clinically meaningful.
 We are only interested in whether episiotomy can reduce third degree tears, but indifferent to whether it
increases tears.
 We will use the common p=0.05 to determine statistical significance, and a power of 0.8.
 What the sample size for each group, if our existing third degree tear rate for
primiparous normal vaginal delivery is 2%, 1%, 0.5%, and 0.2%
A 2. Click to show contents
This is a one tail study, and the sample size per group are as in the bottom row of the following table
Proportion 1  0.02  0.01  0.005  0.002 
Proportion 2  0.01  0.005  0.0025  0.001 
Sample size per group(2 tail)  2319  4673  9383  23510 
Sample size per group(1 tail)  1826  3681  7391  18519 
Parity  Hrs Labour  BWt 
Multipara  6  4199 
Primipara  9  3381 
Multipara  3  4006 
Primipara  7  2591 
Primipara  14  3511 
Primipara  10  3509 
Primipara  13  3523 
Primipara  9  3300 
Multipara  8  3777 
Multipara  5  3383 
Primipara  3  4106 
Primipara  2  3152 
Primipara  2  3980 
Primipara  2  3543 
Multipara  4  3004 
Primipara  2  3306 
Multipara  5  3294 
Multipara  2  3401 
Primipara  5  4017 
Multipara  4  3317 
Multipara  8  3824 
Multipara  5  3744 
Multipara  2  3785 
Multipara  8  3856 
Primipara  4  3415 
Multipara  3  4049 
Primipara  4  3463 
Multipara  5  3241 
Multipara  5  3918 
Multipara  2  3066 
Primipara  11  3094 
Multipara  2  4253 
Multipara  4  2840 
Multipara  6  4161 
Primipara  15  3413 
Primipara  6  3435 
Primipara  5  3274 
Primipara  10  3450 
Multipara  3  3799 
Multipara  5  3473 
Multipara  3  4078 
Multipara  3  3603 
Primipara  11  3477 
Primipara  16  3226 
Multipara  8  3852 
Multipara  5  3943 
Multipara  6  3863 
Multipara  4  3586 
Primipara  17  3393 
Primipara  3  3420 
Q 3.
The table to the right shows parity, duration of labour(hrs) and birth weight(g) of 50 babies.
Q 3a. Estimate the mean, Standard Deviation, and Standard Error, and their 95%
confidence intervals of the birth weights in the two parity groups, and plot the data to show the difference
A 3a. Click to show contents
Grp  n  mean  SD  95%CI Values  SE  95%CI Mean 
Primipara  23  3434  312  2787  4080  65  3299  3569 
Multipara  27  36784  380  2898  4459  73  3528  38289 

Q 3b. Assuming birth weight is normally distributed, estimate the 95% confidence interval of the difference in
birth weight in the 2 parity groups
A 3b. Click to show contents
Difference mean(primiparamultipara) = 244g Standard Error of difference = 99
95% confidence interval of difference (two tails) = 444g to 44g, statistically significance
Birth weight of primiparas on average weigh 44g to 444g lighter than that of multiparas
Q 3c. Plot the percentage of different duration of labour as bar charts in the two parities
A 3c. Click to show contents
Hrs  Primip(%)  multip(%) 
2  17.4  14.8 
3  8.7  18.5 
4  8.7  14.8 
5  8.7  25.9 
6  4.3  11.1 
7  4.3  0.0 
8  0.0  14.8 
9  8.7  0.0 
10  8.7  0.0 
11  8.7  0.0 
13  4.3  0.0 
14  4.3  0.0 
15  4.3  0.0 
16  4.3  0.0 
17  4.3  0.0 

Q 3d. Estimate whether the duration of labour is different in the two parities
A 3d. Click to show contents
MannWhitney U Test : MannWhitney U = 1.8666, p=0.031
Statistically significant
On the whole primiparas laboured longer than multiparas


Group1  VD 
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Q 4.
In a trial on whether controlled breathing for pain relieve during labour can reduce
risk of Caesarean Section, we randomly allocate primiparas into two group during antenatal classes.
 Group 1 received antenatal education plus special breathing exercise for pain control
 Group 2 received antenatal education only.
 The data obtained is presented on the table to the right.
Q 4a. Produce 2 Pie charts, one for each group, the show distribution of Caesarean Section
A 4a. Click to show contents
Q 4b. Estimate the 95% confidence interval of the difference in Caesarean Section rate in the two groups
A 4b. Click to show contents
 CS  VD  CS_ Rate(%) 
Experimental Grp  7  19  26.9 
Control Grp  7  17  29.2 
Risk Difference rd=0.0278 SE = 0.1574
95% CI(one tail) = ∞ to 0.2312
On average, Breathing exercise reduces CS rate by 2.8%, with a 95% confidence overlapping the
null (0) value, so this difference is not statistically significant
Q 4c. Estimate how many primiparas need to receive training for breathing exercise to reduce each case of Caesarean Section
A 4c. Click to show contents
Numbers Needed to Treat (NNT) = 1/rd = 1/0.0278 = 35.97, rounded up to 36


 Midwives  Doctors 
Total  1045  982 
Epidural  45  80 
IV fluid  200  300 
Prolonged First Stage  100  70 
C.S.  180  300 
Q 5.
In a study of outcomes of labour in an obstetric unit, comparing normal women in labour cared for by midwives or doctors,
some of the counts of events are shown in the table to the right
Q 5a. Calculate the 95% confidence interval of risk ratio (care by midwives/care by doctors) for these events
A 5a. Click to show contents
 midwives (grp 1)  Doctors (Grp 2)  Risk 1  Risk 2  Log(Risk Ratio)  Standard Error  Risk Ratio  95%CL of Risk Ratio 
Epidural  45 of 1045  80 of 982  0.043  0.082  0.638  0.181  0.529  0.371  0.754 
IV fluid  200 of 1045  300 of 982  0.191  0.306  0.468  0.080  0.627  0.536  0.732 
Prolonged First Stage  100 of 1045  70 of 982  0.096  0.071  0.295  0.149  1.342  1.002  1.799 
C.S.  180 of 1045  300 of 982  0.172  0.306  0.573  0.083  0.564  0.479  0.664 
Those cared by midwives
 Had lower risks of receiving an epidural, 37% to 75% that of those cared by doctors
 Had lower risk of receiving IV fluid, 54% to 73% that of those cared by doctors
 Had higher risk of prolonged first stage, 100% to 180% that of those cared by doctors
 Had lower risk of being delivered by Caesarean Section, 48% to 66% that of those cared by doctors
Q 5b. Create a Forest Plot to show these results
A 5b. Click to show contents 
Q 6. A retrospective, matched paired control study was carried out to examine whether
a baby in excess of 4Kg is associated with shoulder dystocia.
 A search through the medical record found 50 women who had shoulder dystocia at delivery.
 For each of these women, 4 women with no shoulder dystocia were selected who delivered in the same month
and same time of day, had the same age, parity, height, BMI,
and gestation.
 The babies were classified as >4Kg or not.
After reviewing the records
 There were 50 women who had shoulder dystocia during delivery, 35 of them had a baby >4Kg
 There were 200 matched women who did not have shoulder dystocia during delivery 40 of them had
a baby >4Kg.
Calculate the 95% confidence interval of the odds ratio of having a baby >4Kg between those
who had or had not a shoulder dystocia
A 6. Click to show contents
 Shoulder Dystocia  No Shoulder Dystocia 
baby>4Kg  35  40 
Baby<=4Kg  15  160 
Total  50  200 
Results of analysis
 Odd of having a baby >4Kg in shoulder dystocia group O1 = 35/15 = 2.3333
 Odd of having a baby >4Kg in no shoulder dystocia group O2 = 40/160 = 0.25
 Odds Ratio OR = O1/O2 = 2.3333/0.25 = 9.3333.
 Log(OR) = log(9.3333) = 2.2336, Standard Error = 0.3557
 95% Confidence interval of Log(OR)(two tail) = 2.3333±1.96*0.3557 = 1.5365 to 2.9307
 95% Confidence interval of (OR)(two tail) = exp(1.5365) to exp(2.9307) = 4.6484 to 18.7399
 The odd of having a baby >4Kg in those who had a shoulder dystocia is 4.6 to 18.7 times
greater than those who did not have a shoulder dystocia, and this is statistically significant
 We used the two tail model here because we are interested in both whether big or small babies
are associated with shoulder dystocia. The answer is big babies
More Exercises
