Statistical Analysis of Biological Data III

1. Zagazig university
Faculty of Veterinary Medicine
Animal Wealth Development Department
Session#2:
Statistical Analysis of Biological Data
(Comparison of Proportions)
M.Afifi
M.Sc., Biostatistics (Joint Supervision with ISSR, Cairo University)
Ph.D., Candidate (AVC, UPEI, Canada)
E-mail: M.Afifi@zu.edu.eg, Afifi-stat6@hotmail.com
Tel: +201060658185

2. Comparing of 2-Independent proportions
Testing associations
χ2

3. Assumptions
 Categorical variables, and data contained in the cells of the table are frequencies.
 Animal independent in that no animal/ individual may be represented more than once
in the table.
 No more than 20% of the cells of the table should have an expected frequency < 5.
 If necessary:
• we can reduce the contingency table in size by combining appropriate rows
and/or columns.
• Alternatively, Fisher’s exact test on frequencies contained in a

5.  Comparing two proportions
 2 × 2 table, fourfold table, two-way frequency or contingency table
 If there is no association between the outcome and the group, then we would
expect the proportions of successes to be the same in the two groups. Thus, we
can compare the two proportions by investigating the association between the
two factors that define the contingency table (outcome, group)

6.  9.4.4 Example
 The non-obese diabetic (NOD) mouse develops an autoimmune diabetes that can be used as a
model for human juvenile insulin-dependent diabetes. In the colony of Hawkins et al. (1993),
the incidences for male and female NOD mice were 24% and 73%, respectively. Hawkins et
al. investigated the causes of this sex difference by considering the effect of early castration on
the incidence of diabetes in male NOD mice. (The following is based on their findings.)
 Fifty mice were randomly selected from 100 male mice and were castrated 1 day after birth;
they were compared with the remaining 50 sham-operated mice. The mice were maintained
for 140 days, and blood samples were collected biweekly starting at 42 days old. Diabetes was
determined by three consecutive blood glucose levels greater than 200 mg/dl. It was shown
that neonatal castration more than doubled the incidence of diabetes (52%) when compared
with controls (24%) at day 112. But is this difference significant?

8.  Null hypothesis: that the two population proportions are equal orno association
between the two factors of interest. Specify the alternative hypothesis, generally
that the two proportions are not equal or that there is an association between the
two factors.
 Test statistic: Chi-squared test, With Yates’ correction
 P-value

9.  Diabetes.sav
 2 variables:
• Castration : Castrated, Non-castrated
• Diabetes: Diabetic, Non-diabetic

10.  Analyze >>>> Descriptive Statistics >>>> Crosstabs

17.  Yates’ correction, a continuity correction
 included to remove bias. This bias arises because we are assuming the test
statistic approximates the continuous Chisquared distribution although it has a
discrete distribution.
 Computed only for the 2 × 2 table.

18.  The 95% confidence interval for the true difference in the proportions of mice
with diabetes in the two groups is

19.  χ2 value 7.17 with 1 degree of freedom, and (P = 0.007). >>>> reject null
hypothesis that the true proportions of mice with diabetes are equal in the control
and castrated groups.>>>>> There is evidence to indicate that neonatal castration
is linked with the incidence of diabetes in NOD mice.
 Difference in incidence of diabetes in male and female mice may be associated
with the concentration of testosterone in the blood circulation.

20. > 2 proportions

21.  Following staff training in insemination techniques in cattle, an artificial
insemination centre compared three training methods. The cows were
randomly assigned to a particular training method, each cow was inseminated
once and the proportion of cows that became pregnant in each group is given in
Table 9.3. Is there any evidence for believing that the training methods show
different proportions of pregnant animals?

23.  null hypothesis : proportions pregnant are the same for the three methods. In the
same sense no association between the training methods and the pregnancy state
of the cows. The alternative hypothesis is that the proportions pregnant are not
equal.
 Test statistic: Chi-squared test, With Yates’ correction
 P-value

24.  Insemination-Training
 2 variables:
• Insemination Method (I, II, III)
• Pregnancy ( Preg, Non-preg)

25.  Weight Cases

27.  Weight Cases

28.  The Same previous steps for Chi-square

33.  (P = 0.008). we have evidence to reject the null hypothesis that the proportions
pregnant are the same for the three methods.
 In this instance, Further analysis by comparing the proportions of pregnancies
obtained by any two methods shows that Method III has significantly lower
proportion of cows pregnant than Method I (test statistic = 8.66, P = 0.009 after
employing Bonferroni’s correction for multiple comparisons,

34.  Chi-squared test for trend
 test the null hypothesis that there is no trend in the proportions.
 2×C contingency table in which the variable defining the columns comprises c
ordered categories(e.g. body condition score or age categories). We are
interested in comparing the proportions of successes in c ordered groups, and
would expect any differences in the proportions, if they exist, to be related to the
ordering.

35. Comparing two paired proportions
McNemar’s test

36.  Schönmann et al. (1994) compared two methods of culture of Tritrichomonas
foetus in the washings of the prepuce of infected beef bulls to determine the best
method for detection of the organism. In comparing the methods of culture,
Claussen’s medium detected the organism in 61 of 83 amples whereas a
commercial system detected the organism in 73 of the same 83 samples.

38.  Trichomonas.sav
 Variables 2;
 Claussen: 1;+Ve, 2-Ve
 Commercial: 1;+Ve, 2-Ve

39.  Null hypothesis :the true proportions detected are the same using Claussen’s
medium and the commercial system. The alternative hypothesis is that the two
proportions are different.
 Test statistic:Chi-square
 P-value

43.  (P = 0.006).>>>reject the null hypothesis;>>>we conclude that the commercial
system has the ability to detect the greater proportion of organisms.
 the proportions of organisms detected by Claussen’s medium and the
commercial system to be 61/83 = 0.735 and 73/83 = 0.880, respectively.