statistical analysis of biological data (comaprison of means)
TRANSCRIPT
Zagazig universityFaculty of Veterinary Medicine
Animal Wealth Development Department
Session#2:Statistical Analysis of Biological Data
(Comparison of Means)
M.Afifi
M.Sc., Biostatistics(Joint Supervision with ISSR, Cairo University) Ph.D., Candidate (AVC, UPEI, Canada)
E-mail: [email protected], [email protected] Tel: +201060658185
Course Materials are derived from Statistics for Veterinary and Animal Science, Third Edition.(2013)
Very Popular Minimal Mathematical Approach Well Defined Veterinary Examples (Highlighted) Exercise following each chapter Solutions to Exercise P(303) Ready made SPSS outputs
134 Publications
Flowcharts for selection of appropriate tests
Basic rules of any statistical test
Assumption Hypothesis testing
Basic rules of hypothesis testing Hypothesis:
• Null hypothesis, H0:
• Difference in (means, proportions, medians) is not actual (non-sig),
• Difference not due to treatment effect but due to any other reasons (Chance , Error)
• Alternative hypothesis HA : VS H0
Test statistic- value: value calculated from the data (an algebraic expression particular to the
hypothesis we are testing),
t-test >>>> t-value
F-test >>>> F-value
χ2-test >>>>>> χ2 value
P-value: probability value (0-1) (Sig): Attached to each value of the test statistic It
the probability of getting the observed effect (or one more extreme) if the null hypothesis is true
Set Up your probabilities
True
False
Null hypothesis
Reject
Accept
Accept
Reject
Confidence Level95%
Level of Significance5%
P (Type II error)
Power
Comparison of Means
2-Independent sample Means
2-Independent sample Means
Two-sample t-test (unpaired t-test) Compare the means in two independent groups of observations using representative
samples.
Assumptions
Two samples must be independent unrelated
Normality A small departure from Normality is not crucial and leads to only a marginal loss in power
Homoscedastic (equal variances) >>>> Checked by Levene’s test
7.4.4 Example Consider the comparison of the mean body weights at the time of mating in one
group of ewes which have been flushed (put on a high plane of nutrition for 2–3
weeks prior to mating) and another group which have not. Each ewe in a random
sample of 54 ewes is randomly allocated to the flushed or control group. Table
7.2 shows the weights of two samples of 24 flushed and 30 control ewes.
The null hypothesis: mean body weights in the populations of flushed and
control ewes are equal; the two-sided alternative is that they are different.
The test statistic for the two-sample t-test is It is derived as
The P-value (denoted in SPSS by ‘Sig.’)
Statistics for Veterinary and Animal Science-Wiley-Blackwell (2013)\Data File
(Excel-SPSS-Stata)\SPSS\Table 7.2.sav 2 Variables:
Ewe Weight Group: 0;Flushed, 1;control
Analyze >>>> Compare means >>> Independent-Samples T Test
SPSS Output
Descriptives
Test
Levene’s Test
T-test Test
The first line of the t-test result (equal variances assumed) is relevant in this instance
because the result of Levene’s test for the equality of variances indicates that the two
variances are not significantly different (P = 0.617).
How would these results be reported in a scientific journal article?
Tabular Presentation.
Mean ± SD or SEM with Both t-value and P-value
Mean ± SD or SEM with only P-value
Representing P-values with astrikes
Representing P-values with superscripts
Graphical presentation
Simple Bar Box plots
Report your results in words
Your Formal sentence must includes:
Dependent , independent variable
Exact p-value (unless the p value is less than .001). < 0.000 Or < 0.0001
The direction of the effect as evidenced by the reported means, as well as a
statement about statistical significance,
Symbol of the test (t), the degrees of freedom (6), the statistical value (2.95)
There was a significant difference in the mean body weights between the flushed
ewe (M = 6.79, SD = 1.94) and the control one (M = 5.59, SD = 1.81), t(6) =
2.95, p = 0.018.
The mean body weights of the flushed ewe was significantly different, with the
estimated mean ewe body weight in the flushed ewes being 1.59 kg greater than
that of the control ewes. The significantly higher mean body weight of the
flushed ewes implies an effect on metabolism and is expected to be associated
with an optimal ovulation rate
P = 0.018, indicating that the chance of obtaining a difference in means at least
as large as 1.59 kg is only 1.8% if the null hypothesis is true.
Welch's t-test
(Unequal variances t-test) widely used modification of the t-test,
adjusts the number of degrees of freedom when the variances are not equal to
each other.
If the sample sizes are not large,
equal variances not assumed
non-parametric method,
Mann–Whitney U test
Paired (dependent) t-test
FreezingRefrigeration
Methods of pairing: Self-pairing: each animal used as its own control (Before and After)
Natural pairing: each pair of animals is biologically related (e.g. litter mates).
Artificial (matched) pairing: each animal is paired with an animal matched with
respect to one or more factors that affect response.
To avoid allocation bias in an experiment when there is self-pairing, each animal is
randomly allocated to receive one of the two treatments initially; it then receives
the other treatment later.
If there is natural or matched pairing, one member of the pair is randomly allocated
to one of the two treatments and the other member receives the second treatment.
Assumption Paired
we take the difference between the observations in each pair, the set of
differences for all pairs is approximately Normally distributed even though the
original observations in the groups may not be
7.5.4 Example
Nelson et al. (1998) conducted a randomized cross-over trial of two diets in 11 insulin-
dependent diabetic dogs; they measured serum glucose as the variable indicating the quality
of diabetic control. The diets contained either low insoluble fibre (LF) or high insoluble fibre
(HF). Each dog was randomly allocated to receive a particular diet first. The dogs were
adapted to the diet for 2 months and then fed it for 6 months: evaluation was performed at 6-
week intervals. As the study ran over 16 months of each dog’s life, we might expect changes
in the animal’s metabolic responses to diabetes during the course of the trial, irrespective of
diet.
Null hypothesis: true mean difference in the preprandial serum glucose levels
between the low-fibre and high-fibre diets is zero; the two-sided alternative is
that it is not zero.
Test statistic:
P-value
SPSS Table 7.3
2 variables: Low fiber
High Fiber
Analyze >>>> Compare means >>> Paired Samples T test
SPSS computer output for Paired t-test
Descriptives
Test
Paired Vs. Independent Test
If the sample sizes are not large,
equal variances not assumed
non-parametric method,
Wilcoxon rank test
F-test
ANOVA
Comparing more than two means
Suppose, for example, we have four groups. >>>>> compare using a two-
sample t-test) for every combination of pairs of groups >>> six possible t-tests
Principle
Total variability in a data set is partitioned into a different source of variation.
The sources of variation comprise one or more factors, each explained by the
levels or categories of that factor (e.g. the two levels, ‘male’ and ‘female’, defining
the factor ‘sex’, or three dose levels for a given drug factor), and also unexplained
or residual variation which results from uncontrolled biological variation and
technical error.
We can assess the contribution of the different factors to the total variation by
making the appropriate comparisons of these variances.
The variation is expressed by its variance
The analysis of variance encompasses a broad spectrum of experimental
designs ranging from the simple to the complex.
One-way analysis of variance Single factor with several levels or categories where each level comprises a group
of observations.
For example, the levels may be:
Feed formula for dogs: dry feed formula, a tinned feed and a raw meat
Different treatment dose levels of a drug, one of which is a placebo representing
simply the drug vehicle, while the others are, say, 50%, 100% and 200% of the
presumed effective dose. Consider the simple case >>> only one factor , 2 sources of variation:
Between the group means
Within the groups
In the experimental situation, the animals should be randomly allocated to one of
the levels of the factor, i.e. to one of the groups, in order to avoid allocation bias
(see Section 5.6).
Assumptions:
results are reliable only if the assumptions on which it is based are satisfied
samples representing the levels are independent
Observations in each sample come from a Normally distributed population with
variance σ2; this implies that the group variances are the same. Approximate
Normality may be established by drawing a histogram; moderate departures
from Normality have little effect on the result.
Constant variance, the more important assumption, may be established by
Levene’s test
8.6.4 Example Dogs were fed a dry diet coated with different agents that were believed to
affect the build-up of calculus on the teeth. Calculus accumulation was
measured by an index that combined estimates of both the proportion of the teeth
covered by the deposit and the thickness of the deposit. Twenty-six dogs were
randomly allocated to three treatments: control, soluble pyrophosphate (P2O7)
and sodium hexametaphosphate (HMP). The calculus accumulation index was
measured on each dog 4 weeks after it received treatment.
Null hypothesis: mean calculus indices in the three treatment groups are equal;
the alternative hypothesis is that they are not all equal.
Test statistic: >>>>>F-ratio (F-value)
P-value = 0.005 >>>>Further examination by post hoc multiple comparisons
SPSS Table 8.1: 2 variables: Caculus
Group (1,2, 3)
Analyze >>>> Compare Means >>>> One-way ANOVA
(Unequal variances ANOVA) Welch ANOVA
Brown-Forsythe test
SPSS computer output for analysis of variance of calculus data on three groups of dogs.
Descriptives
(P = 0.44).>>>>> variances of the observations in the three groups are not
significantly different
Which group means Differs?????
Post-hoc Test
Multiple Comparison
Post-hoc testMultiple Comparisons of Means
Multiple comparisons
Conducting a number of tests, but the more tests that we perform, the more
likely it is that we will obtain a significant P-value on the basis of chance alone.
We have to approach this problem of multiple comparisons in such a way that
we avoid spurious P-values.
Adjusted p-values are simply the unadjusted p-values multiplied by the number of possible comparisons (six in this case);
If multiplying a p-value by the number of comparisons produces a value greater than one, the probability is given as 1.00.
Most Common Multiple comparisons
Least significant difference (LSD)
Duncan’s multiple range test, (DMRT)
Tukey’s (HSD)
Newman–Keuls tests,
Bonferroni’s correction
Scheffe’s
. Be aware: they often produce slightly different results!
Bonferroni's
Reporting results in table:
Automatic Multiple comparison using Bonferroni’s test.
Duncan Multiple range Test
ab
If the sample sizes are not large,
equal variances not assumed
non-parametric method,
kruskal wallis test
Data Transformation
Variance within groups (obtain homogenous Individuals (data values)) e.g.:
Titer Data (Virus – HI Titer). Bacterial Count
Skewness (Normalize Data)
Transforming Data Transform menu >> Compute Variable
Essential types of transformationLog transformation: (Log to the base 10)An Example. The bacterial count: 10, 100 and 1000 simple average (arithmetic mean) (10+100+1000 = 1110/3) is 370when converted to logs become 1, 2 and 3 >>>> average (1+2+3 = 6/3) is 2 or converted back >>>>100. Thus the geometric mean of 10, 100 and 1000 is 100 whereas the simple average (arithmetic mean) (10+100+1000 = 1110/3) is 370. Note that the high number of 1000 in the example skews the simple average upwards making it much higher than the geometric mean.
Enter the name of the target Variable for example Logvirus