Normal Distribution. The normal distribution was first introduced by the French mathematician La Place (1749-1827). It is highly useful in the field of statistics. The graph of this distribution is called normal curve or bell - shaped curve. - PowerPoint PPT Presentation
Normal Distribution • The normal distribution was first introduced by the French mathematician La Place (1749-1827). • It is highly useful in the field of statistics. The graph of this distribution is called normal curve or bell - shaped curve. • In normal distribution, observations are more clusters around the mean. Normally almost half the observations lie above and half below the mean and all observations are symmetrically distributed on each side of the mean. • The normal distribution is symmetrical around a single peak so that mean median and mode will coincide. It is such a well-defined and simple shape, a great deal is known about it. The mean and standard
Transcript
Normal Distribution
• The normal distribution was first introduced by the French mathematician
La Place (1749-1827).
• It is highly useful in the field of statistics. The graph of this distribution is
called normal curve or bell-shaped curve.
• In normal distribution, observations are more clusters around the mean.
Normally almost half the observations lie above and half below the mean
and all observations are symmetrically distributed on each side of the
mean.
• The normal distribution is symmetrical around a single peak so that mean
median and mode will coincide. It is such a well-defined and simple shape,
a great deal is known about it. The mean and standard deviation are the
only two values we need to know o be able to describe a normal curve
completely.
Normal Distribution
Characteristics :
• The curve is symmetrical
• It is a bell shaped curve.
• Maximum values at the center and
decrease to zero symmetrically on each side
• Mean, median and mode coincide
Mean = Median = Mode
It is determined by mean and standard deviation.
Mean1SD limits, includes - 68% of all observations
Mean 2SD - ,, ,, - 95% ,, ,,
Mean 3SD - ,, ,, - 99% ,, ,,
Normal Distribution
Normal Distribution
• Almost all statistical tests (t-test, ANOVA etc)
assume normal distributions. These tests work
very well even if the distribution is only
approximately normally distributed.
• Some tests (Mann-whitney U test, Wilcoxon W
test etc) work well even with very wide deviations
from normality.
Group Distribution is normal Distribution is not normal
(Parametric tests) (Nonparametric tests)
Normal Distribution
One Mean±SD Median (Range)
Two t-test
Unpaired t-test
Paired t-test
Non-parametric t-test (or Rank Test)
The Mann-Whitney U test
The Wilcoxon Matched-Pairs Signed-
Ranks TestThree or more
The Kruskal-Wallis One-way ANOVA by Ranks
The Friedman’s Test
One Way ANOVA
The Repeated Measures ANOVA
Relationship between two variables
The Correlation Coefficient
Simple Linear Regression
The Spearman Rank Correlation Coefficient
Nonparametric Regression Analysis
Tests of Significance (testing hypothesis)
Definition:
A significance test is a statistical test (mathematical methods)
that estimates that the observed difference between two or
more groups is due to chance or not.
Statistical tests (such as t-test, F-ratio, and chi-square)
ultimately lead to a p-value which is used to make a
statement/conclusion about the statistical significance of the
results that is for example: a difference between means of
experimental and control groups are statistically
significantly differ or not.
The concept of statistical significance is based on the
assumption that events which occur rarely by chance
are "significant". Traditionally, events which occur less
than 5 out of 100 (or p –value <0.05) are said to be
statistically significant.
Methods of determining the significance of difference
(test of significance) are used to draw inference and
conclusions.
Based on distribution (whether normal distribution or
not) test of significance are two types. (1) Parametric test: t-test, ANOVA etc (2) Nonparametric test: Mannwhitney-U test,
Wilcoxon-W test, test etc
Tests of Significance
Hypothesis
A hypothesis is an assumption to be tested.
Hypothesis tests are widely used for making decisions i.e. very often we
make decision about population on the basis of sample information For example: we may decide on the basis of sample data whether a new
medicine is really effective in curing a disease.
Two types of hypothesis: Null hypothesis: H0 and alternative hypothesis: H1
Null hypothesis (Hypothesis of indifference): A hypothesis which states that
there is no difference between two samples. The null hypothesis is often
the ‘’straw man’’ that we wish to reject. Philosophers of science tell us
that ‘we never prove things conclusively; we can only disprove theories’.
Alternative hypothesis: is the opposite of null hypothesis (or any hypothesis
other than null hypothesis), i.e., there is difference between the two
samples.
Stages in performing a test of significance
The first step in hypothesis testing is to establish the
hypothesis to be tested.
State the null hypothesis of no difference and the alternative
before test of significant, eg. Vit A and D makes no
difference in growth or alternatively they play a positive or
significant role in promoting growth.
Choose the desired level of significance, denoted by ‘’
(usually = 0.05 or 0.01).
Choose a appropriate test of statistics (eg t test, ANOVA, chi- square etc) to test the null hypothesis.
Stages in performing a test of significance (continuation)
Determine p value, the probability of occurrence of the estimate by
chance, i.e., accept or reject the null hypothesis. If p is less than a
certain amount (by convention 5% or o.o5), we consider null
hypothesis to be unlikely and we reject it.
Draw conclusion on the basis of p value, either accepting null
hypothesis or rejecting it, i.e., decide whether the difference observed
is due to chance or play of some external factors on the sample under
study.
For example: H0: there is no association between personal hygiene
and occurrence of diarrhoea. After doing test of significance (chi-
square test) it was figured out that the calculated value (say 11.18)
was more than the tabulated value (say 3.84) indicating that p<0.05;
so the conclusion will be null hypothesis is rejected and alternative
hypothesis is accepted, i.e. there is association between personal
hygiene and occurrence of diarrhea.
P value
P value stands for probability of by chance.
P value is quoted widely in the medical
literature. They are scattered through articles
in medical research, clinical trials,
epidemiological studies, sample surveys and
laboratory work.
P values are standard device for reporting
quantitative results in research where
variability plays a large role. Statement like
“blood glucose level of type 1 diabetes is
significantly (p<0.05) higher than type 2
diabetes.
P value
P value is used to assess the degree of
dissimilarity between two or more groups of
observations. It indicates the probability of
occurrence of the difference by chance.
P value measures the chance of dissimilarity
between two or more set of measurements or
between one set of measurements and a
standard.
P = 0.05 means - the difference between two
different groups is significant at the level of
5%.
If p<0.05, the test hypothesis is rejected and
p>0.05 it is accepted.
P value
“P value is the probability of being
wrong when asserting that a true
difference exist between the
different treatment groups.”
If p<0.05, the result is regarded as statistically
significant If p<0.01, ,, ,, ,, moderately ,, If p<0.001 highly ,,
If p> 0.10, the result is regarded as no
statistically significant
Significance of Difference in Two Means
If the statistics observed in the two means, say and , the
question refer to whether they differ significantly or the
difference found is due to chance error of sampling.
If on the basis of a test, we find that the difference is too large
to attribute to chance errors, we will conclude that the
sample means are unequal or statistically significant
different.
1x 2x
Significance of Difference in Two Means
Depending on the size of sample the test of
significance of difference in two means are
two types: z-test (for large samples, >30) t-test (for small samples).
Z-test (for large samples, >30)
To test the difference between two population means, the test statistic is
Where and are variances of two samples
Example:
The haemoglobin (Hb) level of children was measured in
143 girls and 127 boys. The results were as follows:
Number (n) Mean (Hb%) SD
Girls 143 11.2 1.4
Boys 127 11.0 1.3
2
22
1
21
21 )(
ns
ns
xxz
22s
21s
The test procedure is as follows:
1. State the null hypothesis: mean Hb is same in girls as
well as in boys.
2. Find z value using mean difference (11.2 – 11.0 = 0.2),
, , n1 and n2.
3. Refer the z value to find the probability of the observed
difference by chance (p value) form table.
If p-value is large, we can not reject the null
hypothesis, so we have to conclude that these data give
no evidence of a real sex difference.
21s
22s
Z-test (for large samples, >30)
Here n1 = 143, =11.2, s1 = 1.4
n2 = 127, = 11.0, s2 = 1.3
Therefore, = 1.22
Since z<1.96, p>0.05, the difference is not
statistically significant, i.e. it could easily have
a. the case in which variances are equal, i.e., 12 = 2
2
b. the case in which variances are not equal, i.e., 12 =
22
t-test (for small samples, <30)
Unpaired t-test (in case of equal variance )
When population variances (though unknown) can be
assumed equal then the appropriate test statistic is
Student’s t defined by
Where
21
21
11
)(
nns
xxt
With degree of freedom = n1+n2-2
2
)()(
2
))1()1(
21
222
211
21
222
211
nn
xxxx
nn
snsns
Degree of freedom = freedom of choice
Unpaired t-test (in case of equal variance):
ExampleTwo different types of drugs A and B were tried on certain patients
for increasing weight, 5 persons given drug A and 7 persons given
drug B. Do the drugs differ significantly with regard to their effect
in increasing weight? Null hypothesis: H0: there is no difference in the efficacy of
the two drugs.Sl. No. x1 x2
1 8 -1 1 10 0 0
2 12 +3 9 8 -2 4
3 13 +4 16 12 +2 4
4 9 0 0 15 +5 25
5 3 -6 36 6 -4 16
6 8 -2 47 11 +1 1
Total 45 0 62 70 0 54
)( 11 xx 211 )( xx )( 22 xx 2
22 )( xx
95/451 x 107/702 x
5.0708.1406.3
1
7
1
5
1406.3
109
11
)(
21
21
nns
xxt
d.f. = n1+n2-2 = 5+7-2 = 10
406.375
5462
2
)()(
21
222
211
nn
xxxxs
Unpaired t-test (in case of equal variance):
Example
Interpretation:
The calculated value (0.5) of ‘t’ is less than
the table value (1.81), ie, p>0.05.
Therefore, our hypothesis is accepted.
Hence, we conclude that there is no
significant difference in the efficacy of the
two drugs in case of increasing weight.
t-test (for small samples, <30)
Unpaired t-test (in case of unequal variance )
Sometime it is not reasonable to assume equality of
the population variance and hence we cannot apply
normal test as well because of the smallness of the
sample size. In such case, we apply t-test, defined as
follows:
Where degree of freedom is
2
22
1
21
21 )(
ns
ns
xxt
1)/(
1)/(
.
2
22
2
1
12
1
2
2
22
1
21
nns
nns
ns
ns
fd
t-test (for small samples, <30)
Paired t-test:
Paired t-test is applied for paired data. Typically
paired data refers to repeated or multiple
measurements on the same subjects.
Application: Paired t-test is used to compare
difference between means of paired samples, e.g.,
cross over trials, different time in the same group
(‘’before-after studies’’ and ‘’twin studies’’)
Example: To compare a measurement of the level of
pain before and after administration of an analgesic
drug.
Criteria:1. The data are quantitative 2. Normal distribution 3. paired data
t-test (for small samples, <30)
Paired t –test is defined by
Where = mean of difference, SE ( ) = SE of the mean
difference.
n
SDd
dSE
dt
)(
d
1
)( 2
n
ddSD
d
Paired t-test (Example)
Problem: Antihypertensive effect of a drug was tested on 15 individuals. The recording of diastolic blood pressure are shown in the table
Interpret: The data justifies application of paired t-test.
Step 1: Null hypothesis: There is no difference in the diastolic B.P. before and after the drug treatment. The observed difference is due to sampling variation.
Alternative hypothesis: The observed difference is due to drug.
Step 2: Calculate mean difference in the pairs.
Step 3: Calculate standard deviation (SD) and standard error
Step 4: Calculate t:
Step 5: calculate degree of freedom (df): [df = n-1]Step 6: find out probability (p value)Step 7: Interpretation.
n
SDd
dSE
dt
)(
Paired t-test (Example)
S.No. BP (before drug treatment)
BP (after drug treatment) Difference (d) d - (d - )2
1 96 90 6 -7.6 57.76
2 98 92 6 -7.6 57.76
3 110 100 10 -3.6 12.96
4 112 100 12 -1.6 2.56
5 118 98 20 6.4 40.96
6 120 100 20 6.4 40.96
7 140 100 40 26.4 696.96
8 102 90 12 -1.6 2.56
9 98 88 10 -3.6 12.96
10 124 126 -2 -15.6 243.36
11 118 120 -2 -15.6 243.36
12 120 100 20 6.4 40.36
13 122 100 22 8.4 70.56
14 120 98 22 8.4 70.56
15 98 90 8 -5.6 31.36
Total 204 = 1625.6
Mean =13.6
Problem: Antihypertensive effect of a drug was tested on 15 individuals. The recording of diastolic blood pressure are shown in the table
dd
d 2)( dd
Paired t-test (Example)
n
SDd
dSE
dt
)(
13.6d
77.1011.116115
6.1625
1
)( 2
n
ddSD
89.478.2
6.13t d.f. = n – 1 = 15 – 1 = 14
Step 7: Interpret: The table value of t at 5% level of significance for 14 d.f. is 1.76. Computed value is 4.89.Since the computed value (4.89) is much more higher than the table value (1.76) at 5%, ie, the p value is less than 0.05. Hence, null hypothesis is rejected, alternative hypothesis is accepted, and the difference in before & after values is considered statistically significant.
Sampling errors: Type I & Type II error
Type I error:
A Type I error occurs when the null hypothesis
is rejected when, in fact, it is true.
Type II error:
A Type II error occurs when the null
hypothesis is not rejected when it is false.
Sampling error:
Sampling error are errors due to chance and
concern incorrect rejection or acceptance of a
null hypothesis. It depends on selection of
level of hypothesis.
Why not just use the t-test to compare more than 2 means?
The t-test tells us if the variation between two
groups is "significant", But how do we interpret significant variation
among the means of 3 or more groups? Why
not just do t-tests for all the pairs of means.
Multiple t-tests are not the answer because as
the number of groups grows, the number of
needed pair comparisons grows quickly. For 7
groups there are 21 pairs [No of paired to be
tested = k(k-1)/2, k = no. of groups].
Why not just use the t-test to compare more than 2 means?
When performing k multiple independent significance tests each at the
level, the probability of making at least one Type 1 error (rejecting the
null hypothesis inappropriately) is 1-(1-)k.
For example, with k (pairs of comparisons) =10 and (level of
significance) = 0.05, there is a 40% chance of at least one of the ten tests
being declared significant under the null hypothesis. So, when there is a
significant result among the ten tests, there is 40% chance that something
will turn out significant, ie, group-wise Type 1 error rate is actually 40% -
a far cry from the 5% that may have thought it was.
The difficulty here is that multiple significance testing (by
t-test) gives a high probability of finding a significant
difference just by chance.
Why not just use the t-test to compare more than 2 means?
ANOVA puts all the data into one number (F value) and
gives us one P for the null hypothesis and controlling type
1 error rate at no more than 5%.
= 1-(1-)k.
cp
05.0
0.05
0.017
0.008
0.005
0.003
0.002
0.0017
0.0013
0.0011
ANOVA (Aanalysis of Variance)
ANOVA or `F’ test used to analyze the variance for judging
the significance of more than 2 means. ANOVA replace the
multiple t- test with a single ‘F’ test.
Hypothesis: All the groups have a common population
mean or do the groups have equal means.
22
21
s
sFor
groupsthewithinVariance
groupsthebetweenVarianceF
The calculated value of F is compared with the table value
of F for the given degree of freedom at a certain critical
level. If the calculated value of F is greater than the table
value of F, it indicates that the difference in sample means
is significant or, in the other words the samples do not
come from the same population. On the other hand, if the
calculated value of F is less than the table value, the
difference is not significant and hence could have arisen
due to fluctuations of random sampling.
ANOVA -Calculation
1. Calculation of the variance between the groups:
For calculating variance between groups, we take the sum
of the square of the deviations of the means of the
various groups from the grand grand mean and divide
this sum by the degree of freedom.
Steps:
a. Calculate the mean of each group, eg.
b. Calculate the grand mean
c. Take the difference between the means of the various
groups and grand mean.
d. Square these deviations and obtain the sum which will
give the sum of square between the groups and divide
this by the degree of freedom (df = No. of groups – 1).
3X&
2X,
1X
X
)3(N3
X2
X1
XX
N = number of groups
ANOVA -Calculation
2. Calculation of the variance within the groups:
For calculating variance within the groups, we take the
sum of the square of the deviations of various
observations from the mean of the respective group and
divide this sum by the degree of freedom (number of
observation in each group).
Steps:
a. Calculate the mean of each group, eg.
b. Take the deviations of the various observation from the
means of the respective group.
c. Square these deviations and obtain the sum which gives
the sum of square within the groups and divide this by
the degree of freedom (df = No of observation of each
group – 1)
d. Add all these variances and divide by the number of
groups/variance.
3. Calculate the F-ratio:
3X&
2X,
1X
22
21
s
sFor
groupsthewithinVariance
groupsthebetweenVarianceF
Variance is computed as the sum of squared deviations from
the overall mean, divided by n-1 (sample size minus one, ie.,
degree of freedom).
Thus given a certain n, the variance is a function of the sums
of (deviation) squares, or SS for short.
Calculation of degree of freedom (df):df for between groups = Number of groups minus one = column -1 =
(c-1)
df for within groups = Number of cases in each group minus one
and added, (1st group – 1) + (2nd group – 1) + (3rd group – 1) = total
number – column = (n-c)
1NN
2)x(2x
1N
2)xx(Variance
22
21s
sFor
groupsthewithinsquaresofsumMean
groupsthebetweensquaresofsumMean
groupsthewithinVariance
groupsthebetweenVarianceF
ANOVA –Calculation (Alternative way)
Df Sum of squares Mean sum of squares F(variance) (MS)
Between groups (c – 1) SSB SSB / (c-1) Within groups (n – c) SSW SSW / (n-c)
MSW
MSBF
ANOVA - Exercise
Problem: Three different treatment are given to 3 groups of patients with
anemia. Increase in Hb% level was noted after one month and is given below.
Find whether the difference in improvement in 3 groups is significant or not.
Group A Group B Group C Group A Group B Group C
3 3 3 9 9 9
1 2 4 1 4 16
2 2 5 4 4 25
0 3 4 0 9 16
1 1 2 1 1 4
2 3 2 4 9 4
2 2 4 4 4 16
1x 2x 3x 21x
22x
23x
= 11+16+24 = 51
23
22
21
2 xxxX = 23+40+90 = 153
321T xxxX N = 7+7+7= 21
111 x 162 x 243 x 23x 21 402
2 x 9023 x
ANOVA - Exercise
Steps:
1. Calculation of total sum of squares (SST) (for all values)
2. Calculation of ‘between’ sum of square (SSB) (ie, between group)
(T= Total)
3. Calculation of ‘within’ sum of squares (SSW) (ie, within group)SSw = SST – SSB = 29.14 - 12.28 = 16.86
14.2986.12315321
)51(153
2)(22
N
XXSS T
T
12.2886.123)29.6257.3628.17(21
2)51(
7
2)24(
7
2)16(
7
2)11(
2)()()()(
3
23
2
22
1
21
N
X
n
x
n
x
n
xB
SS T
df Sum of squares Mean sum of squares F(variance)
Between groups 2 (3-1) 12.28 6.14Within groups 18 [(7-1)+ (7-1)+ (7-1)] 16.86 0.94
6.53
(N= total numbers in all groups)
ANOVA - Excercise
Df Sum of squares Mean sum of squares F
Between groups 2 12.28 6.14 6.53Within groups 18 16.86 0.94
Interpretation:
F- value in the table at 2, 18 df (n1=2, n2=18) is 3.55 (at 5% significance level).
Our calculated value is 6.53, which is more than table value (3.55). So it is significant
Ordinary ANOVA compares all groups and confirms difference among the groups but it does not tell us which group differs from which others.
Multiple comparison methods (eg Bonferroni, Dunnett, Duncan’s Tukey tests etc.) solve this problem and control over all type 1 error rate at no more than 5%. They are also known as post-hoc tests.
We can find all the pair wise difference among the group means and test them individually for significance. Basically they adjust the p value for declaring statistical significance.
Bonferroni test
It is based on the the t-statistic and thus can be considered
a form of t-test.
Bonferroni test uses the same criterion of setting the alpha
error rate to the experiment-wise error rate (usually =
0.05) divided by the total number of comparisons to control
for Type I error when multiple comparisons are being made.
Where, = 0.05 and k = number of comparisons = 10. So if
we apply a significant level of 0.005 to each of the 10
tests/comparisons, there is now only a 5% chance that any
of them will be declared significant under null hypothesis.
This modification is referred to as the Bonferroni correction.
k
Bonferroni test
The t test for treatment means following analysis of variance is given by
= the within-treatments variance
= the mean of treatment group number
i
= the number of individuals in treatment
group number i
21
2
21
11
nns
xxt
w
2ws
ix
in
Bonferroni test
For small number of comparisons (say up to 5)
its use is reasonable, but for large numbers it is
