#### Transcript Lecture 1

```Lecture 1
Interpretation of data
1.1 A study in anorexia nervosa
1.2 Testing the difference between the samples
1.3 Confidence intervals for treatment effects
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1.1 A study in anorexia nervosa
“Sufferers from anorexia nervosa, even those whose bodies
have become severely emaciated, often maintain that their
bodily dimensions are quite normal”
 Are anorexics able to judge their own bodily dimensions?
 Are they worse at doing so than healthy controls?
8 anorexics and 11 controls participated in a study of these
questions
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The apparatus
Two lights on a horizontal beam
Move them together:
“Say stop when the distance apart is the same as the width of
Repeat while they move apart, and then average the two
measurements to give the perceived width
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Body perception index
perceived width
BPI 
100
actual width
Let
mA = mean BPI for anorexics
mC = mean BPI for controls
Null hypothesis is H0: mA = mC
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The data
Anorexics: 130, 194, 160, 120, 152, 144, 120, 141
Controls: 202, 140, 168, 160, 147, 133, 229, 172, 130, 206, 153
Summary:
Overall (n = 19):
mean = 157.95, standard deviation = 30.689
Anorexics (nA = 8):
mean = 145.13, standard deviation = 24.398
Controls (nC = 11):
mean = 167.27, standard deviation = 32.426
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Formulae
mean:
x1  ...  x n
x
n
standard deviation (a measure of the spread of the data):
S
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 x1  x 
2
 ...   x n  x 
2
n 1
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Notes
 Here we have means x A for anorexics and x C for controls
and standard deviations SA for anorexics and SC for controls
 These are sample means and sample standard deviations:
they vary from sample to sample
 The population means are mA for anorexics and mC for
controls and the population standard deviations are sA for
anorexics and sC for controls: these are fixed truths that will
never be known precisely
 x A and x C are estimates of mA and mC
SA and SC are estimates of sA and sC
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1.2 Testing the difference between the samples
The two group means are different from one another
Are they significantly different?
Or might the difference just be due to chance?
We will use a t-test to find out
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The t-statistic
t
xA  xC
 1
1 
S 


n
n
C 
 A
where
S
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 n A  1 S2A   n C  1 SC2
nA  nC  2
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Notes
 We begin with an estimate of the difference between the
means: x A  x C
 This is standardised by dividing by S: S2 is a weighted
average of S2A and SC2
 Standardisation ensures that t is unit-free


 Division by  1  1  is a matter of convention,
 nA nC 
but it does ensure that values are not too greatly affected by
sample sizes
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Calculation
7  24.3982  10  32.4262
S
 29.377
17
so that
t
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145.13  167.27
1 1 
29.377   
 8 11 
 1.622
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Theory
Suppose that
 the BPIs of anorexics follow the
normal distribution with mean mA
and standard deviation s
 the BPIs of controls follow the
normal distribution with mean mC
and the same standard deviation s
Then, if mA = mC, the statistic t follows Student’s t-distribution
on 17 degrees of freedom (17 = 19 – 2 = n  # parameters)
 a similar shape (slightly fatter) centred on 0
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The t-distribution
The probability that a random variable following Student’s
t-distribution on 17 degrees of freedom is  1.627 is 0.061
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Interpretation
If the null hypothesis H0: mA = mC is true (and the populations
have the same standard deviation), then t is unusually negative
The chance of it being so negative (or even more so) is 0.061
This is the p-value against the one-sided alternative H1: mA < mC
The value 0.061 is not so small that one would wish to reject H0
and conclude that there is a significant difference – it shows a
trend, but does not constitute strong evidence
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Caution!
The investigators sought evidence that anorexics had a poorer
perception of their bodily dimensions than controls
– that mA > mC
The trend is in the opposite direction!
“So maybe the anorexics have a better perception, being so
obsessed by their bodies”
Investigators are going to wish to interpret the data, whichever
direction the difference, so use a two-sided p-value
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Two-sided p-value
Double the one-sided p-value to give the two-sided p-value:
p = 0.122
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Convention
 A two-sided p-value  0.05 is usually taken to represent
strong evidence of an effect
 This goes back to Fisher in the 1930s
 It is rather arbitrary, but it is a useful yardstick
 A one-sided p-value  0.025 is usually taken to represent
strong evidence of an effect – this avoids “cheating” by
choosing the direction of the difference once the data have
been observed
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1.3 Confidence intervals for treatment effects
We have used mA to denote the population mean of the BPIs
for anorexics and sA to denote their population standard
deviations
These are estimated by the sample mean x A = 145.13 and by
the sample standard deviation SA = 24.398 respectively
How good an estimate of mA is 145.13?
How big or small might mA actually be?
A confidence interval will answer this question
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Another t-distributed random variable
Let
tA
x A  mA 


nA
SA
Note that you cannot calculate tA as it depends on the unknown mA
If the BPI observations are normally distributed, then tA follows
Student’s t-distribution with (nA – 1) df
Now, a t7 random variable lies between 2.365 and 2.635 with
probability 0.95
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A confidence interval for mA
It follows that, with probability 0.95,
x A  mA 

2.365 
SA
nA
 2.365
which is
2.365 SA
n A  x A  m A  2.365 SA
nA
which is
x A  2.365 SA
n A  m A  x A  2.365 SA
nA
which is
145.13  2.365 24.398
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8  m A  145.13  2.365 24.398
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A confidence interval for mA
So, with probability 0.95,
124.73  mA  165.53
We say that (124.73, 165.53) is a 95% confidence interval for mA
The upper and lower limits are random, while mA is fixed
The limits capture the true value of mA with probability 0.95
It could well be that the true mean BPI for anorexics is as low as
124.73, it could also be as high as 165.53
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A confidence interval for mC
For the controls, nC = 11, x C = 167.27 and SC = 32.426
The 97.5% point of the t distribution on 10 df is 2.228
Hence, the 95% confidence interval for mC is
(167.27  2.228 32.426/11)  (145.49, 189.05)
Note that the confidence intervals for mA and mC overlap
What about a 95% confidence interval for d = mA  mC?
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A confidence interval for d = mA  mC
Now
td 
 xA  xC   d
 1
1 
S 


n
n
C 
 A
follows Student’s t-distribution with (n – 2) df
The 97.5% point of the t distribution on 17 df is 2.110
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A confidence interval for d = mA  mC
It follows that, with probability 0.95
 1
 1
1 
1 
 x A  x C   2.110 S     d   x A  x C   2.110 S   
 nA nC 
 nA nC 
so that the 95% confidence interval for d is

 1
1 
  x A  x C   2.110  S 




nA nC  




1 1 
  22.14  2.110  29.377    


8
11




  50.94,6.66 
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Interpretation
0 lies within the confidence interval
 consistent with lack of significant evidence against
H0: d = 0 at the 5% level (2-sided), as found from
the t-test
The mean BPI for anorexics could be substantially lower that that
for controls (by more than 50), or slightly higher
Larger sample sizes would reduce the width of the confidence
intervals, and make it easier to determine whether there really is a
difference
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