Statistics Review for Scientists
July 22, 2026
Normal distribution
2.15%
µ − 3σ
13.6%
µ − 2σ
34.1%
µ−σ
34.1%
µ
13.6%
µ+σ
2.15%
µ + 2σ
µ + 3σ
68.2%
95.4%
99.7%
Normal distribution: X ∼ N (µ, σ 2 )
Center location: mean (µ), E[X ] = µ
Variability (spread): standard deviation (σ), variance (σ 2 ), Var[X ] = σ 2
The 68 − 95 − 99.7 rule: 68%, 95% or 99.7% of the observations fall within 1, 2 or 3
σ of the mean µ.
Statistics Review for Scientists
Statistics of normal distribution
Sample mean:
n
X̄ =
1X
X1 + . . . + Xi + . . . + Xn
Xi =
n i=1
n
X̄ ∼ N (µ,
σ2
)
n
Sample variance:
s2 =
n
1 X
(Xi − X̄ )2 ,
n − 1 i=1
(n − 1)s 2
∼ χ2n−1 ∗
σ2
s is sample standard deviation.
Standard error of the mean (SEM):
s
se = √
n
s → variability (spread) of observation X , fluctuates around σ;
se → variability (spread) of sample mean X̄ , n ↑⇒ se ↓.
∗ 2
χ distribution is defined in slide 18
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Standardized variable & confidence interval
standardized variable (subtract mean then divide standard deviation)
X ∼ N (µ, σ 2 ) ⇒
X −µ
∼ N (0, 1),
σ
X̄ ∼ N
σ2
X̄ − µ
µ,
⇒
√ ∼ N (0, 1)
n
σ/ n
density of N(0, 1)
Area =
α
2
Area = 1 − α
zα
= − z1− α
2
2
0
Area =
α
2
z1− α
2
zα : α percentile of N (0, 1), which means α% observations from N (0, 1) are less than
zα .
The 100(1 − α)% confidence interval (CI) for the mean µ of a normal population with
known variance σ 2 based on a random sample of size n is derived by:
X̄ −µ
√ < z1−α/2 ) = 1 − α
Pr(−z1−α/2 < σ/
n
⇒
⇒
Pr(X̄ − z1−α/2 √σn < µ < X̄ + z1−α/2 √σn ) = 1 − α
95% CI of µ : X̄ − z1−α/2 √σn , X̄ + z1−α/2 √σn
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A simulation on confidence interval
Sample
10 observations from N (0, 1) 100
times, then calculate the 95% CI for the group mean of 10 observations
by X̄ − z.975 √1 , X̄ + z.975 √1
, the results are shown in the following figure.
10
10
CI covers 0
CI doesn't cover 0
Point estimate
True mean
−1.5
−1.0
−0.5
0.0
0.5
1.0
1.5
There are 92 CIs cover the true mean 0. If sample more and more times, the percentage cover true mean 0 should
be more and more closer to 95%.
Typically, 95% CI is used to represent the likely interval of true µ.
Statistics Review for Scientists
Students t-distribution
When the population variance is unknown,
t≜
X̄ − µ
√ ∼ tn−1
s/ n
which is known as Student’s t-distribution, with n − 1 degrees of freedom (df).
df
tdf, 1−.05/2
z1−.05/2
2
5
10
20
4.3
2.57
2.23
2.09
1.96
1.96
1.96
1.96
N(0, 1)
t (df=2)
t (df=5)
t (df=10)
t (df=20)
df ↑⇒ center area ↑, tail area ↓ (more closer to normal distribution).
A 100(1 − α)% confidence interval for the mean µ of a normal population with
estimated standard deviation s, based on a random sample of size n is
s
X̄ − tn−1,1− α √ ,
2
n
s
X̄ + tn−1,1− α √
2
n
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Hypothesis testing, α level and p value
Null hypothesis H0 : X is from normal distribution N (0, 32 ); Alternative hypothesis
H1 : X is not from N (0, 32 ). Alternative hypothesis typically is associated with a
theory one would like to provide evidence to prove.
Observation: x = 6.5.
density of N(0, 32)
1.51%
observations < −6.5
1.51%
observations > 6.5
−6.5
0
6.5
Evidence (p-value): there is only 2 × 1.51% chance that observations from N (0, 32 )
are more extreme (far away from center 0) than 6.5. .0302(2 × .0151) is the p-value
(the probability of obtaining a test statistic at least as extreme as the one that was
actually observed, assuming that the null hypothesis is true).
Criterion (α level): Before see x, pre-specify a significance level α, (e.g., α = .05).
When p < α(.0302 < .05), we accept H1 and conclude x = 6.5 is not from N (0, 32 ).
(With α = .05, we will wrongly reject 5% N (0, 32 ) observations belong to N (0, 32 ).)
Statistics Review for Scientists
One sample inference for the population mean
Hypothesis
Null H0 : µ = µ0
µ0
region to conclude µ ≠ µ0
Type I error rate: α
Alternative
two-sided H1 : µ ̸= µ0
µ0
greater H1 : µ > µ0
region to conclude µ > µ0
Type I error rate: α
less H1 : µ < µ0
µ0
region to conclude µ < µ0
Type I error rate: α
Test statistic and sampling distribution under null
test statistic
∗
distribution
under null
σ known
√0
z = X̄σ/−µ
n
z ∼ N (0, 1)
σ unknown
√0
t = X̄s/−µ
n
t ∼ tn−1
∗
95% CI of µ
X̄ − z.975 √σn , X̄ + z.975 √σn
X̄ − tn−1,.975 √s n , X̄ + tn−1,.975 √s n
observations−expectation under null
test statistic is represented as: standardized distance from center ≜ mean ofspread
of mean of observations
Statistics Review for Scientists
Two sample inference for the population means
Linearity of normal distribution
If X ∼ N (µ, σ 2 ) and c is a constant, then cX ∼ N (cµ, c 2 σ 2 )
If 2 independent random variables X1 ∼ N (µ1 , σ12 ) and X2 ∼ N (µ2 , σ22 ) then
X1 ± X2 ∼ N (µ1 ± µ2 , σ12 + σ22 )
Hypothesis
Null H0 : µ1 = µ2
(µ1 − µ2 = 0)
Alternative
two-sided H1 : µ1 ̸= µ2
greater H1 : µ1 > µ2
less H1 : µ1 < µ2
(µ1 − µ2 ̸= 0)
(µ1 − µ2 > 0)
(µ1 − µ2 < 0)
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Two sample inference for the population means (cont.)
two sample two-sided inference for the population means
σ unknown
σ known
σ 2 estimate
equal variance
s 2 = n +n1 −2
1
2
ni
2 P
P
unequal variance
!∗
(Xij − X̄i )2
s12 = n 1−1
1
i=1 j=1
s22 = n 1−1
2
test statistic
z =
(X̄1 −X̄2 )−(µ1 −µ2 )
q
σ 2 /n1 +σ 2 /n2
1
2
t =
(X̄1 −X̄2 )−(µ1 −µ2 )
√
s 1/n1 +1/n2
t =
z ∼ N (0, 1)
t ∼ tn1 +n2 −2
ν =
95% CI of
∗
†
(X1i − X̄1 )2
(X2i − X̄2 )2
i=1
(X̄1 −X̄2 )−(µ1 −µ2 )
q
s 2 /n1 +s 2 /n2
1
2
t ∼ tν †
sampling
distribution
under null
µ1 − µ2
n1
P
i=1
n2
P
z.975
qX̄1 − X̄2 ±
σ12 /n1 + σ22 /n2
X̄1 − X̄2 ±
p
1/n1 + 1/n2
tn1 +n2 −2,.975 s
pooled sample variance
Welch’s approximation
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!2
s12 s22
+
n1 n2
(s 2 /n1 )2 (s 2 /n2 )2
1
+ 2
n1 −1
n2 −1
tν,.975
X̄
q1 − X̄2 ±
s12 /n1 + s22 /n2
T test case study (with big sample size)
c(0, 1)
−2
0
2
4
6
Raw data
group 1
c(−1, 1)
c(0, 1)
c(0, 1)
Descriptive
statistics
c(0, 1)
c(0, 1)
group 2
x2 (95% CI)
3.23 (2.45, 4.02)
c(−1, 1)
Hypothesis
testing
x1 (95% CI)
1.83 (0.96, 2.69)
c(−1, 1)
x1 − x2 (95% CI)
−1.41 (−2.49, −0.33)
p = .01
−4
c(−1, 1)
−2
0
2
4
c(−1, 1)
source data
exploratory data analysis
descriptive statistics
N
x̄(95%CI)
s
pooled s
Hypothesis testing (T test)
x̄1 − x̄2 (95%CI)
p-value
X1 ∼ N (2, 1)
X2 ∼ N (3.5, 1)
by checking raw data, it looks like both groups follow norm distribution, no obvious outliers
9
1.83 (0.96, 2.69)
1.13
8
3.23 (2.45, 4.02)
0.94
1.04
−1.41(−2.49, −0.33)
.01
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T test case study (with small sample size)
c(0, 1)
−2
0
2
4
6
Raw data
group 1
c(−1, 1)
c(0, 1)
c(0, 1)
Descriptive
statistics
c(0, 1)
c(0, 1)
group 2
x2 (95% CI)
3.27 (1.77, 4.76)
c(−1, 1)
Hypothesis
testing
x1 (95% CI)
2.20 (0.86, 3.53)
c(−1, 1)
x1 − x2 (95% CI)
−1.07 (−2.74, 0.59)
p = 0.18
−4
c(−1, 1)
−2
0
2
4
c(−1, 1)
source data
exploratory data analysis
descriptive statistics
N
x̄(95%CI)
s
pooled s
Hypothesis testing (T test)
x̄1 − x̄2 (95%CI)
p-value
X1 ∼ N (2, 1)
X2 ∼ N (3.5, 1)
by checking raw data, it looks like both groups follow norm distribution, no obvious outliers
5
2.20 (0.86, 3.53)
1.07
5
3.27 (1.77, 4.76)
1.20
1.14
−1.07(−2.74, 0.59)
.18
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T test case study (cont.)
Practical significance: should consider from practical point of view, is x̄1 − x̄2 a
meaningful difference given the observation variance and cost of intervention.
Statistical significance: to check whether 95% confidence interval (CI) of x̄1 − x̄2
covers 0 and p value is significant or not.
Sample size vs significance
sometimes, there is a practical but not statistical significant difference due to
small sample size.
sometimes, there is a statistical but not practical significant difference just dues
to large sample size.
need conduct study design to choose a right sample size to detect a prespecified
difference with certain power.
Statistics Review for Scientists
Paired t test
1.0
0.5
0.0
d = yPost − yPre
y
−1.5
−1.0
−0.5
−1.5
−1.0
−0.5
0.0
0.5
1.0
There is a pair of pre and post treatment measurement for every subject. The left side tests Ypost = Ypre vs.
Ypost ̸= Ypre . The right side first calculate d = Ypost − Ypre for each subject as treatment effect, then test d = 0
vs. d ̸= 0.
pre
post
T test
descriptive statistics
ȳ (95%CI)
s
pooled s
T test
ȳpost − ȳpre (95%CI)
p-value
Paired t test
ypre
ypost
.14 (-.56, .84)
.83
-.69 (-1.45, .07)
.91
d = ypost − ypre
.87
-.84 (-1.77, 0.10)
.076
descriptive statistics
¯
d(95%CI)
s
Paired t test
p-value
-.84 (-1.26, -0.41)
.51
.002
Generally, there is a subject effect (subject has high ypre tends to have high ypost ). By
eliminating this subject effect, paired t test is much easier (more powerful) to detect
treatment effect.
Statistics Review for Scientists
Sample size and power
accept X from N(µ1, σ2)
density of x1 ≈ N(µ1,
σ2
n
claim X from N(µ2, σ2)
)
density of x2 ≈ N(µ2,
area = β
µ1
σ2
n
)
area = α
z1−α
σ
n
xα
z1−β
σ
n
µ2
H0 : X ∼ N (µ1 , σ 2 ), H1 : X ∼ N (µ2 , σ 2 ) (µ1 < µ2 , 1-sided). With sample size n, we can have above study
design figure.
Type I error rate (α): probability of observations from N (µ1 , σ 2 ) be wrongly claimed into N (µ2 , σ 2 )
Type II error rate (β): probability of observations from N (µ2 , σ 2 ) be wrongly accepted into N (µ1 , σ 2 )
Power (1 − β): probability of observations from N (µ2 , σ 2 ) be correctly claimed into N (µ2 , σ 2 )
Sample size and power calculation (given µ1 , µ2 , σ and either β or n)∗

(z1−α +z1−β )σ 2



n =
µ2 −µ1
σ
µ2 − µ1 = z1−α + z1−β √ ⇒

n

√


(µ2 −µ1 ) n
z1−β =
− z1−α
σ
∗
for sample size calculation
for power calculation
z1−α should be changed to z1−α/2 for two-sided test
Statistics Review for Scientists
Binomial distribution, proportion estimation and comparison
A binomial variable X with distribution B(n, p) can be interpreted as the total number of ”successes” in n
independent and identical Bernoulli trials (two outcomes, 0:failure, 1:success with success probability p).
Pr(X = k) =
n!
k
p (1 − p)
k!(n − k)!
n−k
,
Var[X ] = np(1 − p)
E[X ] = np,
For large n,
X ∼ N (np, np(1 − p)),
p̂ =
x
n
∼N
p(1 − p)
p,
n
s

and 95% CI for p

s
p̂ − 1.96
p̂(1 − p̂)
n
, p̂ + 1.96
p̂(1 − p̂)
n
∗
For two proportions from independent samples with large sample sizes n1 , n2
Test statistic for Null H0 ( p1 = p2 = p):
z = r
p̂1 − p̂2
h
i,
p̂(1 − p̂) n1 + n1
1
p̂ =
n1 p̂1 + n2 p̂2
,
n 1 + n2
Z ∼ N (0, 1)
2
95% CI for proportion difference (use p̂1 , p̂2 instead of p̂ because not assume H0 is true):
s
p̂1 − p̂2 − 1.96
∗
p̂1 (1 − p̂1 )
n1
+
p̂2 (1 − p̂2 )
n2
s
, p̂1 − p̂2 + 1.96
p̂1 (1 − p̂1 )
n1
+
1.96 = z1−.05/2
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p̂2 (1 − p̂2 )
n2
!
Contingency table
Randomized trial/prospective study (the numbers of subjects in the treatment groups
are fixed). Physicians Health Study (New Engl. J. Med. 1998, 318:262-264): a 5-year
randomized study.
Placebo
Aspirin
Total
Myocardial Infarction
Yes
No
189
10,845
104
10,933
293
21,778
Total
11,034
11,037
22,071
Case-control study (the numbers of subjects in the outcome groups are fixed). One of
the first studies of smoking and lung cancer (Doll and Hill, British Med. J. Sep 30,
1950, pp. 739-748).
Smoker
Yes
No
Total
Lung Cancer
Cases
Controls
688
650
21
59
709
709
Total
1338
80
1418
Cross-sectional study (total number of subjects is fixed). Records of accidents in 1998
by the Department of Highway Safety and Motor Vehicles in Florida.
Seat belt
No
Yes
Total
Fatal
1601
510
2111
Injury
Nonfatal
162,527
412,368
574,895
Total
164,128
412,878
577,006
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Chi-square (χ2 ) test for contingency table
χ2 distribution: for n independent zi ∼ N (0, 1), n independent wi ∼ N (µi , 1) and let λ ≜
n
X
2
µi
i=1
n
X
2
2
zi ∼ χn
2
(χ distribution with df n),
i=1
n
X
2
2
wi ∼ χn (λ)
2
(noncentral χ distribution with df n)
i=1
Prospective study data can be summarized as
Event (observed (expected))
Yes
No
s)
f )
Control
s1 (s1′ = n1 n
f1 (f1′ = n1 n
f )
s)
f2 (f2′ = n2 n
Treatment
s2 (s2′ = n2 n
Total
s = s1 + s2
Total
n1
n2
f = f1 + f2
n = n1 + n2
Hypothesis: Null H0 : Event rates are the same (p1 = p2 = p), then p̂ = ns ; Alternative H1 : Event rates are
different (p1 ̸= p2 ).
Under H0 , expected numbers can be calculated with p̂ and subtotal numbers of n1 and n2 . The difference between
expected and observed numbers can be quantified by the following statistics
X
2
=
(s1 − s1′ )2
s1′
+
(s2 − s2′ )2
s2′
+
(f1 − f1′ )2
f1′
+
(f2 − f2′ )2
f2′
Under H0 , X 2 ∼ χ21 . Then X 2 > χ21 (1 − α) indicates there is significant difference between expected and
observed numbers, therefore concludes H1 .
For a contingency table with r rows and c columns
X
2
=
r X
c (Observed − Expected )2
X
ij
ij
i=1 j=1
Expectedij
2
∼ χ(r −1)·(c−1)
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Contingency table case study & definitions of relative risk and odds ratio
Prospective study example can be analyzed by 2 approaches:
χ2 test (see slide 18)
Control
Event (observed (expected))
Yes
No
189 (146.48)
10,845 (10,887.52)
Total
11,034
Treatment
104 (146.52)
10,933 (10,890.48)
11,037
293
21,778
22,071
Total
X
2
=
(189 − 146.48)2
146.48
+
=
+
(104 − 146.52)2
146.52
25.02 (p < .0001)
Proportion estimation and comparison
(see slide 16)
p̂c (95%CI)
p̂t (95%CI)
p̂(95%CI)
p̂c − p̂t (95%CI)
z
.017 (.015, .020)
.009 (.008, .011)
.013 (.012, .015)
.008 (.005, .011)
5.001 (p < .0001)
(10845 − 10887.52)2
2
10887.52
2
z = 5.001 = X
+
2 approaches have the same test result!
10890.48
Risk (R) and relative risk (RR) are defined as:
RTreatment ≜ Pr(Yes|Treatment),
RR ≜
RTreatment
RControl
≜
Pr(Yes|Treatment)
Pr(Yes|Control)
The odds and odds ratio (OR) are defined as:
OddsTreatment ≜
RTreatment
1 − RTreatment
2
(10933 − 10890.48)2
,
OR ≜
OddsTreatment
OddsControl
≜
RTreatment /(1 − RTreatment )
RControl /(1 − RControl )
Statistics Review for Scientists
Analysis of Variance (ANOVA)
Model (m groups and n observations per group):
Yij
=
Yij
=
µ
↑
Y ··
+
+
Yij : response j in group i,
τi
↑
(Y i· − Y ·· )
+
+
Y i· : average response in group i,
∗
iid
ϵij ∼ N (0, σ 2 )
i ∈ 1, . . . , m
j ∈ 1, . . . , n
ϵij
↑
(Yij − Y i· )
Y ·· : overall average response.
Hypothesis: H0 τ1 = · · · = τm = 0 versus H1 τi ̸= 0 for some i(s).
Sum of squares (SS) decomposition and ANOVA table
m X
n
X
2
(Yij − Y ·· )
i=1 j=1
=
=
|
{z
SST
}
m X
n
X
2
(Yij − Y i· + Y i· − Y ·· )
i=1 j=1
m X
n
m
X
2 X
2
(Yij − Y i· ) +
n(Y i· − Y ·· )
i=1 j=1
i=1
{z
}
|
{z
} |
SSB
SSW
One-way ANOVA Table
Source of
Variation
between
groups
SSB =
within
groups
SSW =
total
SST =
m
X
n(Y i· − Y ·· )
i=1
m X
n
X
2
(Yij − Y i· )
i=1 j=1
m X
n
X
(Yij − Y ·· )
i=1 j=1
∗
Degrees of
Freedom (df)
Sum of Square (SS)
2
2
Mean Square (MS)
m−1
MSB =
m(n − 1)
MSW =
F
SSB
MSB
m−1
MSW
SSW
m(n − 1)
mn − 1
= m − 1 + m(n − 1)
independent and identically distributed
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Analysis of Variance (ANOVA) (Cont.)
Let λ ≜
m
X
2
nτi , then
i=1
2
SSW
(
SSB
2
∼ χm(n−1) ,
σ2
E[MSB ] = σ +
χ2m−1
χ2m−1 (λ/σ 2 )
∼
σ2
λ
2
E[MSW ] = σ ,
m−1
if H0 ✓
if H1 ✓
,
⇒
MSB
MSW
= σ2
> σ2
(
∼
if H0 ✓
if H1 ✓
Fm−1,m(n−1) ∗
Fm−1,m(n−1) (λ/σ 2 )
if H0 ✓
if H1 ✓
Therefore, MSB /MSW is used to test H0 versus H1 .
Example
A
H0 is true
B
C
All
H1 is true
yij
yi.
y..
60
62
64
66
68
70
60
62
64
x
Source
between
groups
within
groups
Total
∗
66
68
70
x
SS
df
MS
F
p-value
2.97
2
1.48
.21
.81
84.40
12
7.03
87.37
14
Source
between
groups
within
groups
Total
SS
df
MS
F
p-value
113.2
2
56.6
8.05
.006
84.4
12
7.03
197.6
14
X /m
F distribution: if X1 ∼ χ2m is independent of X2 ∼ χ2n , then X1 /n ∼ Fm,n ; or if X1 ∼ χ2m (ν), then
2
X1 /m
∼ Fm,n (ν)
X2 /n
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Factorial Designs
Model (factors A and B have a and b levels respectively, and n replicates for each combination of factor level):
Yijk =
µ
+
τi
+
βj
+
(τ β)ij
+
ϵijk
↑
↑
↑
↑
Yijk = Y ··· + (Y i·· − Y ··· ) + (Y ·j· − Y ··· ) + (Y ij· − Y i·· − Y ·j· + Y ··· ) + (Yijk − Y ij· )
↑
iid
ϵijk ∼ N (0, σ 2 )
i ∈ 1, . . . , a, j ∈ 1, . . . , b
k ∈ 1, . . . , n
Hypothesis
Main effect A: H0 τ1 = · · · = τa = 0 versus H1 τi ̸= 0 for some i(s).
Main effect B: H0 β1 = · · · = βb = 0 versus H1 βj ̸= 0 for some j(s).
Interaction effect: H0 (τ β)ij = 0 for all i and j versus H1 (τ β)ij ̸= 0 for some i(s) and j(s).
Sum of squares (SS) and degree of freedom (df) decomposition
a
b
a X
b
a b
n
X
X
X
2
2
2
2 XX X
2
(Y i·· − Y ··· ) + an
(Y ·j· − Y ··· ) + n
(Y ij· − Y i·· − Y ·j· + Y ··· ) +
(Yijk − Y ij· )
(Yijk − Y ··· ) = bn
i=1
i=1 j=1
i=1 j=1 k=1
i=1 j=1 k=1
j=1
|
{z
}
|
{z
}
|
{z
} |
{z
} |
{z
}
SSA
SST
SSB
SSAB
SSE
a X
b X
n
X
abn − 1
(a − 1)
=
+
(b − 1)
(a − 1)(b − 1)
+
+
ab(n − 1)
Two-factor factorial ANOVA Table
Source of
Variation
Sum of Square (SS)
Degrees of
Freedom (df)
Mean Square (MS)
E[MS]
A factor
SSA
a−1
A
MSA = a−1
B factor
SSB
b−1
B
MSB = b−1
Interaction
SSAB
(a − 1)(b − 1)
AB
MSAB = (a−1)(b−1)
P
2
bn a
i=1 τi
σ2 +
a−1
P
2
an b
j=1 βj
σ2 +
b−1
P
Pb
2
n a
i=1
j=1 (τ β)ij
σ2 +
(a−1)(b−1)
SSE
MSE = ab(n−1)
σ2
Error
SSE
ab(n − 1)
total
SST
abn − 1
SS
SS
SS
Statistics Review for Scientists
F
MSA
MSE
MSB
MSE
MSAB
MSE
linear regression
95% prediction interval
for individuals
yi
Unexplained deviation
45
SXX ≜
n
X
2
(xi − x) ,
i=1
Total deviation
40
Explained deviation
SXX
y
α̂
β̂

∼N
α̂ = y − β̂x
,
α
β

1 + x2
n
SXX
−x
SXX
2
,σ 
−x
SXX
1
SXX
30
35
n
X
(xi − x)(yi − y )
i=1
SXY
β̂ =
y^i
Variance estimation
95% confidence interval
25
y
SXY ≜
RSS(α̂, β̂)
10
12
14
16
18
σ2
20
2
∼ χn−2 ,
2
σ̂ =
RSS(α̂, β̂) a
n−2
x
Sum of square decomposition:
Data: (x1 , y1 ), . . . , (xn , yn )
Model:
iid
yi = α + βxi + ϵi , ϵi ∼ N (0, σ 2 )
Least square estimation: chooses α̂ and β̂ to
make residual sum of squares (RSS)
minimum
i=1
i=1
Residual SS
i=1
P
i=1
2
(ŷ −y )
Reg SS
R2 = P i
2 = Total SS : the proportion
RSS(α,β)
(yi −y )
z
}|
{


n
X
2
(yi − α − βxi ) 
(α̂, β̂) = argmin 
α,β
Regression SS
Total SS
z
}|
{
z
}|
{ z
}|
{
n
n
n
X
X
X
2
2
2
(yi − y ) =
(ŷi − y ) +
(yi − ŷi )
of the variance of y that is explained by x.
a
2 degree of freedom lost due to 2 parameters
Statistics Review for Scientists


Data transformation
If data are not normally distributed you cannot use any of the tests that assume that
it is (e.g. ANOVA, t test, regression analysis). But it is often possible to normalize it
by transforming it.
Different transformations work for different data types:
Logarithms (log (x)): Growth rates, potency estimates from
concentration-response assays are often exponential and log transforms will often
normalize them. Ratio (fold change) also suggested to be log transformed.
Reciprocal (1/x) : Reciprocal transformation is often used for enzyme reaction
rate data.
√
Square root ( x): This transform is often of value when the data are counts,
e.g. blood cells on a haemocytometer or woodlice in a garden. Carrying out a
square root transform will convert data with a Poisson distribution to a normal
distribution.
√
Arcsine (sin−1 ( x)) : This transformation is especially useful for percentages
and proportions.
Statistics Review for Scientists
Index
Normal distribution (go to slide 2)
t-distribution (go to slide 6)
χ2 distribution (go to slide 18)
F distribution (go to slide 21)
Confidence interval (go to slide 4)
Hypothesis testing (go to slide 7)
α level (go to slide 7)
p value (go to slide 7)
Statistics Review for Scientists
Reference and useful web links
BMJ statistics notes series.
Karl Broman, Statistics for laboratory scientists.
Sample size and power freeware UIOWA, WISC
Statistics Review for Scientists