FORMULAS FOR THE 4TH EXAM
One-sided (One-tailed) test:
Lower tailed (Left-sided)
Upper tailed (Right-sided)
H0: population characteristics  claimed constant value H0: population characteristics  claimed constant value
Ha: population characteristics < claimed constant value Ha: population characteristics > claimed constant value
Two-sided (Two-tailed) test: H0: population characteristics = claimed constant value
Ha: population characteristics  claimed constant value
Hypothesis testing for:
A.
Independent Samples
I. Population characteristics: Difference between two population means, 1-2.
0 is the claimed constant.
1 and  2
_
sample means for X's and Y's, respectively. m and n are the sample sizes for X's and Y's, respectively.
the population variances for X's and Y's, respectively.
Assumption
both popn. distributions are normal and
are known
 12 ,  22
 ,
2
2
both popn. distributions are normal and at least one
sample size is small where unknown
 ,
2
1
2
2
are
assumed to be different (    ) and degrees of
2
1
2
2
2
 s12 s 22 
  
m n
freedom, v 
2
2
s12 / m
s 22 / n

m 1
n 1

 
are
100(1-)% confidence
Interval
_ _
 x y    0

z
 12  22
_ _
x

y

z


  /2
m
n



 22
n


 x y    0

z
2
s1 s 22

m n
_ _
 x y    0

t
2
s1 s 22

m n
s2 s2
_ _
 x  y   z / 2 1  2
m n


_ _
 x y    0

,
t
1 1
sp

m n
(m  1) s12  (n  1) s 22
2
sp 
mn2
1 1
_ _

 x  y   t / 2 ;v s p
m n


_
_
s12 s 22
_ _

 x  y   t / 2 ; v
m n



both popn. distributions are normal and at least one
sample size is small where unknown
 12 ,  22
are
assumed to be the same (  1 =  2 ) and degrees of
2
 12 and  22
Test statistics
m
2
1
y are the
s12 and s 22 are the sample variances for X's and Y's, respectively.
 12
large sample size (m>40 and n>40) and
are unknown
_
are the population means for X's and Y's, respectively. x and
2
freedom, v  m  n  2 is used to look up the
critical values.
II.
Population characteristics: Difference between two population proportions, p1-p2.
^
^
p0 is the claimed constant. p 1 and p 2 are the sample proportions for X's and Y's, respectively.
p1 and p 2 are the
population proportions for X's and Y's, respectively. m and n are the large sample sizes for X's and Y's, respectively. The
^
estimator for p is p 
X Y
m ^
n ^

p1 
p2 .
mn mn
mn


^
^
Test statistics:


100(1-)% large sample confidence Interval:  p1  p 2   z / 2
z
^
^

p

p
 1
2   p0


,
^
^

 1 1 
p1  p   

 m n 
^
^
^
^




p1 1  p1  p 2 1  p 2 

 

m
n
 12 /  22 or standard deviations,  1 /  2 .
 12 and  22 are the population variances for X's and Y's,
III. Population characteristics: Ratio of the two population variances,
X and Y's are random sample from a normal distribution.
2
respectively. s1 and
Y's, respectively.
Test statistics:
B.
F
s 22 are the sample variances for X's and Y's, respectively. m and n are the sample sizes for X's and
s12 / s 22
 12
s12 / s 22
s12
2
2


.
100(1-)%
confidence
Interval
for
:

/

1
2
F / 2;m 1,n 1  22 F1 / 2;m 1,n 1
s 22
Dependent Samples- Paired Data
Population characteristics: Difference between two population means, D =1-2.
0 is the claimed constant. Assumption: the difference distribution should be normal.
_
Test statistics: t 
d  0
sD / n
_
where D=X-Y and d and
s D are the corresponding sample average and the standard
deviation of D. Both X and Y must have n observations. The degrees of freedom to look at the table is v=n-1
_
100(1-)% confidence Intervals with the same assumptions: d  t / 2;n 1
sD
n
Decision can be made in one of the two ways in Parts I, II, and III for comparing two populations:
a. Let z* or t* be the computed test statistic values.
if test statistics is z if test statistics is t
Lower tailed test P-value = P(z<z*)
P-value = P(t<t*)
Upper tailed test P-value = P(z>z*)
P-value = P(t>t*)
Two-tailed test
P-value = 2P(z>|z*|) P-value = 2P(t > |t*| )
In each case, you can reject H0 if P-value   and fail to reject H0 (accept H0) if P-value > 
Rejection region for level  test:
test statistics is z
test statistics is t
Lower tailed test z  -z
t  -t;v
Upper tailed test z  z
t  t;v
Two- tailed test
z  -z/2 or z  z/2 t  -t/2;v or t  t/2;v
Do not forget that, F1-/2;m-1,n-1 = 1 / F/2;n-1,m-1 for the F-table
b.
Model:
X ij   i   ij
treatment).
X ij : observations ,
or
X ij     i   ij
 i : ith treatment mean,
test statistics is F
F  F1-;m-1,n-1
F  F;m-1,n-1
F  F1-/2;m-1,n-1 or F  F/2;m-1,n-1
Single Factor ANOVA :
where i=1,...,I (number of treatments), j=1,...,J (number of observations in each
 i   i   : ith treatment effect.
 ij : errors which are normally distributed with mean, 0 and the constant variance,  2 .
Assumptions:
X ij 's are independent (  ij 's are independent).  ij 's are normally distributed with mean, 0 and the constant variance,
 2 . X ij 's are normally distributed with mean,  i
Hypothesis:
Or
H 0 : 1  ...   I 1   I
H0 :i  0
for all i versus
versus
and the constant variance,
2.
H a : at least one  i   j
for
i j
where
 i .and  j 's are treatment means.
H a :  i  0 for at least one i where  i is the ith treatment effect.
I
Analysis of Variance Table:
n   Ji
and j=1,…,Ji . n is used when different number of samples obtained from each population.
i 1
Source
Treatments
Error
df
SS
MS
F
Prob > F
I-1
SSTr
MSTr = SStr / dftreatment
MSTr / MSE
P-value
I(J-1)
SSE
MSE = SSE / dferror
or n-I
Total
IJ-1 or SSTotal
n-I
where df is the degrees of freedom, SS is the sum of squares, MS is the mean square. Reject H 0 if the P-value  or if the test statistics
F > F;I-1,error df. If you reject the null hypothesis, you need to use multiple comparison test such as Tukey-Kramer
Confidence Interval for
   ci  i
:
c
MSE  ci2
_
i
x i  t / 2; I ( J 1)
J
i
i
Simple Linear Regression and Correlation
PEARSON’S CORRELATION COEFFICIENT (r) :measures
X and Y must be numerical variables,
H 0 :  XY  0
(the true correlation is zero) versus
The formal test for the correlation has the test statistics
the strength & direction of the linear relationship between X and Y.
H a :  XY  0
t
r n2
1 r
(the true correlation is not zero).
with n-2 degrees of freedom.
2
Minitab gives you the following output for simple linear regression:
parameters
0
and
Predictor (y)
Coef
SE Coef
T
p-value for
Ha : 0  0
p-value for
H a : 1  0
Constant
b0
sb0
Independent(x)
b1
s b1
b1
s b1
Analysis of Variance
Source
DF
Residual Error
Total
where n observations included, the
 1 are constants whose "true" values are unknown and must be estimated from the data.
b0
s b0
Regression
Y   0  1 x  e ,
SS
P
MS
F
1
SSR
MSR=SSR/1
n-2
n-1
SSE
SST
MSE=SSE/(n-2)
MSR/MSE
P
p-value for
H a : 1  0
Coefficient of Determination, R2 : Measure what percent of Y's variation is explained by the X variables via the regression model. It
tells us the proportion of SST that is explained by the fitted equation. Note that SSE is the proportion of SST that is not explained by the
model.
SSR
SSE
 1
.
SST
SST
H 0 : 1  10
R2 
H a : 1  10
Only in simple linear regression,
R 2  r 2 where r is the Pearson’s correlation coefficient.
Test statistics:
t
b1   10
s b1
where
 10 is the value slope is compared with.
Decision making can be done by using error degrees of freedom for any other t test we have discussed before (either using the P-value or
the rejection region method).
The 100(1-)% confidence interval for
 1 is b1  t / 2;df sb
1