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Transcript
```Formulas for the Final Exam
y
y=∑
∑ ( y − y )2
∑ ( x − x )2
x
x=∑
sy =
sx =
n
n −1
n
n −1
∑ zxz y z = x − x z = y − y
r=
x
y
n −1
sx
sy
sy
b=r
a = y − bx yˆ = a + bx residual = y − yˆ
sx
Sampling Distribution of p̂ :
Sampling Distribution of y :
Mean: p
Mean: μ
Standard Deviation: SD( pˆ ) =
p(1 − p )
n
Standard Error: SE( y ) =
Single sample (Categorical Variable)
Confidence interval for p :
pˆ − z *
pˆ (1 − pˆ )
to pˆ + z *
n
pˆ (1 − pˆ )
n
pˆ − po
p0 (1 − p0 )
n
z=
Test Statistic:
s
s
to y + t *
n
n
t=
y − μo
s
n
Two independent samples
Confidence interval for μ1 − μ2 :
( y1 − y2 ) − t *
df = n – 1
Test Statistic:
s12 s22
s2 s2
+
to ( y1 − y2 ) − t * 1 + 2
n1 n2
n1 n2
Paired samples
Confidence interval for μd :
s
s
d − t * d to d + t * d
nd
nd
n
Test Statistic:
Single sample (Numerical Variable)
Confidence interval for μ:
y − t*
s
t=
( y1 − y2 ) − 0
s12 s22
+
n1 n2
Test Statistic:
t=
d − μd
sd
nd
df = nd – 1
1
```
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