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BUSS1020 Final Exam Formula sheet

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BUSS1020 Final Examination
Formula Sheet
Sample Covariance
Range
n ¡
X
Range = X largest − X smallest
C ov(X , Y ) =
X i − X̄
¢¡
Yi − Ȳ
¢
i =1
n −1
Z-score
Empirical Rule
X − X̄
Z=
S
If distribution is bell-shaped, then approximately:
• 68% of data is within ±1 standard deviation of
the mean.
Population Variance
N ¡
X
σ2 =
• 95% of data is within ±2 standard deviation of
the mean.
¢2
Xi − µ
i =1
N
• 99.7% of data is within ±3 standard deviation
of the mean.
Population Standard Deviation
v
u N
uX¡
¢2
u
Xi − µ
u
t i =1
σ=
N
Chebyshev’s Rule
Regardless of how the data is distributed, at least:
µ
¶
1
1 − 2 × 100%, k > 1
k
Sample Variance
n ¡
X
S2 =
X i − X̄
will fall within k standard deviations of the mean.
¢2
i =1
n −1
Counting Rules
• Multiplication rule: (k 1 )(k 2 )...(k n )
• Repetition rule: k n
Sample Standard Deviation
v
uX
n
¢2
u ¡
X i − X̄
u
t i =1
S=
n −1
• Factorial: n! = (n)(n − 1)(n − 2)...(2)(1)
• Permutations: nP k =
n!
(n − k)!
• Combinations: nC k =
Coefficient of Variation
µ ¶
S
CV =
· 100%
X̄
Sample Coefficient of Correlation
r=
Sample Mean
C ov(X , Y )
S X SY
General Addition Rule
n
X
X̄ =
n!
k!(n − k)!
Xi
i =1
n
P (A or B ) = P (A) + P (B ) − P (A and B )
X 1 + X 2 + X 3 + ... + X n
=
n
Page 1 of 5
BUSS1020 Final Examination
Conditional Probability
P (A|B ) =
Binomial Distribution Formula
P (A and B )
P (B )
P (X = x|n, π) = nC x πx (1 − π)n−x
n!
=
πx (1 − π)n−x
x!(n − x)!
Independence
A and B are independent if:
Mean and Standard deviation:
µ = nπ
p
σ = nπ(1 − π)
P (A|B ) = P (A)
Poisson Distribution Formula
Marginal Probability
P (A) =
k
X
P (A|B i )P (B i )
i =1
= P (A|B 1 )P (B 1 ) + P (A|B 2 )P (B 2 )
Mean and Standard deviation:
+ P (A|B 3 )P (B 3 ) + ... + P (A|B k )P (B k )
µ=λ
p
σ= λ
Bayes Theorem
Hypogeometric Distribution Formula
P (A|B i )P (B i )
P (B i |A) =
P (A)
=
A
P (X = x|n, N , A) =
P (A|B i )P (B i )
k
X
C x · N −AC n−x
NC
n
Mean and Standard deviation:
nA
µ=
N
s
P (A|B i )P (B i )
i =1
=
e −λ λx
x!
P (X = x|λ) =
σ=
P (A|B i )P (B i )
P (A|B 1 )P (B 1 ) + ... + P (A|B k )P (B k )
n A(N − A) N − n
·
N2
N −1
Sum of Two Random Variables: Expected Value
Expected Value
E (X + Y ) = E (X ) + E (Y )
µ = E (X ) =
N
X
x i P (X = x i )
E (a X + bY ) = E (a X ) + E (bY ) = aE (X ) + bE (Y )
i =1
Variance
Sum of Two Random Variables: Variance
2
2
σ = E (X ) − [E (X )]
= E [X − E (X )]
=
N
X
V ar (X + Y ) = σ2X +Y
2
= σ2X + σ2Y + 2σ X ,Y
2
V ar (a X + bY ) = σ2aX +bY
[x i − E (X )]2 P (X = x i )
= σ2aX + σ2bY + 2σaX ,bY
i =1
= a 2 σ2X + b 2 σ2Y + 2abσ X ,Y
Covariance
σX Y =
N
X
£
¤
[x i − E (X )] y i − E (Y ) P (x i , y i )
i =1
= E (X Y ) − [E (X )][E (Y )]
Sum of Two Random Variables: Standard Deviation
q
σ X +Y = σ2X +Y
Page 2 of 5
BUSS1020 Final Examination
t-score for Sampling Distribution
Z-score
Z=
X −µ
σ
t=
Assessing Normality
IQR ∼
= 1.33S
Confidence Interval - µ (σ known)
σ
X̄ ± Zα/2 p
n
Uniform Distribution Formula
P (X < x) =
x −a
b−a
Confidence Interval - µ (σ unknown)
Mean and Standard deviation:
a +b
µ=
s2
(b − a)2
12
σ=
Exponential Distribution Formula
P (X < x) = 1 − e −λx
Mean and Standard deviation:
1
µ=
λ
1
σ=
λ
S
X̄ ± t α/2, n−1 p
n
Confidence Interval - proportions
s
p(1 − p)
p ± Zα/2
n
Sample Size for the Mean
n=
n=
X̄ − µ
σ
p
n
e2
2
Zα/2
π(1 − π)
e2
Pooled Variance - σ1 and σ2 equal
Z-score for Sampling Distribution
Z=
2
Zα/2
σ2
Sample Size for Proportion
Standard Error of the Mean
σ
σ X̄ = p
n
X̄ − µ
S
p
n
S P2 =
(n 1 − 1)S 12 + (n 2 − 1)S 22
(n 1 − 1) + (n 2 − 1)
Test Statistic - σ1 and σ2 equal
¢ ¡
¢
X¯1 − X¯2 − µ1 − µ2
= s µ
¶ , d . f = n1 + n2 − 2
1
1
S P2
+
n1 n2
¡
Standard Error of Population Proportion
s
σp =
π(1 − π)
n
Z-score for Proportions
Z=r
t st at
p −π
π(1 − π)
n
Confidence Interval - σ1 and σ2 equal
s µ
¶
¡
¢
1
1
2
X¯1 − X¯2 ± t α/2, n1 +n2 −2 S P
+
n1 n2
Page 3 of 5
BUSS1020 Final Examination
Test statistic - σ1 and σ2 not equal
¢ ¡
¢
¡
X¯1 − X¯2 − µ1 − µ2
t st at =
s
S 12 S 22
+
n1 n2
Sum of Squares
=
X
n1
v = Ã !2
S 12
n1
n1 − 1
+
+
S 22
(X − X̄ )(Y − Ȳ )
P
P
( X )( Y )
XY −
n
X¡
SS X X =
S 12
!2
n2
à !2
S 22
Slope Coefficients
n2
b1 =
n2 − 1
SS X Y
SS X X
P
Y b1 X
b 0 = Ȳ − b 1 X̄ =
−
n
n
P
Confidence Interval - σ1 and σ2 not equal
s
¡
¢
S 12 S 22
X¯1 − X¯2 ± t α/2,v
+
n1 n2
Measures of Variation
SST =
SSR =
Test statistic - Paired Difference
t st at =
X
¢2
X − X̄
P
X 2 ( X )2
= X −
n
With degree of freedom:
Ã
SS X Y =
D̄ − µD
SD
p
n
SSE =
n
X
i =1
n
X
i =1
n
X
(Yi − Ȳ )2
(Yˆi − Ȳ )2
(Yi − Yˆi )2
i =1
Coefficient of Determination
Confidence Interval - Paired Difference
r2 =
SD
D̄ ± t α/2,n−1 p
n
Standard Error of Estimate
s
s
P
SSE
(Yi − Yˆi )2
SY X =
=
n −2
n −2
Pooled Proportion Estimate
p̄ =
SSR
SST
X1 + X2
n1 + n2
Standard Error of Regression Slope
SY X
SY X
S b1 = p
= pP
SS X
(X i − X̄ )2
Z-score for Two Sample Proportions
¡
¢
p 1 − p 2 − (π1 − π2 )
Z=s
µ
¶
¡
¢ 1
1
p̄ 1 − p̄
+
n1 n2
Test Statistic - regression slope
t st at =
Confidence Interval for Two Sample Proportions
s
¡
¢
p 1 (1 − p 1 ) p 2 (1 − p 2 )
p 1 − p 2 ± Zα/2
+
n1
n2
b 1 − β1
r −ρ
=s
S b1
1−r2
n −2
F Test for Overall Significance
F st at
Page 4 of 5
M SR
=
=
M SE
SSR
k
SSE
n −k −1
BUS
USS10
S1020
20 Final Examina
xamination
tion
Confidence Interval for the average Y given X
𝑌𝑌� ± 𝑡𝑡𝛼𝛼⁄2 𝑆𝑆𝑆𝑆𝑆𝑆�ℎ 𝑖𝑖
Prediction Interval for an individual Y given X
𝑌𝑌� ± 𝑡𝑡𝛼𝛼⁄2𝑆𝑆𝑆𝑆𝑆𝑆�1 + ℎ 𝑖𝑖
Where
ℎ 𝑖𝑖 =
(𝑋𝑋𝑖𝑖 − 𝑋𝑋�)2
1 (𝑋𝑋𝑖𝑖 − 𝑋𝑋�)2
+
=
𝑛𝑛
𝑆𝑆𝑆𝑆𝑆𝑆
∑(𝑋𝑋𝑖𝑖 − 𝑋𝑋�)2
Page 5 of 5
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