𝑃(𝐵) = 𝑃(𝐵 ∩ 𝐴) + 𝑃(𝐵 ∩ 𝐴′ ) = 𝑃(𝐵|𝐴)𝑃(𝐴) + 𝑃(𝐵|𝐴′ )𝑃(𝐴′)
𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐵|𝐴)𝑃(𝐴) = 𝑃(𝐴|𝐵)𝑃(𝐵)
𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵) ≤ 1
𝐸1 ∩ 𝐸2 = 𝐸1 ∩ 𝐸1′ = 0 – Mutually exclusive
𝐸(𝜃̂) − 𝜃 = 0 – Unbiased
𝑉(𝑋) = 𝐸(𝑋 2 ) − (𝐸(𝑋))2
Two events are independent if:
(1) 𝑃(𝐴|𝐵) = 𝑃(𝐴)
(2) 𝑃(𝐵|𝐴) = 𝑃(𝐵)
(3) 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)𝑃(𝐵)
Probability density function:
(1) 𝑓(𝑥) ≥ 0
∞
(2) ∫−∞ 𝑓(𝑥)𝑑𝑥 = 1
Joint probability mass function:
(1) 𝑓𝑋𝑌 (𝑥, 𝑦) ≥ 0
(2) ∑𝑥 ∑𝑦 𝑓𝑋𝑌 (𝑥, 𝑦) = 1
(3) 𝑓𝑋𝑌 (𝑥, 𝑦) = 𝑃(𝑋 = 𝑥, 𝑌 = 𝑦)
𝐸(𝑋𝑖 ) = 𝑛𝑝𝑖 , 𝑉(𝑋𝑖 ) = 𝑛𝑝𝑖 = (1 − 𝑝𝑖 )
Marginal probability mass function:
𝑓𝑋 (𝑥) = 𝑃(𝑋 = 𝑥) = ∑ 𝑓𝑋𝑌 (𝑥, 𝑦)
𝑦
𝑓𝑌 (𝑦) = 𝑃(𝑌 = 𝑦) = ∑ 𝑓𝑋𝑌 (𝑥, 𝑦)
𝑥
(3) 𝑃(𝑎 ≤ 𝑋 ≤ 𝑏) = ∫𝑎 𝑓(𝑥)𝑑𝑥
Probability mass function:
(1) 𝑓(𝑥𝑖 ) ≥ 0
(2) ∑𝑛𝑖=1 𝑓(𝑥𝑖 ) = 1
(3) 𝑓(𝑥𝑖 ) = 𝑃(𝑋 = 𝑥𝑖 )
Mean and variance of the discrete rv:
Conditional probability mass function of Y given X=x is:
𝑓𝑋𝑌 (𝑥, 𝑦)
𝑓𝑌|𝑥 (𝑦) =
,
𝑓𝑜𝑟 𝑓𝑋 (𝑥) > 0
𝑓𝑋 (𝑥)
Because a conditional probability mass function 𝑓𝑌|𝑥 (𝑦) is a probability mass
function, the following prop. are satisfied:
(1) 𝑓𝑌|𝑥 ( 𝑦) ≥ 0
(2) ∑𝑦 𝑓𝑌|𝑥 (𝑦) = 1
(3) 𝑃(𝑌 = 𝑦|𝑋 = 𝑥) = 𝑓𝑌|𝑥 (𝑦)
Conditional mean, variance:
𝜇 = 𝐸(𝑋) = ∑ 𝑥𝑓(𝑥) = ∫ 𝑥𝑓(𝑥)𝑑𝑥
𝜇𝑌|𝑥 = 𝐸(𝑌|𝑥) = ∑ 𝑦𝑓𝑌|𝑥 (𝑦)
𝑏
∞
𝑥
−∞
∞
𝜎 2 = 𝑉(𝑋) = ∑ 𝑥 2 𝑓(𝑥) − 𝜇 2 = ∫ 𝑥 2 𝑓(𝑥)𝑑𝑥 − 𝜇 2
𝑥
−∞
Discrete uniform distribution, all n has equal probability:
1
𝑏+𝑎
(𝑏 − 𝑎 + 1)2 − 1
𝑓(𝑥𝑖 ) = , 𝐸(𝑋) =
, 𝑉(𝑋) =
𝑛
2
12
Continuous uniform distribution:
1
(𝑎 + 𝑏)
(𝑏 − 𝑎)2
𝑓(𝑥) =
, 𝐸(𝑋) =
, 𝑉(𝑋) =
(𝑏 − 𝑎)
2
12
A Bernoulli trial (Binomial distribution, success or failure)
𝑛 𝑥
𝑓(𝑥) = ( ) 𝑝 (1 − 𝑝)𝑛−𝑥 (𝑔𝑜𝑜𝑑 𝑓𝑜𝑟 𝑛𝑝 > 5 𝑎𝑛𝑑 𝑛(1 − 𝑝) > 5)
𝑥
𝐸(𝑋) = 𝑛𝑝,
𝑉(𝑥) = 𝑛𝑝(1 − 𝑝)
Geometric distribution:
𝑓(𝑥) = (1 − 𝑝)𝑥−1 𝑝,
𝑥 = 1,2, …
1
2
𝜇 = 𝐸(𝑋) = ,
𝜎 = 𝑉(𝑥) = (1 − 𝑝)/𝑝2
𝑝
Negative binomial distribution nr of trials until r successes:
𝑥−1
) (1 − 𝑝)𝑥−𝑟 𝑝𝑟 , 𝑟 = 1, 2, … , 𝑥 = 𝑟, 𝑟 + 1, 𝑟 + 2, …
𝑓(𝑥) = (
𝑟−1
𝜇 = 𝐸(𝑋) = 𝑟/𝑝, 𝜎 2 = 𝑉(𝑋) = 𝑟(1 − 𝑝)/𝑝2
Hypergeometric distribution:
N objects contains K objects = successes, N-K objects = failures. A sample of size
n, X = # of successes in the sample. p=K/N, (good for n/N < 0.1)
(𝐾)(𝑁−𝐾)
𝑓(𝑥) = 𝑥 𝑁𝑛−𝑥 , 𝑥 = max{0, 𝑛 + 𝐾 − 𝑁} 𝑡𝑜 min{𝐾, 𝑛}
(𝑛 )
𝑁−𝑛
)
𝜇 = 𝐸(𝑋) = 𝑛𝑝, 𝜎 2 = 𝑉(𝑋) = 𝑛𝑝(1 − 𝑝) (
𝑁−1
Poisson distribution:
𝑒 −𝜆 𝜆𝑥
𝑓(𝑥) =
, 𝑥 = 0, 1, 2, … (𝑔𝑜𝑜𝑑 𝑓𝑜𝑟 𝜆 > 5)
𝑥!
𝜇 = 𝐸(𝑋) = 𝑉(𝑋) = 𝜆
Cumulative distribution function:
𝑥
𝐹(𝑥) = 𝑃(𝑋 ≤ 𝑥) = ∫ 𝑓(𝑢)𝑑𝑢,
𝑓𝑜𝑟 − ∞ < 𝑥 < ∞
−∞
Expected value of a function of a continuous random value:
∞
𝐸[ℎ(𝑥)] = ∫ ℎ(𝑥)𝑓(𝑥)𝑑𝑥
−∞
Normal distribution:
−(𝑥−𝜇)2
1
𝑓(𝑥) =
𝑒 2𝜎2 , 𝑁(𝜇, 𝜎 2 )
√2𝜋𝜎
Standardizing to calculate a probability:
𝑋−𝜇 𝑥−𝜇
) = 𝑃(𝑍 ≤ 𝑧)
𝑃(𝑋 ≤ 𝑥) = 𝑃 (
≤
𝜎
𝜎
Normal approximation to the binomial distribution:
𝑋 − 𝑛𝑝
𝑍=
√𝑛𝑝(1 − 𝑝)
is approximately a standard normal rv. To approximate a binomial probability
with normal distribution
𝑥 + 0.5 − 𝑛𝑝
)
𝑃(𝑋 ≤ 𝑥) = 𝑃(𝑋 ≤ 𝑥 + 0.5) ≅ 𝑃 (𝑍 ≤
√𝑛𝑝(1 − 𝑝)
and
𝑥 − 0.5 − 𝑛𝑝
𝑃(𝑋 ≤ 𝑥) = 𝑃(𝑥 − 0.5 ≤ 𝑋) ≅ 𝑃 (
≤ 𝑍)
√𝑛𝑝(1 − 𝑝)
The approximation is good for 𝑛𝑝 > 5 and 𝑛(1 − 𝑝) > 5
Normal approximation to the Poisson distribution:
𝑋−𝜆
𝑍=
,
𝜆>5
√𝜆
Exponential distribution:
𝑓(𝑥) = λe−λx , for 0 ≤ x ≤ ∞
𝜇 = 𝐸(𝑋) = 1/𝜆, 𝜎 2 = 𝑉(𝑋) = 1/𝜆2
𝑦
2
2
𝜎𝑌|𝑥
= 𝑉(𝑌|𝑥) = ∑ 𝑦 2 𝑓𝑌|𝑥 (𝑦) − 𝜇𝑌|𝑥
𝑦
For discrete random variable X & Y, if one of the following is true, the other are
also true and X, Y are independent:
(1) 𝑓𝑋𝑌 (𝑥, 𝑦) = 𝑓𝑋 (𝑥)𝑓𝑌 (𝑦), 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 & 𝑦
(2) 𝑓𝑌|𝑥 (𝑦) = 𝑓𝑌 (𝑦), 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 & 𝑦 𝑤𝑖𝑡ℎ 𝑓𝑋 (𝑥) > 0
(3) 𝑓𝑋|𝑦 (𝑥) = 𝑓𝑋 (𝑥), 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 & 𝑦 𝑤𝑖𝑡ℎ 𝑓𝑌 (𝑦) > 0
(4) 𝑃(𝑋 ∈ 𝐴, 𝑌 ∈ 𝐵) = 𝑃(𝑋 ∈ 𝐴)𝑃(𝑌 ∈ 𝐵)
Covarience:
∑ 𝑥𝑖
𝑐𝑜𝑣(𝑋, 𝑌) = 𝜎𝑋𝑌 = 𝐸[(𝑋 − 𝜇𝑋 )(𝑌 − 𝜇𝑌 )] = 𝐸(𝑋𝑌) − 𝜇𝑋 𝜇𝑌 , 𝜇𝑋 =
𝑛
Correlation:
𝑐𝑜𝑣(𝑋, 𝑌)
𝜎𝑋𝑌
𝜌𝑋𝑌 =
=
, −1 ≤ 𝜌𝑋𝑌 ≤ 1
√𝑉(𝑋)𝑉(𝑌) 𝜎𝑋 𝜎𝑌
If X & Y are independent random variables, 𝜎𝑋𝑌 = 𝜌𝑋𝑌 = 0 (⇍)
𝑛!
Combinations: (𝑛𝑟) =
𝑟!(𝑛−𝑟)!
𝑛!
Permutations: 𝑃𝑟𝑛 = (𝑛−𝑟)!
̂ is its standard deviation, given by:
The standard error of an estimator Θ
̂)
𝜎Θ̂ = √𝑉(Θ
̂ of the parameter θ is:
The mean squared error of the estimator Θ
̂ ) = 𝐸(Θ
̂ − θ)2
𝑀𝑆𝐸(Θ
Likelihood function: 𝐿(𝜃) = ∏𝑛𝑖=1 𝑓(𝑥𝑖 )
Suppose a random experiment consists of a series of n trials. Assume that
1)
The result of each trial is classified into one of k classes.
2)
The probability of a trial generating a result in class 1 to class k is
constant over trials and equal to p1 to pk.
3)
The trials are independent
The random variables X1, X2,…,Xn that denote the number of trials that result in
class 1 to class k have a multinomial distribution and the joint probability mass
function is
𝑛!
𝑥
𝑥
𝑥
𝑃(𝑋1 = 𝑥1 , 𝑋2 = 𝑥2 , … , 𝑋𝑘 = 𝑥𝑘 ) =
𝑝 1 𝑝 2 … 𝑝𝑘 𝑘
𝑥1 ! 𝑥2 ! … 𝑥𝑘 ! 1 2
𝑓𝑜𝑟 𝑥1 + 𝑥2 + ⋯ + 𝑥𝑘 = 𝑛 𝑎𝑛𝑑 𝑝1 + 𝑝2 + ⋯ + 𝑝𝑘 = 1
𝐸(𝑋𝑖 ) = 𝑛𝑝𝑖 ,
𝑉(𝑋𝑖 ) = 𝑛𝑝𝑖 (1 − 𝑝𝑖 )
If x1, x2, … , xn is a sample of n observations, the sample variance is:
(∑𝑛𝑖=1 𝑥𝑖 )2
𝑛
2
∑𝑛𝑖=1(𝑥𝑖 − 𝑥̅ )2 ∑𝑖=1 𝑥𝑖 −
𝑛
𝑠2 =
=
𝑛−1
𝑛−1
Confidence interval on the mean, variance known:
𝑥̅ − 𝑧𝛼/2 𝜎/√𝑛 ≤ 𝜇 ≤ 𝑥̅ + 𝑧𝛼/2 𝜎/√𝑛
𝑧𝛼/2 𝜎 2
𝑋̅ − 𝜇
) , 𝐸 = |𝑥̅ − 𝜇|
𝑍=
,
𝐶ℎ𝑜𝑖𝑐𝑒 𝑜𝑓 𝑛 = (
𝐸
𝜎/√𝑛
Confidence interval on the mean, variance unknown:
𝑥̅ − 𝑡𝛼/2,𝑛−1 𝑠/√𝑛 ≤ 𝜇 ≤ 𝑥̅ + 𝑡𝛼/2,𝑛−1 𝑠/√𝑛
𝑋̅ − 𝜇
𝑇=
𝑆/√𝑛
Random sample normal disr. mean=µ, var=σ2, S2=sample var.
(𝑛 − 1)𝑆 2
𝑋2 =
ℎ𝑎𝑠 𝜒 2 𝑑𝑖𝑠𝑡. 𝑤𝑖𝑡ℎ 𝑛 − 1 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚
𝜎2
Ci on variance, s2=sample variance, σ2 unknown
(𝑛 − 1)𝑠 2
(𝑛 − 1)𝑠 2
≤ 𝜎2 ≤ 2
𝜒𝛼2,𝑛−1
𝜒1−𝛼,𝑛−1
2
2
Lower and upper confidence bounds on σ2:
(𝑛 − 1)𝑠 2
(𝑛 − 1)𝑠 2
≤ 𝜎 2 𝑎𝑛𝑑 𝜎 2 ≤ 2
2
𝜒𝛼,𝑛−1
𝜒1−𝛼,𝑛−1
Binomial proportion:
If n is large, the distribution of
𝑍=
𝑋 − 𝑛𝑝
√𝑛𝑝(1 − 𝑝)
=
𝑃̂ − 𝑝
√𝑝(1 − 𝑝)
𝑛
is approximately standard normal.
Ci on binomial proportion (obs, lower, upper change zα/2 to zα):
𝑝̂ (1 − 𝑝̂ )
𝑝̂ (1 − 𝑝̂ )
≤ 𝑝 ≤ 𝑝̂ + 𝑧𝛼/2 √
𝑛
𝑛
Sample size for a specified error on binomial proportion:
𝑧𝛼 2
𝑛 = ( 2 ) 𝑝(1 − 𝑝), 𝑛 𝑖𝑠 max 𝑓𝑜𝑟 𝑝 = 0.5
𝐸
Prediction int. on a single future observation from norm. dist:
𝑥̅ − 𝑡𝛼/2,𝑛−1 𝑠√1 + 1/𝑛 ≤ 𝑋𝑛+1 ≤ 𝑥̅ + 𝑡𝛼/2,𝑛−1 𝑠√1 + 1/𝑛
Ci, difference in mean, variances known:
𝑝̂ − 𝑧𝛼/2 √
𝜎12 𝜎22
𝜎12 𝜎22
𝑥̅1 − 𝑥̅ 2 − 𝑧𝛼/2 √ +
≤ 𝜇1 − 𝜇2 ≤ 𝑥̅1 − 𝑥̅ 2 + 𝑧𝛼/2 √ +
𝑛1 𝑛2
𝑛1 𝑛2
for one-sided, change zα/2 to zα.
Sample size for a c.i. on difference in mean, variances known:
𝑧𝛼/2 2 2
) (𝜎1 + 𝜎22 )
𝑛=(
𝐸
Ci Case 1, difference in mean, variance unknown & equal:
1
1
𝑥̅1 − 𝑥̅ 2 − 𝑡𝛼/2,𝑛1+𝑛2−2 𝑠𝑝 √ + ≤ 𝜇1 − 𝜇2
𝑛1 𝑛2
1
1
≤ 𝑥̅1 − 𝑥̅ 2 + 𝑡𝛼/2,𝑛1+𝑛2−2 𝑠𝑝 √ +
𝑛1 𝑛2
(𝑛1 − 1)𝑆12 + (𝑛2 − 1)𝑆22
𝑛1 + 𝑛2 − 2
Ci Case 2, difference in mean, variance unknown, not equal:
𝑆𝑝2 =
𝑠12 𝑠22
𝑠12 𝑠22
𝑥̅1 − 𝑥̅ 2 − 𝑡𝛼/2,𝜈 √ + ≤ 𝜇1 − 𝜇2 ≤ 𝑥̅1 − 𝑥̅ 2 + 𝑡𝛼/2,𝜈 √ +
𝑛1 𝑛2
𝑛1 𝑛2
2
𝑠2 𝑠2
( 1 + 2)
𝑛1 𝑛2
𝜈= 2
(𝑠1 /𝑛1)2 (𝑠22 /𝑛2 )2
+
𝑛1 − 1
𝑛2 − 1
ν is degrees of freedom for tα/2,if not integer, round down.
Ci for µD from paired samples:
𝑑̅ − 𝑡𝛼/2,𝑛−1 𝑠𝐷 /√𝑛 ≤ 𝜇𝐷 ≤ 𝑑̅ + 𝑡𝛼/2,𝑛−1 𝑠𝐷 /√𝑛
Approximate ci on difference in population proportions:
𝑝̂1 (1 − 𝑝̂1 ) 𝑝̂ 2 (1 − 𝑝̂ 2 )
𝑝̂1 − 𝑝̂ 2 − 𝑧𝛼/2 √
+
≤ 𝑝1 − 𝑝2
𝑛1
𝑛2
𝑝̂1 (1 − 𝑝̂1 ) 𝑝̂ 2 (1 − 𝑝̂ 2 )
≤ 𝑝̂1 − 𝑝̂ 2 + 𝑧𝛼/2 √
+
𝑛1
𝑛2
Hypothesis test:
1.
Choose parameter of interest
2.
H0:
3.
H1:
4.
α=
5.
The test statistic is
6.
Reject H0 at α= … if
7.
Computations
8.
Conclusions
Test on mean, variance known
𝑋̅ − 𝜇0
𝐻0 : 𝜇 = 𝜇0 ,
𝑍0 =
𝜎/√𝑛
Alternative hypothesis
Rejection criteria
H1: µ ≠ µ0
z0 > zα/2 or z0 < -zα/2
H1: µ > µ0
z0 > zα
H1: µ < µ0
z0 < -zα
Test on mean, variance unknown
𝑋̅ − 𝜇0
𝐻0 : 𝜇 = 𝜇0 ,
𝑇0 =
𝑆/√𝑛
Alternative hypothesis
Rejection criteria
H1: µ ≠ µ0
t0 > tα/2,n-1 or t0 < -tα/2, n-1
H1: µ > µ0
t0 > tα,n-1
H1: µ < µ0
t0 < -tα,n-1
Test in the variance of a normal distribution:
(𝑛 − 1)𝑆 2
𝐻0 : 𝜎 2 = 𝜎02 ,
𝜒02 =
𝜎02
Alternative hypothesis
Rejection criteria
2
2
H1: σ2 ≠𝜎02
𝜒02 > 𝜒𝛼/2,𝑛−1
𝑜𝑟 𝜒02 < −𝜒𝛼/2,𝑛−1
2
H1: σ2 > 𝜎02
𝜒02 > 𝜒𝛼,𝑛−1
2
H1: σ2 < 𝜎02
𝜒02 < −𝜒𝛼,𝑛−1
Approximate test on a binomial proportion:
𝑋 − 𝑛𝑝0
𝐻0 : 𝑝 = 𝑝0 ,
𝑍0 =
√𝑛𝑝0 (1 − 𝑝0 )
Alternative hypothesis
Rejection criteria (**)
H1: p ≠ p0
z0 > zα/2 or z0 < -zα/2
H1: p > p0
z0 > zα
H1: p < p0
z0 < -zα
App. Sample size for a 2-sided test on a binomial proportion:
2
𝑧𝛼/2 √𝑝0 (1 − 𝑝0 ) + 𝑧𝛽 √𝑝(1 − 𝑝)
] , 𝑓𝑜𝑟 1 − 𝑠𝑖𝑑𝑒𝑑 𝑢𝑠𝑒 𝑧𝛼
𝑛=[
𝑝 − 𝑝0
Test on the differens in mean, variance known
𝑋̅1 − 𝑋̅2 − ∆0
𝐻0 : 𝜇1 − 𝜇2 = ∆0 ,
𝑍0 =
𝜎2 𝜎2
√ 1+ 2
𝑛1 𝑛2
Alternative hypothesis
Rejection criteria
H1: 𝜇1 − 𝜇2 ≠ ∆0
z0 > zα/2 or z0 < -zα/2
H1: 𝜇1 − 𝜇2 > ∆0
z0 > zα
H1: 𝜇1 − 𝜇2 < ∆0
z0 < -zα
Sample size, 1-sided test on difference in mean, with power of at least 1-β,
n1=n2=n, variance known:
2
(𝑧𝛼 + 𝑧𝛽 ) (𝜎12 + 𝜎22)
𝑛=
(∆ − ∆0 )2
Tests on diff. in mean, variances unknown and equal:
𝑋̅1 − 𝑋̅2 − ∆0
𝐻0 : 𝜇1 − 𝜇2 = ∆0 ,
𝑇0 =
1
1
𝑆𝑝 √ +
𝑛1 𝑛2
Alternative hypothesis
Rejection criteria
H1: 𝜇1 − 𝜇2 ≠ ∆0
𝑡0 > 𝑡𝛼/2,𝑛1+𝑛2−2 𝑜𝑟
𝑡0 < −𝑡𝛼/2,𝑛1+𝑛2−2
H1: 𝜇1 − 𝜇2 > ∆0
𝑡0 > 𝑡𝛼,𝑛1 +𝑛2−2
H1: 𝜇1 − 𝜇2 < ∆0
𝑡0 < −𝑡𝛼,𝑛1+𝑛2−2
Tests on diff. in mean, variances unknown and not equal:
𝑋̅1 − 𝑋̅2 − ∆0
If 𝐻0 : 𝜇1 − 𝜇2 = ∆0 is true, the statistic 𝑇0 =
𝑆2 𝑆2
√ 1+ 2
𝑛1 𝑛2
is distributed app. as t with ν degrees of freedom, ~𝑡(𝜈)
Paired t-test:
̅ − 𝛥0
𝐷
𝜇𝐷
𝜇1 − 𝜇2
𝐻0 : 𝜇𝐷 = 𝜇1 − 𝜇2 = Δ0 ,
𝑇0 =
,
𝑑=
=
𝜎𝐷 √𝜎12 + 𝜎22
𝑆𝐷 /√𝑛
Alternative hypothesis
Rejection criteria
H1: 𝜇𝐷 ≠ Δ0
t0 > tα/2,n-1 or t0 < -tα/2, n-1
H1: 𝜇𝐷 > Δ0
t0 > tα,n-1
H1: 𝜇𝐷 < Δ0
t0 < -tα,n-1
Approximate tests on the difference of two population proportions:
𝑃̂1 − 𝑃̂2
𝑋1 + 𝑋2
𝐻0 : 𝑝1 = 𝑝2 , 𝑍0 =
, 𝑃̂ =
, 𝑠𝑒𝑒 (∗∗) ↑
𝑛1 + 𝑛2
1
1
√𝑃̂ (1 − 𝑃̂ ) ( + )
𝑛1 𝑛2
Goodness of fit:
𝑘 (𝑂 − 𝐸 )2
𝑖
𝑖
2
𝑋02 = ∑
> 𝜒𝛼,𝑘−𝑝−1
𝐸𝑖
𝑖=1
Expected frequency: 𝐸𝑖 = 𝑛𝑝𝑖 , 𝑝𝑖 = 𝑃(𝑋 = 𝑥) = 𝑓(𝑥)
The power of a test: = 1 − 𝛽
Δ − Δ0
𝛽 = Φ zα/2 −
σ2
√ 1
n1
+
Δ − Δ0
− Φ −zα/2 −
σ22
σ2 σ2
√ 1+ 2
n1 n2
n2
(
)
(
)
The P-value is the smallest level of significance that would lead to rejection of
the null hypothesis H0 with the given data.
2[1 − Φ(|𝑧0 |)] 𝑓𝑜𝑟 𝑎 𝑡𝑤𝑜 − 𝑡𝑎𝑖𝑙𝑒𝑑 𝑡𝑒𝑠𝑡: 𝐻𝑜 : 𝜇 = 𝜇0
𝐻1 : 𝜇 ≠ 𝜇0
𝐻1: 𝜇 > 𝜇0
𝑃 = {1 − Φ(z0 ) 𝑓𝑜𝑟 𝑎 𝑢𝑝𝑝𝑒𝑟 − 𝑡𝑎𝑖𝑙𝑒𝑑 𝑡𝑒𝑠𝑡: 𝐻𝑜 : 𝜇 = 𝜇0
Φ(z0 )
𝑓𝑜𝑟 𝑎 𝑙𝑜𝑤𝑒𝑟 − 𝑡𝑎𝑖𝑙𝑒𝑑 𝑡𝑒𝑠𝑡: 𝐻𝑜 : 𝜇 = 𝜇0
𝐻1 : 𝜇 < 𝜇0
Fitted or estimated regression line: 𝑦̂ = 𝛽̂0 + 𝛽̂1 𝑥
𝑛
𝑛
1
1
𝛽̂0 = 𝑦̅ − 𝛽̂1 𝑥̅ ,
𝑦̅ = ( ) ∑ 𝑦𝑖 ,
𝑥̅ = ( ) ∑ 𝑥𝑖 , 𝑆𝑆𝑇 = 𝑆𝑆𝑅 + 𝑆𝑆𝐸
𝑛
𝑛
𝑖=1
𝑖=1
(∑𝑛𝑖=1 𝑦𝑖 )(∑𝑛𝑖=1 𝑥𝑖 )
𝑛
𝑛
𝑆𝑥𝑦 ∑𝑖=1 𝑦𝑖 𝑥𝑖 −
𝑛
𝛽̂1 =
=
, 𝑒𝑖 = 𝑦𝑖 − 𝑦̂𝑖 , 𝑆𝑆𝐸 = ∑ 𝑒𝑖2
𝑛
2
(∑
)
𝑥
𝑆𝑥𝑥
𝑖
𝑖=1
∑𝑛𝑖=1 𝑥𝑖2 − 𝑖=1
𝑛
𝑛
𝑛
𝑆𝑆𝐸
𝜎̂ 2 =
, 𝑆𝑆𝐸 = 𝑆𝑆𝑇 − 𝛽̂1 𝑆𝑥𝑦 , 𝑆𝑆𝑇 = ∑ (𝑦𝑖 −𝑦̅)2 = ∑ 𝑦𝑖2 − 𝑛𝑦̅ 2
𝑛−2
𝑖=1
𝑖=1
Ci: 𝛽1 ∈ (𝛽̂1 ± 𝑡𝛼/2,𝑛−2 √
̂2
𝜎
𝑆𝑥𝑥
𝑅2 =
𝑆𝑆𝑅
𝑆𝑆𝑇
=1−
𝑆𝑆𝐸
𝑆𝑆𝑇
1
𝑥̅ 2
𝑛
𝑆𝑥𝑥
) , 𝛽0 ∈ (𝛽̂0 ± 𝑡𝛼/2,𝑛−2 √𝜎̂ 2 [ +
, 𝑆𝑆𝑅 = 𝛽̂1 𝑆𝑥𝑦 , 𝐹0 =
𝑆𝑆𝑅
1
𝑆𝑆𝐸
𝑛−2
=
𝑀𝑆𝐸
𝑀𝑆𝐸
])