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Guide e consigli
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Statistica: Quartili, Correlazione, Intervalli e Test, Formulari di Statistica

Formulario con i principali concetti e formule di statistica descrittiva e inferenziale

Tipologia: Formulari

2020/2021

Caricato il 20/04/2021

Giada_Scarabello
Giada_Scarabello 🇮🇹

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FORMULARIO
QUARTILES
IQR : Q3 – Q1
SAMPLE CORRELATION COEFFICIENT :
by CENTRAL LIMIT THEOREM:
P(z>(...))
1.1) E[x]=p
MODE = the value which compares the most;
MEAN = ( 1+ …. + n)/;
MEDIAN = the central value of a DATA SET.
DISPARI : MEDIAN is the central value
PARI : MEDIAN: ( 2 central values / 2 )
IMPORTANT! FOR BOTH CASES ORDER THE DATA SET IN INCREASING ORDER!
1st : 0.25 * n
2nd : 0.50 * n it corresponds to the MEDIAN.
3rd : 0.75 * n
( Interquartile Range )
∑xy-nx(bar)y(bar)/√(∑x^2-nx(bar)^2)(∑y^2-ny(bar)^2)
1) r<0,3 : none correlation or very weak;
2) 0,3 < r < 0,5 : weak corr.;
3) 0,5 < r < 0,7: moderate corr.;
4) r>0,7 : strong corr.
DISTRIBUTION of the SAMPLE MEAN :
(x(bar)-µ)/(o/√n)
P{X<=a} = P{(x(bar)-µ/(o/√n) <= (a-µ)/(o/√n)} # 1-(P<=z) when
I must calculate
Z : standardized
variable
SAMPLING PROP. From a finite POP:
1) E[x] = np
2) SD[x] = √np(1-p) 2.1) SD[x]= √(p(1-p))/n
X(bar) = sample mean of a sample size “n” from a population
having mean µ
1) E[x(bar)] = µ
2) SD[x(bar)]=(o/√n)
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FORMULARIO

QUARTILES

IQR : Q3 – Q

SAMPLE CORRELATION COEFFICIENT :

by CENTRAL LIMIT THEOREM: P(z>(...)) 1.1) E[x]=p MODE = the value which compares the most; MEAN = ( 1+ …. + n)/; MEDIAN = the central value of a DATA SET. DISPARI : MEDIAN is the central value PARI : MEDIAN: ( 2 central values / 2 ) IMPORTANT! FOR BOTH CASES ORDER THE DATA SET IN INCREASING ORDER! 1 st^ : 0.25 * n 2 nd^ : 0.50 * n it corresponds to the MEDIAN. 3 rd^ : 0.75 * n ( Interquartile Range ) ∑xy-nx(bar)y(bar)/√(∑x^2-nx(bar)^2)(∑y^2-ny(bar)^2)

  1. r<0,3 : none correlation or very weak;
  2. 0,3 < r < 0,5 : weak corr.;
  3. 0,5 < r < 0,7: moderate corr.;
  4. r>0,7 : strong corr. DISTRIBUTION of the SAMPLE MEAN : (x(bar)-μ)/(o/√n) P{X<=a} = P{(x(bar)-μ/(o/√n) <= (a-μ)/(o/√n)} # 1-(P<=z) when I must calculate Z : standardized variable SAMPLING PROP. From a finite POP:
  5. E[x] = np
  6. SD[x] = √np(1-p) 2.1) SD[x]= √(p(1-p))/n X(bar) = sample mean of a sample size “n” from a population having mean μ
  7. E[x(bar)] = μ
  8. SD[x(bar)]=(o/√n)

POINT ESTIMATOR :

CONFIDENCE LEVEL

CONSTRUCTION CONF. INTERVAL:

CONSTRUCTION CONF. INTERVAL:

CONSTRUCTION CONF. INTERVAL

PROPORTION

TEST STATISTICS :

  1. p(hat) = X/n
  2. E[p(hat)] = p
  3. SD[p(hat)] = √(p(1-p))/n alpha alpha/2 z(alpha/2) x(bar)±z(alpha/2)*(o/n)

( known POP. VARIANCE )

  1. LOWER CONF. BOUND: x(bar)-z(alpha)*(o/n)
  2. UPPER CONF. BOUND: x(bar)+z(alpha)+(o/n)
  3. LOWER CONF. BOUND ( of the CONF.INTERVAL ): x(bar)-z(alpha/2)*(o/n)
  4. UPPER CONF. BOUND ( of the CONF. INTERVAL ): x(bar)+z(alpha/2)+(o/n) x(bar)±t(n-1),(alpha/2)*(o/n)

( unknown POP. VARIANCE )

  1. LOWER CONF. BOUND: x(bar)-t(n-1)(alpha)*(o/n)
  2. UPPER CONF. BOUND: x(bar)+t(n-1)(alpha)+(o/n)
  3. LOWER CONF. BOUND ( of the CONF.INTERVAL ): x(bar)-t(n-1)(alpha/2)*(o/n)
  4. UPPER CONF. BOUND ( of the CONF. INTERVAL ): x(bar)+t(n-1)(alpha/2)+(o/n) p(hat)±z(alpha/2)√(p(hat)(1-p))/n ? How LARGE a sample would be necessary to be at most 0,03 ( example)? (( z(alpha/2)√(p(1-p))/0,03)^ ( x(bar)- μ )/o/√n

KNOWN VARIANCE

H0 H

2P{ z≥│ v│ } P{ z≥ v }

H0 H

2P{t(n+m-2) ≥│ v│ } P{t(n+m-2) ≥ v }

HYPOTHESIS TEST concerning 2 populations

Testing EQUALITY of MEANS of two NORMAL POPULATIONS ( Ẋ-Ῡ( μ1- μ2))/√(σ^2x/n)+(σ^2y/n)

p.v.

μx =μy μx ≠μy REJ. │TS │≥Z(alpha/2) NOT REJ. Otherwise μx ≤ μy μ x > μy REJ. TS ≥Z(alpha) NOT REJ. Otherwise TS : ( Ẋ-Ῡ)/√(σ^2x/n)+(σ^2y/n) Testing EQUALITY of MEANS: SMALL-SAMPLE TEST when unknown POP. VARIANCES are equal Sp^2 = [((n-1)/(n+m-2))s^2(x)]+[((m-1)/(n+m-2))s^2(y)] TS : ( Ẋ-Ῡ)/√Sp^2(1/n + 1/m )

p.v.

μx =μy μx ≠μy REJ. │TS │≥{t(n+m-2),(alpha/2) NOT REJ. Otherwise μx ≤ μy μ x > μy REJ. TS ≥{t(n+m-2),(alpha/2) NOT REJ. Otherwise

EMPIRICAL RULE

x(bar) ± s

x(bar) ±2s

x(bar) ±3s

HOW TO FIND STANDARD DEVIATION SD

ABOUT REGRESSION EQUATION

EXAMPLE :

x 0 2 2 5 6 y 6 5 0 -5 1 ? SST, SSR & SSE? when I have to assign the right histogram Data set approximately normal with MEAN X(bar) and sample standard deviation SD

1) 68% of the observations lie within

2) 95% of the observations lie within

3) 99,7% of the observations lie within

1 st^ STEP : Finding ∑x^2 # summation 2 nd^ STEP: Finding ẋ^2 ( x-bar) 3 rd^ STEP : ∑x^2 – n( ẋ^2 )

4 th^ STEP : ( result from 3rd STEP above ) / ( n-1 ) #I obtain the VARIANCE

5 th^ STEP : √( VARIANCE ) # I obtain the SD

  1. y(hat) = (…) - (...)x y(hat) = 5.15-1.25x To find y(hat) I nust
  2. SST= ∑(y-y(bar))^2 substitute x-values into the expression.
  3. SSR = ∑(y(hat)-y(bar))^
  4. SSE = ∑(y-y(hat))^