Estatística - Cap. - 6 Variáveis AleatÃ
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Estatística - Cap. - 6 Variáveis AleatÃ


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z \u2264 
 }= P{ z \u2264 -2,19} = 0,5 \u2013 0,48574 = 0,0143.
 Os demais valores a serem usados na técnica do 
serão obtidos na tabela abaixo:
	Classes
	Z
	P
	Ei = n.P
	oi
	
	5 6
	-2,19
	0,0143
	1,7
	
11,2
	
15
	
1,29
	6 7
	-1,75
	0,0258
	3,1
	
	
	
	7 8
	-1,32
	0,0533
	6,4
	
	
	
	8 9
	-0,88
	0,0960
	11,5
	11,5
	12
	0,0217
	 9 10
	-0,44
	0,1405
	17,0
	17,0
	18
	0,0588
	10 11
	0,00
	0,1700
	20,4
	20,4
	22
	0,1255
	11 12
	0,44
	0,1700
	20,4
	20,4
	18
	0,2824
	12 13
	0,88
	0,1405
	17,0
	17,0
	13
	0,9412
	13 14
	1,32
	0,0960
	11,5
	11,5
	11
	0,0217
	14 15
	1,75
	0,0533
	6,4
	
10,7
	
11
	
0,0084
	15 16
	2,19
	0,0258
	3,1
	
	
	
	 16 17 
	2,63
	0,0099
	1,2
	
	
	
	Total
	-
	
	-
	-
	
	2,7493
Obs: 
1) A freqüência esperada tem que ser sempre maior ou igual a 5 para se poder utilizar 
.
2) Os valores das linhas da coluna P são obtidos pela diferença do valor de P acumulado da linha, subtraído do correspondente valor de P da linha anterior.
3) n = 120
4) Devido à primeira observação, as três linhas dos extremos são agrupadas, gerando um número de classes igual a 8.
5) 
 = 2,7493
 Na tabela da distribuição de 
, o número de graus de liberdade será dado por:
 
onde: K = número de classes (K = 8)
 m = número de parâmetros estimados (m = 2; média e desvio padrão). Quando estes valores forem conhecidos para a distribuição, m = 0. O valor de m será sempre 0, 1 ou 2 (na distribuição Normal será 2, na de Poisson será 1). 
 Logo:
 
.
 Procura-se, então, na tabela da distribuição de
: 
; sendo 
 o valor do risco da afirmação. Na tabela é colocado o valor de P = (1 - 
). Assim, tem-se, por exemplo:
a) 
= 15,1 (risco 1%)
b) 
= 11,1 (risco 5%)
 Em ambas as situações a hipótese de normalidade deve ser aceita.
X = a
X = b
x
f(x)
(1, 1)
P(0 < z < zc)
P(0 \u2264 z \u2264 1,73) = 0,4582
P(- 1,73 \u2264 z \u2264 0) = 0,4582
P(0 \u2264 z \u2264 1,73) = 0,4582
P(z \u22651,73) = 0,0418
P(0 \u2264 z \u2264 - 1,73) = 0,4582
P(z \u2264 - 1,73) = 0,0418
P(0 \u2264 z \u2264 0,47) = 0,1808
P(0,47 \u2264 z \u2264 1,73) = 0,2774
P(0 \u2264 z \u2264 1,73) = 0,4582
1/\ufffd EMBED Equation.3 \ufffd\ufffd\ufffd
n = 1
n = 2
n > 2
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