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User:Deep evil/Sandbox

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< User:Deep evil

Contents

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  • 2 2
  • 3 3
  • 4 4
  • 5 5

1

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H 0 : p 1 = p 2 {\displaystyle H_{0}:p_{1}=p_{2}}  

z = p ^ 1 − p ^ 2 S E D p {\displaystyle z={\frac {{\hat {p}}_{1}-{\hat {p}}_{2}}{\mathrm {SE} _{Dp}}}}  

S E D p = p ^ ( 1 − p ^ ) ( 1 n 1 + 1 n 2 ) {\displaystyle \mathrm {SE} _{Dp}={\sqrt {{\hat {p}}(1-{\hat {p}})({\frac {1}{n_{1}}}+{\frac {1}{n_{2}}})}}}  

p ^ = X 1 + X 2 n 1 + n 2 {\displaystyle {\hat {p}}={\frac {X_{1}+X_{2}}{n_{1}+n_{2}}}}  

R R = p ^ 1 p ^ 2 {\displaystyle \mathrm {RR} ={\frac {{\hat {p}}_{1}}{{\hat {p}}_{2}}}}  


2

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H 0 : μ t = 0.5714 s e c {\displaystyle \mathrm {H} _{0}:\mu _{t}=0.5714\mathrm {sec} }  

H α : μ t > 0.5714 s e c {\displaystyle \mathrm {H} _{\alpha }:\mu _{t}>0.5714\mathrm {sec} }  

s = 0.1429 {\displaystyle s=0.1429}  

t = 2.1287 − 0.5714 0.1429 30 {\displaystyle t={\frac {2.1287-0.5714}{\frac {0.1429}{\sqrt {30}}}}}  

H 0 {\displaystyle H_{0}}  

H 0 : μ m − t = 0 {\displaystyle H_{0}:\mu _{\mathrm {m-t} }=0}  

H α : μ m − t > 0 {\displaystyle H_{\alpha }:\mu _{\mathrm {m-t} }>0}  

t = 59.31 {\displaystyle t=59.31}  

a = .01 {\displaystyle a=.01}  

p = 0 + {\displaystyle p=0^{+}}  

p < a {\displaystyle p<a}  

d = 0.1379 {\displaystyle d=0.1379}  

s = 0.2964 {\displaystyle s=0.2964}  

t = .1379 − 0 0.1964 30 {\displaystyle t={\frac {.1379-0}{\frac {0.1964}{\sqrt {30}}}}}  

t = 2.54 {\displaystyle t=2.54}  

a = 0.01 {\displaystyle a=0.01}  

p = 0.0082 {\displaystyle p=0.0082}  

p > a {\displaystyle p>a}  

H 0 : μ t = μ o {\displaystyle H_{0}:\mu _{t}=\mu _{o}}  

H α : μ t > μ o {\displaystyle H_{\alpha }:\mu _{t}>\mu _{o}}  

x ¯ t = 2.6237 {\displaystyle {\bar {x}}_{t}=2.6237}  

x ¯ o = 2.1187 {\displaystyle {\bar {x}}_{o}=2.1187}  

d f = 57.9848 {\displaystyle df=57.9848}  

s t = 0.1405 {\displaystyle s_{t}=0.1405}  

s o = 0.1428 {\displaystyle s_{o}=0.1428}  

t = 13.8000 {\displaystyle t=13.8000}  

a = 0.01 {\displaystyle a=0.01}  

p = 0 + {\displaystyle p=0^{+}}  

x ¯ = 2.12 {\displaystyle {\bar {x}}=2.12}  

μ t {\displaystyle \mu _{t}}  

μ m − t {\displaystyle \mu _{m-t}}  



3

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χ 2 = ∑ ( o b s e r v e d − e x p e c t e d ) 2 e x p e c t e d {\displaystyle \chi ^{2}=\sum {\frac {\mathrm {(observed-expected)} ^{2}}{\mathrm {expected} }}}  

χ 2 = ( 210 − 224 ) 2 224 + . . . + ( 530 − 558 ) 2 558 {\displaystyle \chi ^{2}={\frac {(210-224)^{2}}{224}}+...+{\frac {(530-558)^{2}}{558}}}  

χ 2 = 6.5722 {\displaystyle \chi ^{2}=6.5722}  

d f = ( 4 − 1 ) ( 2 − 1 ) = 3 {\displaystyle \mathrm {df} =(4-1)(2-1)=3}  

p − v a l u e = .0374 {\displaystyle \mathrm {p-value} =.0374}  

P ( χ 2 > 6.5722 ) = .0374 {\displaystyle \mathrm {P} (\chi ^{2}>6.5722)=.0374}  

χ 2 = ( 505 − 432.2182 ) 2 432.2182 + . . . + ( 47 − 12.6234 ) 2 12.6234 {\displaystyle \chi ^{2}={\frac {(505-432.2182)^{2}}{432.2182}}+...+{\frac {(47-12.6234)^{2}}{12.6234}}}  

χ 2 = 236.0876 {\displaystyle \chi ^{2}=236.0876}  

d f = ( 3 − 1 ) ( 3 − 1 ) = 4 {\displaystyle \mathrm {df} =(3-1)(3-1)=4}  

p − v a l u e = 0 + {\displaystyle \mathrm {p-value} =0^{+}}  


P ( χ 2 > 236.0876 ) = 0 + {\displaystyle \mathrm {P} (\chi ^{2}>236.0876)=0^{+}}  


4

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χ 2   t e s t   o f   i n d e p e n d e n c e {\displaystyle \chi ^{2}\ \mathrm {test\ of\ independence} }  

χ 2 = ∑ ( o b s e r v e d − e x p e c t e d ) 2 e x p e c t e d {\displaystyle \chi ^{2}=\sum {\frac {\mathrm {(observed-expected)} ^{2}}{\mathrm {expected} }}}  

χ 2 = ( 20 − 27.3995 ) 2 27.3995 + . . . + ( 30 − 36.9054 ) 2 36.9054 {\displaystyle \chi ^{2}={\frac {(20-27.3995)^{2}}{27.3995}}+...+{\frac {(30-36.9054)^{2}}{36.9054}}}  

χ 2 = 6.8519 {\displaystyle \chi ^{2}=6.8519}  

p − v a l u e = 0.2319 {\displaystyle \mathrm {p-value} =0.2319}  

d f = ( 6 − 1 ) ( 2 − 1 ) = 5 {\displaystyle df=(6-1)(2-1)=5}  

α = .05 {\displaystyle \alpha =.05}  

χ 2 = ( 6 − 8.1176 ) 2 8.1176 + . . . + ( 7 − 4.5882 ) 2 4.5882 {\displaystyle \chi ^{2}={\frac {(6-8.1176)^{2}}{8.1176}}+...+{\frac {(7-4.5882)^{2}}{4.5882}}}  

χ 2 = 4.9236 {\displaystyle \chi ^{2}=4.9236}  

d f = ( 2 − 1 ) ( 3 − 1 ) = 2 {\displaystyle df=(2-1)(3-1)=2}  

p − v a l u e = .0853 {\displaystyle \mathrm {p-value} =.0853}  

χ 2 = ( 24 − 28.682 ) 2 28.682 + . . . + ( 10 − 14.682 ) 2 14.682 {\displaystyle \chi ^{2}={\frac {(24-28.682)^{2}}{28.682}}+...+{\frac {(10-14.682)^{2}}{14.682}}}  

χ 2 = 4.4252 {\displaystyle \chi ^{2}=4.4252}  

d f = ( 2 − 1 ) ( 2 − 1 ) = 1 {\displaystyle df=(2-1)(2-1)=1}  

p − v a l u e = .0354 {\displaystyle \mathrm {p-value} =.0354}  

5

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μ y = β 0 + β 1 x {\displaystyle \mu _{y}=\beta _{0}+\beta _{1}x}  

( x 1 , y 1 ) , ( x 2 , y 2 ) , . . . , ( x n , y n ) {\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),...,(x_{n},y_{n})}  

y i = β 0 + β 1 x i + ∈ i {\displaystyle y_{i}=\beta _{0}+\beta _{1}x_{i}+\in _{i}}  

e i = o b s e r v e d   r e s p o n s e − p r e d i c t e d   r e s p o n s e {\displaystyle e_{i}=\mathrm {observed\ response-predicted\ response} }  

= y i − y ^ i {\displaystyle =y_{i}-{\hat {y}}_{i}}  

= y i − b 0 − b 1 x i {\displaystyle =y_{i}-b_{0}-b_{1}x_{i}}  

b 0 ± t ∗ S E b 0 {\displaystyle b_{0}\pm t^{*}\mathrm {SE} _{b_{0}}}  

b 1 ± t ∗ S E b 1 {\displaystyle b_{1}\pm t^{*}\mathrm {SE} _{b_{1}}}  

t = b 1 S E b {\displaystyle t={\frac {b_{1}}{\mathrm {SE} _{b}}}}  


y ^ ± t ∗ S E y ^ {\displaystyle {\hat {y}}\pm t^{*}\mathrm {SE} _{\hat {y}}}  

Retrieved from "https://en.wiki.x.io/w/index.php?title=User:Deep_evil/Sandbox&oldid=123494351"
Last edited on 17 April 2007, at 10:39

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