[數撚圍爐區] 全港最多數撚的POST上線啦 (31)
MAEC大師兄
1001 回覆
8 Like
1 Dislike
第 1 頁第 2 頁第 3 頁第 4 頁第 5 頁第 6 頁第 7 頁第 8 頁第 9 頁第 10 頁第 11 頁第 12 頁第 13 頁第 14 頁第 15 頁第 16 頁第 17 頁第 18 頁第 19 頁第 20 頁第 21 頁第 22 頁第 23 頁第 24 頁第 25 頁第 26 頁第 27 頁第 28 頁第 29 頁第 30 頁第 31 頁第 32 頁第 33 頁第 34 頁第 35 頁第 36 頁第 37 頁第 38 頁第 39 頁第 40 頁第 41 頁
想問下如果有個2x2 contingency table應該用fisher exact test定chi square test
同埋value 高係咪代表可以reject null hypothesis
同埋value 高係咪代表可以reject null hypothesis
depends on個expected frequency係咪>=5
係嘅話用邊個都冇所謂, 方便計就得, <5就用返exact去做
chi square test嘅話係
係嘅話用邊個都冇所謂, 方便計就得, <5就用返exact去做
chi square test嘅話係
但n係exponential in input size
仲有計\phi係as least as hard as breaking RSA:
Let n = pq where p, q are distinct primes
\phi(n) = (p-1)(q-1) = n+1 - (p+q), so
p+q = n+1-\phi(n)
pq = n
可以solve p同q
Google左下, 見話計\phi同factoring係equivalent under poly-time randomized reduction (deterministic if assume some version of Riemann Hypothesis)
仲有計\phi係as least as hard as breaking RSA:
Let n = pq where p, q are distinct primes
\phi(n) = (p-1)(q-1) = n+1 - (p+q), so
p+q = n+1-\phi(n)
pq = n
可以solve p同q
Google左下, 見話計\phi同factoring係equivalent under poly-time randomized reduction (deterministic if assume some version of Riemann Hypothesis)
Me數撚做緊野，咁啱要用到hypothesis testing
