Statistics高手来帮忙

whyyie

puangenlun

n=150
x_bar=12/150=0.08
x_0=0.1

x~B(150,0.08)   (I hope so)
s=sqrt(150*0.08*0.92)=3.32264955

(1)
z=(x-x_0)/(s/sqrt(n))
z~N(0,1)
P(|z|<k)=0.99
then k=2.575
so |z|<k
-k<(x-x_0)/(s/sqrt(n))<k
x_0-ks/sqrt(n)<x<x_0+ks/sqrt(n)
0.1-0.698579989<x<0.1+0.698579989
-0.6<x<0.8
but x>=0
so 0<x<0.8

(ii)
z=(x_bar-x_0)/(s/sqrt(n))=(0.08-0.1)/(3.32264955/sqrt(150))=-0.073720978
|z|<2.575
so do not reject

(III)
(i) the probability of x between [0,0.8) is 0.99
(ii) the probability of x=0.1 is 0.99

大致如此，不很確定

whyyie

回复 2# puangenlun

part (ii)我是这样做....不懂哪里错..
至于Part (i)有没有那里可以参考hypothesis testing for confidence interval of binomial proprotion?

H_0 = 0.10; H_1 < 0.10
np = 150 x 0.10 = 15
npq = 150 x 0.10 x 0.90 = 13.5
X~N (15, 13.5)

Reject H_0 if z <- 2.326 at 1% level
z = (11.5 - 15)/sqrt(13.5) = -1.089

Since z > 2.326, do not reject H_0. The customers' claim is correct.

puangenlun

X~B(150,0.08)==>X~N (15, 13.5)

這個可能有問題
因為要求
n很大（符合）
p靠中間（不符合）

Confidence interval 其實從hypothesis test 中來的
詳情見下面的example小結
http://en.wikipedia.org/wiki/Confidence_interval

whyyie

回复 4# puangenlun

The Normal distribution can be used to approximate Binomial probabilities when n is    large and p is close to 0.5.  In answer to the question "How large is    large?", or "How close is close?", a rule of thumb is that the    approximation should only be used when both np>5 and nq>5.

上面是网上找的, 但书没提到 p-> 0.5 .
只是np>5 和 nq>5不可以吗?
说回来.. H_0 = 0.10, H_1 < 0.10 这个对吗?

puangenlun

http://en.wikipedia.org/wiki/Bin ... ormal_approximation
"The approximation generally improves as n increases (at least 20) and is better when p is not near to 0 or 1. ....., n is large enough, and p is far enough from the extremes of zero or one."

H0=0.1
H1<>0.1
therefore it is two tails test

whyyie

为什么是 not equal 不是 <

puangenlun

大于小于都是不等于

whyyie

puangenlun

x~Bernoulli(p=0.35)   (I hope so)
s=sqrt(pq)=0.476969601
1.
P(|z|<k)=1-a
since a=0.05, k=1.96
let |z|=k
|(x_bar-x_0)/(s/sqrt(n))|=k
|x_bar-x_0|<=0.03
sqrt(n)*0.03/s>=k
sqrt(n)>=sk/0.03
n>(sk/0.03)^2=971.071112
min n=972

2. similar with 1.

如果你有答案（没有步骤也可以）
那么请公布出来
很想知道我做的是不是正确的。

x的distribution有时候根本不知道是什么
只可以凭感觉猜
所以做到最后会很心虚

whyyie

回复 10# puangenlun

没有答案

whyyie

是不是 67.5 < X < 72.5?

鸭王之王

是不是 67.5 < X < 72.5?
whyyie 发表于 26-3-2011 02:19 AM

为什么不是67.1784<X<72.8216?

鸭王之王

回复  puangenlun

没有答案
whyyie 发表于 14-3-2011 03:32 PM

有答案了吗?和p兄一样~

puangenlun

X~N(x,6^2)
s=6
a=0.01
n=30
a0=70

since sample is small
so use t-test
t=(x-x0)/(s/sqrt(n))
P(|t|<k)=1-a=0.99
k=2.756
(if you use z-test, k=2.576)

|t|<2.756
|x-x0|/(s/sqrt(n))<2.756
|x-x0|<2.756*s/sqrt(n)
|x-70|<2.756*6/sqrt(30)
66.981 < x < 73.019

三個人的答案竟然都不一樣？

鸭王之王

X~N(x,6^2)
s=6
a=0.01
n=30
a0=70

since sample is small
so use t-test
t=(x-x0)/(s/sqrt(n))
...
puangenlun 发表于 26-3-2011 12:53 PM

30是z-test最低要求哦，所以我用z

puangenlun

如果用z-test
|x-70|<2.756*6/sqrt(30)
67.1781 < x < 72.8219
=鸭王之王的答案

whyyie

回复 17# puangenlun

为什么用2.756? 我用 2.326.

puangenlun

H0=x0
h1<>x0
This is a two tailed test
so P(middle)=1-a

H0=x0
h1<x0 (h1>x0)
This is a one tailed test
so P(right)=1-a
(P(left)=1-a)

http://www.chem.utoronto.ca/cour ... orial/12tailed.html