# Probability and statistic problem

• Jan 5th 2009, 08:29 PM
zorro
Probability and statistic problem
Suppose that the manufacturer a of a new medication wants to test the null hypothesis $\displaystyle \theta$ = 0.90 against the alternative hypothesis $\displaystyle \theta=0.60$ . His test statistic i X, the observed number of suces(recoveries) in 20 trials, and he will accept the null hypothesis if x>14 ; otherwise , he will rejext it.Find $\displaystyle \alpha$and $\displaystyle \beta$

Solution:

$\displaystyle \alpha$ =$\displaystyle P(X \leqq 14; \theta=0.90)=0.0114$

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Please can anybody tell me hw did he get 0.0114 .If posible hw did u read the binomial table .Because i am not getting the answer.
• Jan 5th 2009, 10:57 PM
Constatine11
Quote:

Originally Posted by zorro
Suppose that the manufacturer a of a new medication wants to test the null hypothesis $\displaystyle \theta$ = 0.90 against the alternative hypothesis $\displaystyle \theta=0.60$ . His test statistic i X, the observed number of suces(recoveries) in 20 trials, and he will accept the null hypothesis if x>14 ; otherwise , he will rejext it.Find $\displaystyle \alpha$and $\displaystyle \beta$

Solution:

$\displaystyle \alpha$ =$\displaystyle P(X \leqq 14; \theta=0.90)=0.0114$

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Please can anybody tell me hw did he get 0.0114 .If posible hw did u read the binomial table .Because i am not getting the answer.

$\displaystyle P(X \le 14|\theta=0.9)=1-P(X>14|\theta=0.9)$ $\displaystyle =1-[P(X=15|\theta=0.9)+P(X=16|\theta=0.9)+ ... + P(X=20|\theta=0.9)]$

and according to my table this gives:

$\displaystyle P(X \le 14|\theta=0.9)=1-[0.0319+0.0898+0.1901+0.2852+0.2702+0.1216]=0.0112$

.