Question

a)𝑋1,…,𝑋𝑛∼𝑖.𝑖.𝑑.𝖯𝗈𝗂𝗌𝗌(𝜆) for some unknown 𝜆>0 ;

𝐻0:𝜆=𝜆0 v.s. 𝐻1:𝜆≠𝜆0where 𝜆0>0.

(Type barX_n for 𝑋⎯⎯⎯⎯⎯𝑛 , lambda_0 for 𝜆0 . If applicable, type abs(x) for |𝑥| , Phi(x) for Φ(𝑥)=𝐏(𝑍≤𝑥) where 𝑍∼N(0,1) , and q(alpha) for 𝑞𝛼 , the 1−𝛼 quantile of a standard normal variable, e.g. enter q(0.01) for 𝑞0.01 .)

Asymptotic 𝑝-value=?


b) 𝑋1,…,𝑋𝑛∼𝑖.𝑖.𝑑.𝖯𝗈𝗂𝗌𝗌(𝜆) for some unknown 𝜆>0 ;

𝐻0:𝜆≥𝜆0 v.s. 𝐻1:𝜆<𝜆0where 𝜆0>0.

( Type barX_n for 𝑋⎯⎯⎯⎯⎯𝑛 , lambda_0 for 𝜆0. . If applicable, type abs(x) for |𝑥| , Phi(x) for Φ(𝑥)=𝐏(𝑍≤𝑥) where 𝑍∼N(0,1) , and q(alpha) for 𝑞𝛼 , the 1−𝛼 quantile of a standard normal variable. )

Asymptotic 𝑝-value=?


c)𝑋1,…,𝑋𝑛∼𝑖.𝑖.𝑑.𝖤𝗑𝗉(𝜆) for some unknown 𝜆>0 ;

𝐻0:𝜆=𝜆0 v.s. 𝐻1:𝜆≠𝜆0where 𝜆0>0.

(Type barX_n for 𝑋⎯⎯⎯⎯⎯𝑛 , lambda_0 for 𝜆0. If applicable, type abs(x) for |𝑥| , Phi(x) for Φ(𝑥)=𝐏(𝑍≤𝑥) where 𝑍∼N(0,1) , and q(alpha) for 𝑞𝛼 , the 1−𝛼 quantile of a standard normal variable.)

Asymptotic 𝑝-value=?

Answer 1

a) 2*Phi((sqrt(n)*(barX_n-lambda_0))/sqrt(lambda_0))
b) 1-Phi((sqrt(n)*(lambda_0-barX_n))/sqrt(lambda_0))
c) 2*(1-Phi(sqrt(n)*(barX_n-(1/lambda_0))*lambda_0))