Please help to determine the power of a hypothesis test

Thread starter bryandnk
Start date May 1, 2010
Tags

Power Test

Click For Summary

The discussion focuses on determining the power of a hypothesis test, specifically how to calculate it using the formula involving beta. The user attempts to express beta as the probability related to the Z-score and the difference in means. They suggest that the power is calculated as 1 minus beta. Clarification is sought on whether their approach and calculations are correct. Accurate understanding of these concepts is crucial for effective hypothesis testing.

May 1, 2010

bryandnk

Homework Statement

[PLAIN]http://img19.imageshack.us/img19/7396/66899700.jpg

The Attempt at a Solution

Please help, Would the power test be something like:

beta = P(Z < Z_alpha_2 - delta/(1/m+1/n)^.5)

power = 1 - beta

Last edited by a moderator: May 4, 2017

Physics news on Phys.org

May 1, 2010

bryandnk

so does anyone know if I did this right?

Thread 'Does this series converge uniformly?'

First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...

View full post »