%
% problem 241
% example 211
% 
% Written by:
% -- 
% John L. Weatherwax                2008-02-20
% 
% email: wax@alum.mit.edu
% 
% Please send comments and especially bug reports to the
% above email address.
% 
%-----

clear all; 
close all; 
clc; 

% our alternative H_1 hypothesis:
mualt = 10:120;

% our "critical value" used in determining whether to reject H_0 or not:
cv = 1.645; 
% in terms of physical variables of H_0 this translates into 
mu_0    = 45;
sigma_0 = 15; 
pcv     = mu_0 + cv*sigma_0;

% the probability of a type II error (as a function of \mu) is then: 
pII = normcdf( pcv, mualt, sigma_0 ); 

figure; plot( mualt, pII, '-rx' ); axis tight; grid on; 
title( 'the probability of a type II error' ); 
xlabel( '\mu' ); ylabel( 'probability' ); 
saveas( gcf, 'ex_6_2_p_type_II_error', 'epsc' );