Question

(Mean: 1 sample)A survey had indicated that the average weight of men over 55 years of age with newly diagnosed heart disease was 90 kg. However, it is suspected that the average weight of such men is now somewhat lower. How large a sample would be necessary to test, at the 5% level of significance with a power of 90%, whether the average weight is unchanged versus the alternative that it has decreased from 90 to 85 kg with an estimated standard deviation of 20 kg?

Answer 1

True mean   µ =    90                              
hypothesis mean,   µo =    85                              
                                      
Level of Significance ,    α =    0.05                              
std dev =    σ =    20.000                              
power =    1-ß =    0.1                              
ß=       0.9                              
δ=   µ - µo =    5                              
                                      
Z ( α ) =       1.6449   [excel function: =normsinv(α)                          
                                     
Z (ß) =        1.2816   [excel function: =normsinv(ß)                         
                                      
sample size needed =    n = ( ( Z(ß)+Z(α) )*σ / δ )² = ( (   1.2816   +   1.6449   ) *   20.0   /   5   ) ² =   137.0216
                                      
so, sample size =        138.000      

.................

                       

THANKS

revert back for doubt

please upvote