A recent survey of 120 senior citizens at a local supermarke

Solution

Solution:

where p is population proportion and n is sample size.

We are given p=44%=44/100=0.44

1-p

1-0.44=0.56

n=120

standard error of the sampling distribution of the sample proportion=sqrt(0.44(0.56)/120)

                                                                                      =.sqrt(.2464/120)

                                                                                       =0.045