Question 1136930
will use 1-way
Ho: use of internet <=0.75
Ha: use of internet >0.75
alpha=0.05, p{reject Ho|Ho true)
test stat 1-sample proportion
z=(phat-po)/sqrt(p*(1-p)/n)
reject Ho for z>=2.32
z=(0.792-0.75)/sqrt (0.75*.25/120)
z=0.042/0.0395=1.06
fail to reject Ho at the 1% level
p-value is z>1.06 or 0.145