Question

To attract a higher income from commercials, a new morning soap opera must guarantee advertising agencies that at least 20% of the tele-audience will watch the program. The producers of the new soap opera hired a marketing company to conduct an investigation among 2,000 people in the teleauditorio. Of the 2,000 people, 300 watch the soap opera at least once a week. At the 0.05 level of significance, can it be said that 20% of the audience watches the soap opera? Or, is this sample proportion close enough to the necessary

Answer 1

Given that,
possible chances (x)=300
sample size(n)=2000
success rate ( p )= x/n = 0.15
success probability,( po )=0.2
failure probability,( qo) = 0.8
null, Ho:p=0.2
alternate, H1: p!=0.2
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.15-0.2/(sqrt(0.16)/2000)
zo =-5.59
| zo | =5.59
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =5.59 & | z α | =1.96
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -5.59017 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
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null, Ho:p=0.2
alternate, H1: p!=0.2
test statistic: -5.59
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that 20% of the audience watches the soap opera