Question

A survey stated that 40% of people in Indiana watch TV shows at night.

(a) At α = 0.05, test this claim using a random sample of 100 Indiana residents, out of whom 20 said that they do not watch TV shows at night.

(b) Would the test results be the same if 40 people instead said they watch TV shows at night?

Suppose we want to compare the results with Ohio population results of the survey, where it has been found that 30% of Ohio residents watch TV shows at night.

(c) At α = 0.05, test the claim that fewer people in Ohio with respect to Indiana watch TV shows at night, using also a random sample of 60 Ohio residents, out of whom 30 said that they do not watch TV shows at night.

(d) Would the test results be the same if α would be higher?

Answer 1

ANSWER::

Given that,
possibile chances (x)=20
sample size(n)=100
success rate ( p )= x/n = 0.2
success probability,( po )=0.4
failure probability,( qo) = 0.6
null, Ho:p=0.4
alternate, H1: p!=0.4
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=0.2-0.4/(sqrt(0.24)/100)
zo =-4.0825
| zo | =4.0825
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =4.082 & | 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 != -4.08248 ) = 0.00004
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
a).
null, Ho:p=0.4
alternate, H1: p!=0.4
test statistic: -4.0825
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.00004
b).
we have enough evidence to support the claim that proportion of 40% of people in Indiana watch TV shows at night.
c).
Given that,
possibile chances (x)=30
sample size(n)=60
success rate ( p )= x/n = 0.5
success probability,( po )=0.3
failure probability,( qo) = 0.7
null, Ho:p=0.3
alternate, H1: p!=0.3
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=0.5-0.3/(sqrt(0.21)/60)
zo =3.3806
| zo | =3.3806
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =3.381 & | 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 != 3.38062 ) = 0.00072
hence value of p0.05 > 0.0007,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.3
alternate, H1: p!=0.3
test statistic: 3.3806
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.00072
d).
we have enough evidence to support the claim that proportion of 30% of people in Indiana watch TV shows at night.
No,
the test results are α value is higher.

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