Question

A popular blog reports that 42% of college students use Twitter. The director of media relations at a large university thinks that the proportion may be different at her university. She polls a simple random sample of 200 students, and 101 of them report that they use Twitter. Can she conclude that the proportion of students at her university who use Twitter differs from 0.42? Answer by showing the five steps of signigicance test, allowing a Type I error rate of 0.05.

Answer 1

Given that,
possibile chances (x)=101
sample size(n)=200
success rate ( p )= x/n = 0.505
success probability,( po )=0.42
failure probability,( qo) = 0.58
null, Ho:p=0.42
alternate, H1: p!=0.42
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=0.505-0.42/(sqrt(0.2436)/200)
zo =2.4355
| zo | =2.4355
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =2.436 & | 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 != 2.43554 ) = 0.01487
hence value of p0.05 > 0.0149,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.42
alternate, H1: p!=0.42
test statistic: 2.4355
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.01487
we have enough evidence to support the claim that the proportion of students at her university who use Twitter differs from 0.42
type 1 error is possible because it reject the null hypothesis.