Question

It seems to becoming more common for students to live in off campus housing than to pay to live on campus. A study at a large state university revealed that in a random sample of 50 seniors last year, 56% lived in off campus housing whereas this year, a random sample of 50 seniors showed that 64% lived in off campus housing.

(a) Is this convincing evidence that the true proportion of all seniors that are living in off campus housing last year is less than this year?

(b) Based on your conclusion in part (a), which mistake (a Type 1 or a Type 2 error) could you have made? Interpret the potential error in context.

Answer 1

Given that,
possibile chances (x)=28
sample size(n)=50
success rate ( p )= x/n = 0.56
success probability,( po )=0.64
failure probability,( qo) = 0.36
null, Ho:p=0.64
alternate, H1: p<0.64
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.64
since our test is left-tailed
reject Ho, if zo < -1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.56-0.64/(sqrt(0.2304)/50)
zo =-1.1785
| zo | =1.1785
critical value
the value of |z α| at los 0.05% is 1.64
we got |zo| =1.179 & | z α | =1.64
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: left tail - Ha : ( p < -1.17851 ) = 0.1193
hence value of p0.05 < 0.1193,here we do not reject Ho
ANSWERS
---------------
a.
null, Ho:p=0.64
alternate, H1: p<0.64
test statistic: -1.1785
critical value: -1.64
decision: do not reject Ho
p-value: 0.1193
we do not have enough evidence to support the claim that the true proportion of all seniors that are living in off campus housing last year is less than this year.
b.
in this context,
type 2 error is possible because it fails to reject the null hypothesis.