Question

A plan for an executive travelers’ club has been developed by airline on the premise that 5% of its current customers would qualify for membership. A random sample of 500 customers yielded 40 who would qualify.   

1. Write down the null and alternative hypotheses.

2. Explain whether this is a one-sided or a two-sided test, and why.

3. What is the value of the test statistic for your observed results?

4. What is the p-value for your observed results?

5. What is your conclusion? Please use words that a non-statistics student would understand, and justify your answer. Assume a significance level of .05.

6. Based on your conclusion, what type of error are you risk at?

Answer 1

Given that,
possibile chances (x)=40
sample size(n)=500
success rate ( p )= x/n = 0.08
success probability,( po )=0.05
failure probability,( qo) = 0.95
null, Ho:p=0.05
alternate, H1: p!=0.05
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=0.08-0.05/(sqrt(0.0475)/500)
zo =3.078
| zo | =3.078
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =3.078 & | 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 != 3.07794 ) = 0.00208
hence value of p0.05 > 0.0021,here we reject Ho
ANSWERS
---------------
a.
null, Ho:p=0.05
alternate, H1: p!=0.05
b.
two tailed test because club has been developed by airline on the premise that 5% of its
current customers would qualify for membership.
c.
test statistic: 3.078
critical value: -1.96 , 1.96
decision: reject Ho
d.
p-value: 0.00208
e.
we have enough evidence to support the claim that A plan for an executive travelers’ club has been developed by airline on the premise that 5% of its
current customers would qualify for membership.
f.
In this context, type 1 error is possible because it is reject the null hypothesis.