A fintech startup company claims that its mobile payment application will be accepted by at least...

A fintech startup company claims that its mobile payment application will be accepted by at least 80 percent of hotels in the U.S. A survey, carried out by a consulting firm, of a random sample of 150 American hotels found that the application payment was not accepted in 36 of these hotels.

a. Using a suitable distributional approximation and the 5% significance level, test the validity of the claim made by the company.

Maria went on holiday in Chicago and stayed in 11 different hotels. She found that the mobile application was not accepted in 7 of these hotels.

b. Assuming that these hotels may be regarded as a random sample of all American hotels, test the validity of the claim made by the fintech company, using the 1% significance level.

c. Discuss the conclusions that you reached in parts (a) and (b). Include in your answer two possible reasons why it may be inappropriate to assume that the hotels used by Maria may be regarded as a random sample of all American hotels.

Expert Solution

a.
Given that,
possibile chances (x)=36
sample size(n)=150
success rate ( p )= x/n = 0.24
success probability,( po )=0.8
failure probability,( qo) = 0.2
null, Ho:p=0.8
alternate, H1: p!=0.8
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=0.24-0.8/(sqrt(0.16)/150)
zo =-17.146
| zo | =17.146
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =17.146 & | 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 != -17.14643 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.8
alternate, H1: p!=0.8
test statistic: -17.146
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that the mobile payment application will accepted by at least 80% of hotels in the US.
b.
Given that,
possibile chances (x)=7
sample size(n)=11
success rate ( p )= x/n = 0.636
success probability,( po )=0.8
failure probability,( qo) = 0.2
null, Ho:p=0.8
alternate, H1: p!=0.8
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.63636-0.8/(sqrt(0.16)/11)
zo =-1.357
| zo | =1.357
critical value
the value of |z α| at los 0.01% is 2.576
we got |zo| =1.357 & | z α | =2.576
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -1.3568 ) = 0.17484
hence value of p0.01 < 0.1748,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.8
alternate, H1: p!=0.8
test statistic: -1.357
critical value: -2.576 , 2.576
decision: do not reject Ho
p-value: 0.17484
we do not have enough evidence to support the claim that the mobile payment application will accepted by at least 80% of hotels in the US.
c.
from part (a), it reject the null hypothesis,the mobile payment application will accepted by at least 80% of hotels in the US.
from part (b), it fails to reject the null hypothesis , random sample of all American hotels, test the validity of the claim made by the fintech company.

orchestra answered 2 years ago

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