Answers
10.
Given that,
population mean(u)=1.14
standard deviation, σ =0.25
sample mean, x =1.07
number (n)=36
null, Ho: μ=1.14
alternate, H1: μ<1.14
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 1.07-1.14/(0.25/sqrt(36)
zo = -1.68
| zo | = 1.68
critical value
the value of |z α| at los 5% is 1.645
we got |zo| =1.68 & | z α | = 1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : left tail - ha : ( p < -1.68 ) = 0.05
hence value of p0.05 < 0.05, here we do not reject Ho
ANSWERS
---------------
a.
null, Ho: μ=1.14
alternate, H1: μ<1.14
test statistic: -1.68
critical value: -1.645
decision: reject Ho
p-value: 0.05
we have enough evidence to support the claim that whether there has been reduction in the
population mean fee charged on such transactions.
b.
type 1 error is possible because it reject the null hypothesis.
type 2 error is not possible.
(if type 2 error is possible only it fails to reject the null hypothesis)
11.
Given that,
population mean(u)=32
sample mean, x =33.8
standard deviation, s =6
number (n)=36
null, Ho: μ=32
alternate, H1: μ!=32
level of significance, α = 0.1
from standard normal table, two tailed t α/2 =1.69
since our test is two-tailed
reject Ho, if to < -1.69 OR if to > 1.69
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =33.8-32/(6/sqrt(36))
to =1.8
| to | =1.8
critical value
the value of |t α| with n-1 = 35 d.f is 1.69
we got |to| =1.8 & | t α | =1.69
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 1.8 ) = 0.0805
hence value of p0.1 > 0.0805,here we reject Ho
ANSWERS
---------------
a.
null, Ho: μ=32
alternate, H1: μ!=32
test statistic: 1.8
critical value: -1.69 , 1.69
decision: reject Ho
p-value: 0.0805
b.
we have enough evidence to support the claim that whether the population mean years of life
lost has changed