Question

Calculate the test statistic and p-value for each sample mean and make the statistical decision (3 points each).

1. H0: µ ≤ 16 versus H1: µ > 16, x = 17, s = 5, n = 25, α = 0.05 test statistic___________ p-value___________ Decision (circle one) Reject the H0 Fail to reject the H0

2. H0: µ ≥ 205 versus H1: µ < 205, x = 198, σ = 15, n = 20, α = 0.05 test statistic___________ p-value___________ Decision (circle one) Reject the H0 Fail to reject the H0

3. H0: µ = 26 versus H1: µ<> 26, x = 22, s = 10, n = 30, α = 0.01 test statistic___________ p-value___________ Decision (circle one) Reject the H0 Fail to reject the H0

4. H0: µ ≥ 155 versus H1: µ < 155, x = 145, σ = 19, n = 25, α = 0.01 test statistic___________ p-value___________ Decision (circle one) Reject the H0 Fail to reject the H0

5. H0: µ ≥ 155 versus H1: µ < 155, x = 145, σ = 19, n = 15, α = 0.01 test statistic___________ p-value___________ Decision (circle one) Reject the H0 Fail to reject the H0

6. The mean cost of a basic rental car per week is said to be greater than $125 per week. To determine if this is true, a random sample of 25 rental cars is taken and resulted in an average of $130.50 and an sample standard deviation of $15.40. Test the appropriate hypotheses at  = 0.05 (1 point each). H0: _________________ H1: _________________ Test statistic : _________________ p-value : _________________ Is the mean cost greater than $125/week ? (circle one) YES NO

Answer 1

1.
Given that,
population mean(u)=16
sample mean, x =17
standard deviation, s =5
number (n)=25
null, Ho: μ<=16
alternate, H1: μ>16
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.711
since our test is right-tailed
reject Ho, if to > 1.711
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =17-16/(5/sqrt(25))
to =1
| to | =1
critical value
the value of |t α| with n-1 = 24 d.f is 1.711
we got |to| =1 & | t α | =1.711
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :right tail - Ha : ( p > 1 ) = 0.16364
hence value of p0.05 < 0.16364,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ<=16
alternate, H1: μ>16
test statistic: 1
critical value: 1.711
decision: do not reject Ho
p-value: 0.16364
we do not have enough evidence to support the claim that mean is greater than 16.
2.
Given that,
population mean(u)=205
standard deviation, σ =15
sample mean, x =198
number (n)=20
null, Ho: μ>=205
alternate, H1: μ<205
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 = 198-205/(15/sqrt(20)
zo = -2.087
| zo | = 2.087
critical value
the value of |z α| at los 5% is 1.645
we got |zo| =2.087 & | z α | = 1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : left tail - ha : ( p < -2.087 ) = 0.018
hence value of p0.05 > 0.018, here we reject Ho
ANSWERS
---------------
null, Ho: μ>=205
alternate, H1: μ<205
test statistic: -2.087
critical value: -1.645
decision: reject Ho
p-value: 0.018
we have enough evidence to support the claim that mean is less than 205.
3.
Given that,
population mean(u)=26
sample mean, x =22
standard deviation, s =10
number (n)=30
null, Ho: μ=26
alternate, H1: μ<26
level of significance, α = 0.01
from standard normal table,left tailed t α/2 =2.462
since our test is left-tailed
reject Ho, if to < -2.462
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =22-26/(10/sqrt(30))
to =-2.1909
| to | =2.1909
critical value
the value of |t α| with n-1 = 29 d.f is 2.462
we got |to| =2.1909 & | t α | =2.462
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :left tail - Ha : ( p < -2.1909 ) = 0.01832
hence value of p0.01 < 0.01832,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=26
alternate, H1: μ<26
test statistic: -2.1909
critical value: -2.462
decision: do not reject Ho
p-value: 0.01832
we do not have enough evidence to support the claim that mean is less than 26.
4.
Given that,
population mean(u)=155
standard deviation, σ =19
sample mean, x =145
number (n)=25
null, Ho: μ>=155
alternate, H1: μ<155
level of significance, α = 0.01
from standard normal table,left tailed z α/2 =2.326
since our test is left-tailed
reject Ho, if zo < -2.326
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 145-155/(19/sqrt(25)
zo = -2.632
| zo | = 2.632
critical value
the value of |z α| at los 1% is 2.326
we got |zo| =2.632 & | z α | = 2.326
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : left tail - ha : ( p < -2.632 ) = 0.004
hence value of p0.01 > 0.004, here we reject Ho
ANSWERS
---------------
null, Ho: μ>=155
alternate, H1: μ<155
test statistic: -2.632
critical value: -2.326
decision: reject Ho
p-value: 0.004
we have enough evidence to support the claim that mean is less than 155.
5.
Given that,
population mean(u)=155
standard deviation, σ =19
sample mean, x =145
number (n)=15
null, Ho: μ>=155
alternate, H1: μ<155
level of significance, α = 0.01
from standard normal table,left tailed z α/2 =2.326
since our test is left-tailed
reject Ho, if zo < -2.326
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 145-155/(19/sqrt(15)
zo = -2.038
| zo | = 2.038
critical value
the value of |z α| at los 1% is 2.326
we got |zo| =2.038 & | z α | = 2.326
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : left tail - ha : ( p < -2.038 ) = 0.021
hence value of p0.01 < 0.021, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ>=155
alternate, H1: μ<155
test statistic: -2.038
critical value: -2.326
decision: do not reject Ho
p-value: 0.021
we do not have enough evidence to support the claim that mean is less than 155.
6.
Given that,
population mean(u)=125
sample mean, x =130.5
standard deviation, s =15.4
number (n)=25
null, Ho: μ=125
alternate, H1: μ>125
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.711
since our test is right-tailed
reject Ho, if to > 1.711
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =130.5-125/(15.4/sqrt(25))
to =1.7857
| to | =1.7857
critical value
the value of |t α| with n-1 = 24 d.f is 1.711
we got |to| =1.7857 & | t α | =1.711
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :right tail - Ha : ( p > 1.7857 ) = 0.04339
hence value of p0.05 > 0.04339,here we reject Ho
ANSWERS
---------------
null, Ho: μ=125
alternate, H1: μ>125
test statistic: 1.7857
critical value: 1.711
decision: reject Ho
p-value: 0.04339
we have enough evidence to support the claim that the mean cost greater than $125/week.