Question

I have an uneven coin, and think it will provide more heads then tails.  

H0: p = 1/2 vs H1: p > 1/2, where p= is true probability of getting a Head.
If I observe 18 Heads from 25 tosses, then I should decide:

• A) not to perform the test, as I do not have a large enough sample for the Normality assumption to be valid
• B) to reject H0 at the 10% and 5% levels of significance, but not to reject H0 at the 1% level of significance
• C) to reject H0 at the 10% level of significance, but not at the 5% or 1% levels of significance
• D) to reject H0 at each of the 10%, 5% and 1% levels of significance
• E) not to reject H0 at any of the 10%, 5% or 1% levels of significance

Answer 1

a.
level of significance = 0.1
Given that,
possible chances (x)=18
sample size(n)=25
success rate ( p )= x/n = 0.72
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5
alternate, H1: p>0.5
level of significance, α = 0.1
from standard normal table,right tailed z α/2 =1.28
since our test is right-tailed
reject Ho, if zo > 1.28
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.72-0.5/(sqrt(0.25)/25)
zo =2.2
| zo | =2.2
critical value
the value of |z α| at los 0.1% is 1.28
we got |zo| =2.2 & | z α | =1.28
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: right tail - Ha : ( p > 2.2 ) = 0.0139
hence value of p0.1 > 0.0139,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p>0.5
test statistic: 2.2
critical value: 1.28
decision: reject Ho
p-value: 0.0139

b.
level of significance=0.05
Given that,
possible chances (x)=18
sample size(n)=25
success rate ( p )= x/n = 0.72
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5
alternate, H1: p>0.5
level of significance, α = 0.05
from standard normal table,right tailed z α/2 =1.64
since our test is right-tailed
reject Ho, if zo > 1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.72-0.5/(sqrt(0.25)/25)
zo =2.2
| zo | =2.2
critical value
the value of |z α| at los 0.05% is 1.64
we got |zo| =2.2 & | z α | =1.64
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: right tail - Ha : ( p > 2.2 ) = 0.0139
hence value of p0.05 > 0.0139,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p>0.5
test statistic: 2.2
critical value: 1.64
decision: reject Ho
p-value: 0.0139
we have enough evidence to support the claim that observes the 18 heads from 25 tosses.

c.
level of significance=0.01
Given that,
possible chances (x)=18
sample size(n)=25
success rate ( p )= x/n = 0.72
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5
alternate, H1: p>0.5
level of significance, α = 0.01
from standard normal table,right tailed z α/2 =2.33
since our test is right-tailed
reject Ho, if zo > 2.33
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.72-0.5/(sqrt(0.25)/25)
zo =2.2
| zo | =2.2
critical value
the value of |z α| at los 0.01% is 2.33
we got |zo| =2.2 & | z α | =2.33
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: right tail - Ha : ( p > 2.2 ) = 0.0139
hence value of p0.01 < 0.0139,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p>0.5
test statistic: 2.2
critical value: 2.33
decision: do not reject Ho
p-value: 0.0139
we do not have enough evidence to support the claim that observes the 18 heads from 25 tosses.
Answer:
option:B