Hypothesis Test for a Population Proportion: Suppose that a survey of dentists f

Question

Hypothesis Test for a Population Proportion:

Suppose that a survey of dentists finds that 60% of dentists sampled recommend a certain toothbrush. Assuming that this is a random sample and that the dentists’ opinions are sincere, do these results provide evidence that more than half of all dentists prefer that toothbrush. Test using a level of significance of .05 and using sample sizes of n=25, 50, 100 and 500. What are your conclusions? You will have to perform 4 hypothesis tests, one for each sample size.

Please provide very detailed, step by step work. Thank you!

Explanation / Answer

Q1.
Given that,
sample size(n)=25
success rate ( p )= x/n = 0.6
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5  
alternate, H1: p>0.5
level of significance, = 0.05
from standard normal table,right tailed z /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=0.6-0.5/(sqrt(0.25)/25)
zo =1
| zo | =1
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =1 & | z | =1.64
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: right tail - Ha : ( p > 1 ) = 0.15866
hence value of p0.05 < 0.15866,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p>0.5
test statistic: 1
critical value: 1.64
decision: do not reject Ho
p-value: 0.15866
no evidence to support

Q2.
Given that,
sample size(n)=50
success rate ( p )= x/n = 0.6
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p<0.5  
alternate, H1: p>0.5
level of significance, = 0.05
from standard normal table,right tailed z /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=0.6-0.5/(sqrt(0.25)/50)
zo =1.4142
| zo | =1.4142
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =1.414 & | z | =1.64
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: right tail - Ha : ( p > 1.41421 ) = 0.07865
hence value of p0.05 < 0.07865,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p>0.5
test statistic: 1.4142
critical value: 1.64
decision: do not reject Ho
p-value: 0.07865
no evidence to support

Q3.
Given that,
sample size(n)=100
success rate ( p )= x/n = 0.6
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5  
alternate, H1: p>0.5
level of significance, = 0.05
from standard normal table,right tailed z /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=0.6-0.5/(sqrt(0.25)/100)
zo =2
| zo | =2
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =2 & | z | =1.64
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 2 ) = 0.02275
hence value of p0.05 > 0.02275,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p>0.5
test statistic: 2
critical value: 1.64
decision: reject Ho
p-value: 0.02275

evidence to support

Q4.
Given that,
possibile chances (x)=300
sample size(n)=500
success rate ( p )= x/n = 0.6
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5  
alternate, H1: p>0.5
level of significance, = 0.05
from standard normal table,right tailed z /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=0.6-0.5/(sqrt(0.25)/500)
zo =4.4721
| zo | =4.4721
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =4.472 & | z | =1.64
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 4.47214 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p>0.5
test statistic: 4.4721
critical value: 1.64
decision: reject Ho
p-value: 0
evidence to support

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