meuitdi Study s claim at a 001 level of significance 2) Drug-eluting vs Bare Met

Question

meuitdi Study s claim at a 001 level of significance 2) Drug-eluting vs Bare Metal Stents. A medical study wanted to determine if using drug eluting stents (stents coated with a drug to reduce restenosis or re-narrowing of t were bet drug), 11.9 % of the people with the bare metal stents either died, had a heart attack, had bypass surgery or underwent another stent implantation. 971 of the 10,327 people with the drug-eluting stents either died, had a heart attack, had bypass surgery or underwent another stent implantation. Test the claim at a 0.01 level of significance 5 he artery) tter than those adults who received the bare metal stent (stents not coated with a 3) Stents vs Medicine. A medical study wanted to test the claim that stent implantation was less effective for clearing arteries of plaque than medical therapy alone. 15.6 % of the patients treated with medical therapy alone experienced another cardiovascular event (hear attack, stroke or death) within four years of their surgery. 187 of the 1,083 patients treated with stent implantation experienced another cardiovascular event (heart attack, stroke or death) within four years of their surgery. Test the medical study's claim at a 0.01 level of significance.

Explanation / Answer

2.

Given that,
possibile chances (x)=971
sample size(n)=10327
success rate ( p )= x/n = 0.094
success probability,( po )=0.119
failure probability,( qo) = 0.881
null, Ho:p=0.119  
alternate, H1: p!=0.119
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.58
since our test is two-tailed
reject Ho, if zo < -2.58 OR if zo > 2.58
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.09403-0.119/(sqrt(0.104839)/10327)
zo =-7.838
| zo | =7.838
critical value
the value of |z | at los 0.01% is 2.58
we got |zo| =7.838 & | z | =2.58
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -7.83835 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.119
alternate, H1: p!=0.119
test statistic: -7.838
critical value: -2.58 , 2.58
decision: reject Ho
p-value: 0
we have enough evidence to support the claim

3.
Given that,
possibile chances (x)=187
sample size(n)=1083
success rate ( p )= x/n = 0.173
success probability,( po )=0.156
failure probability,( qo) = 0.844
null, Ho:p=0.156  
alternate, H1: p<0.156
level of significance, = 0.01
from standard normal table,left tailed z /2 =2.33
since our test is left-tailed
reject Ho, if zo < -2.33
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.17267-0.156/(sqrt(0.131664)/1083)
zo =1.512
| zo | =1.512
critical value
the value of |z | at los 0.01% is 2.33
we got |zo| =1.512 & | z | =2.33
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: left tail - Ha : ( p < 1.51174 ) = 0.9347
hence value of p0.01 < 0.9347,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.156
alternate, H1: p<0.156
test statistic: 1.512
critical value: -2.33
decision: do not reject Ho
p-value: 0.9347
we do not have enough evidence to support the claim

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