Arthritis is a painful, chronic inflammation of the joints. An experiment on the

Question

Arthritis is a painful, chronic inflammation of the joints. An experiment on the side effects of pain relievers examined arthritis patients to find the proportion of patients who suffer side effects when using ibuprofen to relieve the pain. 440 subjects with chronic arthritis were given ibuprofen for pain relief. 23 subjects suffered from adverse side effects. If more than 3% of users suffer side effects, the Food and Drug Administration will put a stronger warning label on packages of ibuprofen. Is there sufficient evidence that stronger warning labels are needed? a) Write hypotheses to test this claim; use = 0.05. Explain why you should use a one sided alternative. b) You may assume any necessary conditions have been met. Perform your test. c) If your conclusion was found to be incorrect, what type of error did you make? d) In this situation, do you think a Type I error or Type II error is more serious? Explain. e) Give a 95% confidence interval for the proportion of subjects who will suffer side effects.

Explanation / Answer

PART A.
Given that,
possibile chances (x)=23
sample size(n)=440
success rate ( p )= x/n = 0.0523
success probability,( po )=0.03
failure probability,( qo) = 0.97
null, Ho:p<0.03
alternate, H1: p>0.03
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=0.05227-0.03/(sqrt(0.0291)/440)
zo =2.7388
| zo | =2.7388
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =2.739 & | z | =1.64
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 2.73876 ) = 0.00308
hence value of p0.05 > 0.00308,here we reject Ho

ANSWERS
---------------
null, Ho:p<0.03
alternate, H1: p>0.03
test statistic: 2.7388
critical value: 1.64
decision: reject Ho
p-value: 0.00308

PART E.
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
No. of success(x)=23
Sample Size(n)=440
Sample proportion = x/n =0.03
Confidence Interval = [ 0.03 ±Z a/2 ( Sqrt ( 0.03*0.97) /440)]
= [ 0.03 - 1.96* Sqrt(0) , 0.03 + 1.96* Sqrt(0) ]
= [ 0.014,0.046]

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