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 from part b) 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

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
---------------
a.
null, Ho:p=0.03
alternate, H1: p>0.03
b.
test statistic: 2.7388
critical value: 1.64
c.
decision: reject Ho
p-value: 0.00308
d.
In this situation, i think a Type I error is use because reject the null hypothesis,there is more than 3% of users suffer side effects, the Food and Drug Administration will put a stronger warning label on packages of ibuprofen. Is sufficient evidence that stronger warning labels are needed
e.
TRADITIONAL METHOD
given that,
possibile chances (x)=23
sample size(n)=440
success rate ( p )= x/n = 0.052
I.
sample proportion = 0.052
standard error = Sqrt ( (0.052*0.948) /440) )
= 0.011
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.645
margin of error = 1.645 * 0.011
= 0.017
III.
CI = [ p ± margin of error ]
confidence interval = [0.052 ± 0.017]
= [ 0.035 , 0.07]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possibile chances (x)=23
sample size(n)=440
success rate ( p )= x/n = 0.052
CI = confidence interval
confidence interval = [ 0.052 ± 1.645 * Sqrt ( (0.052*0.948) /440) ) ]
= [0.052 - 1.645 * Sqrt ( (0.052*0.948) /440) , 0.052 + 1.645 * Sqrt ( (0.052*0.948) /440) ]
= [0.035 , 0.07]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ 0.035 , 0.07] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion

95% confidence interval for the proportion of subjects who will suffer side effects

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