Treatment Sham Researchers conducted a study to determine whether magnets are ef

Question

Treatment Sham Researchers conducted a study to determine whether magnets are effective in treating back pain. The results are shown in the table for the treatment (with magnets) group and the sham (or placebo) group. The results are a measure of reduction in back pain. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below Use a 0.05 significance level for both parts 0.56 0.51 0.38 1.08 a. Test the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment What are the null and altemative hypotheses? Hy: H122 The test statistic t is (Round to two decimal places as needed ) The P-value is (Round to three decimal places as needed.) State the conclusion for the test the null hypothesis There | | sufficient evidence to support the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment Is it valid to argue that magnets might appear to be effective if the sample sizes are larger? Since the that magnets might appear to be effective if the sample sizes are larger b. Construct a confidence interval suitable for testing the claim that those treated with magnets have a greater mean reduction in pain than those given a sham for those treated with magnets is | the sample mean for those given a sham treatment- valid to argue Round to three decimal places as needed )

Explanation / Answer

PART A.
Given that,
mean(x)=0.56
standard deviation , s.d1=0.51
number(n1)=12
y(mean)=0.38
standard deviation, s.d2 =1.08
number(n2)=12
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.1
from standard normal table,right tailed t /2 =1.36
since our test is right-tailed
reject Ho, if to > 1.36
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =0.56-0.38/sqrt((0.2601/12)+(1.1664/12))
to =0.522
| to | =0.522
critical value
the value of |t | with min (n1-1, n2-1) i.e 11 d.f is 1.36
we got |to| = 0.52207 & | t | = 1.36
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:right tail - Ha : ( p > 0.5221 ) = 0.30599
hence value of p0.1 < 0.30599,here we do not reject Ho
ANSWERS
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a. null, Ho: u1 = u2 alternate, H1: u1 > u2
b. test statistic: 0.52
c. p-value: 0.306
d. failed to reject, there is not

we dont have evdence that those treated with magnets have greater mean reduction

PART B.

CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 0.56-0.38) ± t a/2 * sqrt((0.26/12)+(1.166/12)]
= [ (0.18) ± t a/2 * 0.345]
= [-0.439 , 0.799]
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interpretations:
1. we are 95% sure that the interval [-0.439 , 0.799] 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

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