Construct a confidence interval suitable for testing the claim that those treate
ID: 3310276 • Letter: C
Question
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 treatment. Using a ti-84 calculator.
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 1 13 x 0.52 S 0.74 13 0.37 1.33 The P-value is .363 (Round to three decimal places as needed.) State the conclusion for the test. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that those treated with magnets have a greater mean reduction in pain tharn 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 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 sample mean for those treated with magnets is greater than the sample mean for those given a sham treatment, it is valid to argue that treatment. H1H2 (Round to three decimal places as needed.)Explanation / Answer
a.
Given that,
mean(x)=0.52
standard deviation , s.d1=0.74
number(n1)=13
y(mean)=0.37
standard deviation, s.d2 =1.33
number(n2)=13
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.782
since our test is right-tailed
reject Ho, if to > 1.782
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =0.52-0.37/sqrt((0.5476/13)+(1.7689/13))
to =0.355
| to | =0.355
critical value
the value of |t | with min (n1-1, n2-1) i.e 12 d.f is 1.782
we got |to| = 0.35534 & | t | = 1.782
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:right tail - Ha : ( p > 0.3553 ) = 0.36425
hence value of p0.05 < 0.36425,here we do not reject Ho
ANSWERS
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null, Ho: u1 =u2
alternate, H1: u1 > u2
test statistic: 0.355
critical value: 1.782
decision: do not reject Ho
p-value: 0.36425 = 0.363
we do not have enough evidence to support that those treated with magnets have a greater mean reduction in pain than those given a sham treatment
b.
TRADITIONAL METHOD
given that,
mean(x)=0.52
standard deviation , s.d1=0.74
number(n1)=13
y(mean)=0.37
standard deviation, s.d2 =1.33
number(n2)=13
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((0.548/13)+(1.769/13))
= 0.422
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table,right tailedand
value of |t | with min (n1-1, n2-1) i.e 12 d.f is 1.782
margin of error = 1.782 * 0.422
= 0.752
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (0.52-0.37) ± 0.752 ]
= [-0.602 , 0.902]
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DIRECT METHOD
given that,
mean(x)=0.52
standard deviation , s.d1=0.74
sample size, n1=13
y(mean)=0.37
standard deviation, s.d2 =1.33
sample size,n2 =13
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 = [( 0.52-0.37) ± t a/2 * sqrt((0.548/13)+(1.769/13)]
= [ (0.15) ± t a/2 * 0.422]
= [-0.602 , 0.902]
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interpretations:
1. we are 95% sure that the interval [-0.602 , 0.902] 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