A study was conducted to measure the effectiveness of hypnotism in reducing pain

Question

A study was conducted to measure the effectiveness of hypnotism in reducing pain. The measurements are centimeters on a pain scale before and after hypnosis. Assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal. Construct a 95% confidence interval for the mean of the 'before-after' differences. Does hypnotism appear to be effective in reducing pain? in this example. mu_d is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the difference in the measurements in centimeters on a pain scale before and after hypnosis. What is the P-value for this hypothesis test? (Round to three decimal places as needed.)

Explanation / Answer

Given that,
population mean(u)=9.2
sample mean, x =9.3
standard deviation, s =11.3
number (n)=8.6
null, H0: Ud < 0
alternate, H1: Ud > 0
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.895
since our test is right-tailed
reject Ho, if to > 1.895
we use Test Statistic
to= d/ (S/n)
Where
Value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 2.09
We have d = 2.09
Pooled variance = Calculate value of Sd= S^2 = Sqrt [ 49.81-(16.7^2/8 ] / 7 = 1.46
to = d/ (S/n) = 4.05
Critical Value
The Value of |t | with n-1 = 7 d.f is 1.895
We got |t o| = 4.05 & |t | =1.895
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
p-value :right tail - Ha : ( p > 4.0452 ) = 0.00245
hence value of p0.05 > 0.00245,here we reject Ho
ANSWERS
---------------
null, H0: Ud < 0
alternate, H1: Ud > 0
test statistic: 4.05
critical value: reject Ho, if to > 1.895
decision: Reject Ho
p-value: 0.00245

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