A research team is interested in the effectiveness of hypnosis in reducing pain.

Question

A research team is interested in the effectiveness of hypnosis in reducing pain. The responses from 8 randomly selected patients before and after hypnosis are recorded in the table below (higher values indicate more pain) Construct a 90% confidence interval for the true mean difference in pain after hypnosis. a) Fill in the missing table cells for the pain level differences. Compute the differences as 'Pre Post b) If the hypnosis treatment is effective in reducing pain, we expect the differences (pre post) to be Note: For (c), (d), and (e) use 3 decimals in your answers. You should use JMP to calculate these values. c) The point estimate for the true average effect that hypnosis has on pain d) The point estimate for the true standard deviation of the effect that hypnosis has on pain perception (i.e. sd)s:- ] e) The standard error for the mean difference in pain scores is: f) For this problem, the sample size is small enough that approximating the critical value as being t = 2 will induce substantial error. It turns out that, for a sample size of n = 8, the 95% t-critical value is about t = 2.4. Using this, this 95% confidence interval for the true mean difference in pain level after hypnosis is: IMPORTANT: don't enter the 95% CIJMP gives you. It will be wrong, because t-critical2.4 has been rounded to the nearest decimal, while JMP uses a more precise value. You need to calculate this yourself using the CI formula. (round your answer to 2 decimals) g) Based on your confidence interval in part (f), does the data seem to suggest strong evidence that this form of hypnosis has an effect on pain?

Explanation / Answer

a)

Pre Post Differnce

8.9 10.2 -1.3

8.1 9.5 -1.4

7.7 6.1 1.6

10 6.7 3.3

7.2 9.8 -2.6

10.3 9.2 1.1

9 8.1 0.9

12.3 8.7 3.6

b) Differences to be positive

c)

Point estimate for the average (Xd) = 0.65

d)

Point estimate for true std.dev (Sd)= 2.251

e) standard error = std.dev / sqrt(n)

= 2.251 / sqrt(8)

= 0.796

f)

CI = Xd + /- 2.4 * SE

= 0.65 +/- 2.4 * 0.796

= (-1.2604 , 2.5604)

-1.2604 < mud < 2.5604

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