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. Does hypnotism appear to be effective in reducing pain? (Table attached) In this example, μd 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 on a pain scale before and after hypnosis. What is the test statistic for this hypothesis test?

Answer 1

Explanation:

Corresponding measurements on a pain scale before and after hypnosis form matched pairs.

The data for the test are the differences between the measurements on a pain scale before and after hypnosis.

μd = the​ measurements on a pain scale before hypnosis minus the​ measurements on a pain scale after hypnosis

Before after diff

6.3 6.5 - 0.2

4 2.5 1.5

9.2 7.7 1.5

9.3 8.4 0.9

11.3 8.6 2.7

Sample mean, xd

= (- 0.2 + 1.5 + 1.5 + 0.9 + 2.7)/5 = 1.28

xd = 1.28

Standard deviation = √(summation(x - mean)²/n

n = 5

Summation(x - mean)² = (- 0.2 - 1.28)^2 + (1.5 - 1.28)^2 + (1.5 - 1.28)^2 + (0.9 - 1.28)^2 + (2.7 - 1.28)^2 = 4.448

Standard deviation = √(4.448/5

sd = 0.94

For the null hypothesis

H0: μd ≥ 0

For the alternative hypothesis

H1: μd < 0

The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 5 - 1 = 4

The formula for determining the test statistic is

t = (xd - μd)/(sd/√n)

t = (1.28 - 0)/(0.94/√5)

t = 3.04

The test statistic for the hypothesis test is 3.04