10.4 Matched or paired samples  (Page 5/1)

Corresponding "before" and "after" values form matched pairs. (Calculate "sfter" - "before").

The data for the test are the differences:{0.2, -4.1, -1.6, -1.8, -3.2, -2, -2.9, -9.6}

The sample mean and sample standard deviation of the differences are: $\overline{x_d}=-3.13$ and $s_d=2.91$ Verify these values.

Let ${\mu }_d$ be the population mean for the differences. We use the subscript $d$ to denote "differences."

Random Variable: $\overline{X_d}$ = the mean difference of the sensory measurements

Distribution for the test: The distribution is a student-t with $\mathrm{df}=n-1=8-1=7$ . Use $t_7$ . (Notice that the test is for a single population mean.)

Calculate the p-value using the Student-t distribution: $\text{p-value}=0.0095$

Graph:

${\overline{X}}_d$ is the random variable for thedifferences.

The sample mean and sample standarddeviation of the differences are:

${\overline{x}}_d=-3.13$

${\overline{s}}_d=2.91$

Compare $\alpha$ and the p-value: $\alpha =0.05$ and $\text{p-value}=0.0095$ . $\alpha >\text{p-value}$ .

Make a decision: Since $\alpha >\text{p-value}$ , reject $H_o$ .

This means that ${\mu }_d<0$ and there is improvement.

Conclusion: At a 5% level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower afterhypnotism. Hypnotism appears to be effective in reducing pain.