9.4 Matched or paired samples -- rrc math 1020  (Page 27/13)

Corresponding "before" and "after" values form matched pairs. (Calculate "after" – "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: $\overline{x_d}=\mathrm{–3.13}$ 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

H ₀ : μ _d ≥ 0

The null hypothesis is zero or positive, meaning that there is the same or more pain felt after hypnotism. That means the subject shows no improvement. μ _d is the population mean of the differences.)

H _a : μ _d <0

The alternative hypothesis is negative, meaning there is less pain felt after hypnotism. That means the subject shows improvement. The score should be lower after hypnotism, so the difference ought to be negative to indicate improvement.

Distribution for the test: The distribution is a Student's t with 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's-t distribution: p -value = 0.0095

Graph:

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

The sample mean and sample standard deviation of the differences are:

${\overline{x}}_d$ = –3.13

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

Compare α and the p -value: α = 0.05 and p -value = 0.0095. α > p -value.

Make a decision: Since α > p -value, reject H ₀ . This means that μ _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 after hypnotism. Hypnotism appears to be effective in reducing pain.