Question

A researcher hypothesizes that a pleasant distraction will change tolerance of a painful stimulus. 10 volunteers are exposed to experimental conditions: pleasant imagery (e.g., walking on the beach in Hawaii) vs. no instructions (the control condition). Each participant experiences both conditions. During each condition, participants perform the cold-pressor task. On this task, the participant sticks his/her arm in a bucket of freezing cold water. The dependent measure is how long the individual can keep his/her arm in before pulling it out. Longer durations indicate better pain tolerance. The data are as follows:

Subject ID #	Distraction	Control
1	20	18
2	35	36
3	10	3
4	45	41
5	50	42
6	25	20
7	30	30
8	40	39
9	45	46
10	30	25
Mean	33.0	30.00
SD	12.52	14.04

a. Conduct a statistical test of the null hypothesis that the population means of the Distraction and Control conditions are equal. Present your calculations, your critical value, and your decision about whether or not to reject the null hypothesis. (7)

b. Form a 95% confidence interval around the population mean for the Distraction condition.  (3)

Answer 1

a) H₀:

    H₁:

The test statistic t = ()/sqrt(s1^2/n1 + s2^2/n2)

                             = (33 - 30)/sqrt((12.52)^2/10 + (14.04)^2/10)

                             = 0.504

df = (s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + (s2^2/n2)^2/(n2 - 1))

   = ((12.52)^2/10 + (14.04)^2/10)^2/(((12.52)^2/10)^2/9 + ((14.04)^2/10)^2/9)

= 18

At 0.05 significance level the critical values are +/- t_0.025,18 = 2.101

Since the test statistic value is not greater than the upper critical value (0.504 < 2.101), so we should not reject the null hypothesis.

b) At 95% confidence interval the critical value is t^* = 2.101

The 95% confidence interval for the difference in population means is

() +/- t^* * sqrt(s1^2/n1 + s2^2/n2)

= (33 - 30) +/- 2.101 * sqrt((12.52)^2/10 + (14.04)^2/10)

= 3 +/- 12.498

= -9.498, 15.498