Question

A researcher carried out an investigation of the extra sleep produced by drug A and drug B. The data from the experiment in hours are shown. Drug A 0.7 −1.6 −0.2 −1.2 −0.1 3.4 3.7 0.8 0.0 2.0 Drug B 1.9 0.8 1.1 0.1 −0.1 4.4 5.5 1.6 4.6 3.4 The data for both treatment groups are not normally distributed. To determine if the mean sleep response (extra hours of sleep produced) differs between drug A and drug B, a permutation test was performed. This was done by pooling both groups of data and randomly reassigning half the observations to the drug A group, and the other half to the drug B group. The difference of means for these two groups, ?⎯⎯⎯⎯?−?⎯⎯⎯⎯? , was then calculated and recorded. This permutation process was repeated a large number of times. The histogram shows the resulting distribution of values for the mean difference for all the permutations. Nearly symmetric histogram centered at 0 with vertical dashed lines at -1.8 and +1.8, separating the critical region from the non-critical region With a null hypothesis of no difference in the amount of extra sleep between the two drugs, and a two‑sided alternative hypothesis, determine if the actual observed difference is statistically significant. The dashed lines represent the critical values at −1.8 and 1.8, and the total area in the tails of the histogram is 0.05 , which represents the critical (rejection) region.

Compute the difference of the means of the two drugs for the observed data. Give your answer to two decimal places.

Answer 1

H0: There is no significant difference in the amount of extra sleep between the two drugs

H1: There is significant difference in the amount of extra sleep between the two drugs

two saple t-test

Drug A	Drug B
0.7	1.9
-1.6	0.8
-0.2	1.1
-1.2	0.1
-0.1	-0.1
3.4	4.4
3.7	5.5
0.8	1.6
0	4.6
2	3.4

	A	B
avg	0.75	2.33
sd	1.789	2.0022
n	10	10

Sp^2 =   ((10-1)*1.789^2+(10-1)*2.0022^2)/(10+10+-2)
=3.6047
      
t =   (0.75-2.33)/SQRT(3.6047*((1/10)+(1/10)))
   = -1.8608

df = 10+10-2 =18

P-value =   2*(1-T.DIST(abs(-1.8608),18,1))

   =0.0792

With t=-1.8608 and p-value =0.0792 > 0.05 we do not reject H0 at the 5% significance level

There is no significant difference between mean amount of extra sleep between the two drugs

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