Question

In this exercise involving paired differences, consider that it is reasonable to assume the populations being compared have approximately the same shape and that the distribution of paired differences is approximately symmetric.

A sample of 10 men was used in a study to test the effects of a relaxant on the time required to fall asleep. Suppose data for 10 subjects showing the number of minutes required to fall asleep with and without the relaxant follow.

Subject	Relaxant
	No	Yes
1	16	10
2	13	10
3	23	12
4	9	10
5	10	9
6	7	6
7	9	11
8	9	7
9	13	12
10	8	6

Use a 0.05 level of significance to determine whether the relaxant reduces the median time required to fall asleep.

State the null and alternative hypotheses.

H₀: Median time without relaxant − Median time with relaxant ≠ 0
H_a: Median time without relaxant − Median time with relaxant = 0H₀: Median time without relaxant − Median time with relaxant = 0
H_a: Median time without relaxant − Median time with relaxant ≠ 0     H₀: Median time without relaxant − Median time with relaxant ≥ 0
H_a: Median time without relaxant − Median time with relaxant < 0H₀: Median time without relaxant − Median time with relaxant > 0
H_a: Median time without relaxant − Median time with relaxant = 0H₀: Median time without relaxant − Median time with relaxant ≤ 0
H_a: Median time without relaxant − Median time with relaxant > 0

Find the value of the test statistic.

T⁺ =

Find the p-value. (Round your answer to four decimal places.)

p-value =

What is your conclusion?

Reject H₀. There is sufficient evidence to conclude that the relaxant reduces the median time required to fall asleep.Reject H₀. There is not sufficient evidence to conclude that the relaxant reduces the median time required to fall asleep.     Do not reject H₀. There is sufficient evidence to conclude that the relaxant reduces the median time required to fall asleep.Do not reject H₀. There is not sufficient evidence to conclude that the relaxant reduces the median time required to fall asleep.

Answer 1

wilcoxon SIGN rank test

sample 1	sample 2	difference=sample1-sample 2	absolue difference	rank	rank if positive	rank if negative
16	10	6	6	9	9
13	10	3	3	8	8
23	12	11	11	10	10
9	10	-1	1	2.5		2.5
10	9	1	1	2.5	2.5
7	6	1	1	2.5	2.5
9	11	-2	2	6		6
9	7	2	2	6	6
13	12	1	1	2.5	2.5
8	6	2	2	6	6

H0: Median time without relaxant − Median time with relaxant ≤ 0
Ha: Median time without relaxant − Median time with relaxant > 0

Level of Significance , α =    0.05
  
number of non zero difference , n =    10
  
sum of positive ranks, W+ =    46.5
sum of negative ranks , W- =   8.5
  
T+ = W+   46.5
  
mean ,µ = n(n+1)/4 =    27.5
  
std dev ,σ = √(n(n+1)(2n+1)/24) =   9.811
  
Z-stat = (T - µ)/σ =    1.937
  
  
P-value = P(Z>1.937) = 0.0264 [excel function =normsdist(-1.937) ]
Conclusion:     P-value<α , Reject null hypothesis

so, conclusion is

Reject H₀. There is sufficient evidence to conclude that the relaxant reduces the median time required to fall asleep